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 ELEMENTS OF
FINITE DIFFERENCES,
ALSO
SOLUTIONS TO QUESTIONS
SET FOR PART I OF THE EXAMINATIONS OF
THE INSTITUTE OF ACTUARIES.
Part I. FINITE DIFFERENCES.
Part 2. QUESTIONS.
Part 3. SOLUTIONS.
Part 4. ANSWERS, WITHOUT SOLUTIONS
BY
J. BURN, F.I.A., AND E. H. BROWN, F.I.A.,
OF THE PRUDENTIAL ASSURANCE COMPANY.
LONDON :
CHARLES AND EDWIN LAYTON,
56, FARRINGDON STREET, E.G.
1902.
•
^ ' PREFACE
Part I. of the following work is mainly intended as a TextBook
of Finite Differences for Students preparing for Part I. of the
Examinations of the Institute of Actuaries. It has been so
arranged that the Student need not necessarily have any
knowledge of trigonometry or conic sections^ and it will be
noticed that this has in some cases necessitated somewhat
lengthy arguments.
Parts II. and III. deal with 250 questions set during the
last sixteen years in the Actuarial Examination papers.
The solutions are not given in all cases in such form as
would be most suitable as answers in an examination paper,
since it has been thought advisable to give a full discussion
in any case where it is possible that a Student might otherwise
find a difficulty in following the solution.
It will be seen that there are 40 solutions of Finite
Difference questions, which practically form an appendix to
Part I.
The answers are given separately in Part TV. in order that
the advantages to be gained by independent thought may not
be forfeited by premature reference to the solution.
Unless otherwise stated, the references are to be found ,in
the fourth edition of Hall & Knight^s Higher Algebra.
We have much pleasure in acknowledging the assistance
rendered by Mr. C. W. Kenchington, F.I. A., not only in the
work of revision and correction, but also for several useful
suggestions.
J. B.
E. H. B.
September, 1902.
261599
CONTENTS.
PART I.
Page
Chapter I.— (1) to (6) Definitions, (7) to (9) Examples in Differencing,
(11) Equidistant Terms, (12) to (17) Separation of Symbols, (18) to
(20) Constant nth. Difference, (10), (21) to (26) Applications of Finite
Differences ... ... ... ... ... ... ... ... ... 1
Chapter II.— (27) to (34) Elementary Formulas, (35) to (43) Factorials,
(44) to (46) Products and Quotients of Two Functions, (47) to (52)
Differences of 0... ... ... ... ... ... ... ... ••• 12
Chapter III.— (53) to (56) Interpolation, (57) by Leading Differences, (58) to
(61) nth. Difference Zero, (62) to (66) Subdivision of Intervals, (67) to
(77) Lagrange's Theorem .... 21
Chapter IV.— (78) to (84) Central Differences 33
Chapter V.— (85) Definitions, (86) to (89) Summation, (90) to (94)
"Integration, (95) Integration between Limits, (96) to (100) Indefinite
Integrals, (101) to (104) Approximate Summation 37
PARTS 2, 3, AND 4.
Chapter Subject.
I. — Algebraical Examples
II. — Equations
III. — Theory of Equations
IV. — Arithmetical Examples
V. — Variation and Ratio
VI. — Problems
VII. — Combinations and Permutations
VIII.— Binomial Theorem
IX. — Probabilities
X. — Finite Differences
XI. — Recurring Series
XI [. — Arithmetic, Geometric and Harmonic
Progressions
XIII. — Logarithmic and Exponential Series
XIV. — Logarithms
XV. — Interest
FESTIONS.
Page
45
Solutions.
Page
89
Answers.
Page
253
48
100
253
50
118
254
51
120
254
53
126
254
55
131
254
60
152
255
63
162
255
66
173
256
73
200
257
80
220
258
81
225
258
84
239
259
86
246
259
87
250
259
THE ELEMENTS OF FINITE DIFFERENCES.
CHAPTER I.
1. The Calculus of Finite Differences deals with the changes in value
which arise in functions as a result of finite changes in the value of the
variables on which the functions depend. Thus, to take a simple
illustration, suppose the function to be dealt with is ^^, where the
variable is x. If this variable be increased by the finite quantity h, the
function becomes (iv\hy, that is, ic^ + SxhiSxh^ + h^, so that a finite
increase of h in the variable results in an increase of (Sa:Vi + S.vh^ + h^)
in the given function.
2. By means of the Calculus of Finite Differences we investigate
the nature of various functions, and thereby obtain various relationships
which we can express by means of formulas which can easily be used
in practice for the purpose of summing or interpolating. It will be
necessary first to give a brief explanation of the symbols generally
employed.
3. It will be remembered that a function of a? is any expression
which involves £c, and whose value is dependent on that of .r.
Functions of a: are usually denoted by symbols of the form tix, Vxt
fx. Fa;, Q^,, &c.* Similarly, a function involving (^ + 1) is denoted by
.»+!? ^a;+i> &c, and generally a symbol of the form Ux\ii, ^a;+/t5 &c.,
n
denotes a function involving (a: + A).
For example, if Ux denotes —
{ax—}})(dx — c)
{})x—d){cx—d)
Hall and Knight's Higher Algebra, page 93.
2' TfiE ELEMENTS OF FINITE DIFFEEENCES.
then will iix\n denote —
{ax \h—h) {dx \h—c)
(bx \ h—a) (ex + h — d)
Here we see that x is the variable, {x + Ji) is the increased value of the
variable, therefore h is the increment of the variable.
4. Again zix denotes the function of x, Ux+h denotes the increased
value of that function, therefore Ux+h—i^x denotes the increment of that
function, and is denoted by the symbol Aux, whence
^t^x=Ux+h — Ux (1)
This relation defines the operation denoted by A, and Mi^ is known
as the first difference of u^.
Similarly, we may have
A(^Ux)=^tlx+h — ^Ux=^^Ux
and /^^Uxi or the first difference of the first diff'erence of Ux^ is known
as the second difference of Ux'
Whence generally we have the ^th difference of ttx denoted by
A^^^^=A(A«%,^0 = A^%a?+7iA»^%a; .... (2)
5. The student will now be in a better position to grasp the
definition given in Boole's Finite Differenees where the Calculus of Finite
Differences has been defined as the science of the laws of the operation
A, when A prefixed to the expression of any function of x denotes the
operation of taking the increment of that function corresponding to a
given constant increment of the variable x.
6. The simple case where Aa;=l is more generally met with in
practice since the calculus deals chiefly with the terms of series ; but it
must be remembered that ^x or h may have any positive value, either
fractional or integral.
Examples in Differencing.
7. When ^x=h.
1. — Let a\{x—V)d be denoted by Ux^ then
= hd.
■ THE ELEMENTS OF FINITE DIFFERENCES. 6
2. — Let ar^~^ be denoted by Uxy then
= ar^^(rf' — l)
3. — Let cc'^ he denoted by u^, then
When ^cc=l.
1. — If 2ix denotes «+(a7— !)«? then Aux=d
2.— „ „ ar^^ „ „ =«r^^(rl)
3.— „ „ 0!^ „ „ =(^+l)^.T^
8. The successive terms of a series and their differences are usually
tabulated as follows : —
Variable
Function
1st
Difference
2nd
Difference
&c.
cc + h
x\2h
Ux
Ux+2h
AUx
^Ux+7i
^^Ux
When Aa?=l, this scheme becomes
Variable
Function
1st
Difference
2nd
Difference
&c.
a?
ICx
^Ux
^''Ux
...
■
0:41
nx\.i
Aux+i
A2z^.r+i
...
■
a:{2
Ux+2
AUx+2
^^Ux+2
...
L
orfS
...
...
...
B 2
4 THE ELEMENTS OF FINITE DIFFEBENCES.
9. Now let x=l and 2^0;= a^, then, replacing the headings by
their symbols,
X
«a:
A«:,
Ahi^
A^Ux
1
1
7
12
6
2
8
19
18 ^^
6
3
27
37
24
6
4
64
61
30
...
5
125
91
...
...
6
216
...
...
...
10. Here we see that the third differences are the same for any
particular value of the function, therefore the fourth and all higher
differences are equal to zero. Then, given the numerical values of any
term in the series and the differences of that term, we can construct a
complete table of the i\xnciion,hy addition instead of multiplication;
since
A^ Wa; = A%a?+ 1 — A%a;
or A%;P+l = A2«^a; + A%a;
i.e.f if we have given u^, ^u^^ ^^u^, and A^u^, when Ua;=o^^, we can
complete the column A^Ux by the continued addition of A^Ux . Similarly,
the columns Aux and Ux can be formed by the addition of the successive
values of A.^Ux, Ahtx+i, &c., and Aw^?? ^^x+u &c., respectively. The
first term of a series and the differences of that term are called the
leading term and the leading differences.
11. The values of a function corresponding to equidifferent values of
the variable are said to be eq^ui distant, whatever their numerical values
may be. Thus a?^ {x^hy, {x\2hY, {x^Zhy, or x'^, (^ + 1)^,
(a? 4 2)^, are said to be equidistant, although the differences between
the terms are not equal. Briefly stated, equidistant terms are not
necessarily equidifferent.
12. In the same way as A is employed to denote the operation of
taking the increment of the function to which it is prefixed, so we employ
the letter E to denote the increased value of the function to which it is
prefixed. Thus, as
AUx = Ux^h — l('x
THE ELEMENTS OF FINITE DIFFEEENCES. 5
SO does ^Ux^=Ua;+h
similarly 'EiCj;+h= 'El'Eit'x=^^Ux=Ua:+2fi
We also have the relation —
^Ux=Ux+^Ua; (3)
13. It will have been noticed that we have applied the law of indices
to both A and E, ^.e., in the same way as we write a?^ to denote the
operation of multiplying x into x, so we write ^^Ux to denote the
operation of taking the increment of the increment of u^ .
14. Now it is important to note that, when symbols of operation
obey the following laws, they can be treated algebraically as though
they were symbols of quantity, and can be separated from the symbol
expressing the function, provided they are ultimately read in connection
with the symbol denoting the function to which they relate, and are
then interpreted as symbols of operation. This principle is briefly
described as the Law of Separation of Symbols. The law^s referred to
above are —
(1) The Law of Distribution.
(2) The Law of Commutation.
(3) The Law of Indices. '
15. It remains to prove that A and E obey these three laws :
(1) The symbol A is distributive in its operation, for —
^(t^x\Vx+Wx+. . .) = {Ux+h + 'Vx+h^Wx+h\') — (Ux\'Vx\tOx+...)
= Ux+h—Ux+Vx+Ji — Vx + Wx+7i — Wx\&n.
= ^Ux + ^Vx\^tOx + &G (4)
Similarly E is distributive, for —
^(Ux + Vx + Wx+ . . ')={Ux^Vx\Wx+ . . ')i^(Ux\Vx + Wx+ . . .)
= U,v + ^Ux + Vx + ^Vx + Wx + AtOa; + &C.
= E?^.i; + EiJ^ + Et^.^ + &c.
(2) The symbol A is commutative in its operation, for, if c
denote any constant quantity,
A (cUx) = CUx^h — CUx
= c(ux+h—Ux)
= c^Ux (5)
6 THE ELEMENTS OF FINITE DIFFEEENCES.
i
Similarly E is commutative, for —
E (cUx) = CUa; + ^CUa:
= C^Ua
(3) The symbol A obeys the law of indices, for when m and n
are positive integers, ^^^^Ux denotes the operation of taking
the increment of the increment . . . (repeated m times) . . .
of the function, or —
A^Ux= (AAA . . . m times) u^
,', A»»A*»e^a;= (AA . . . n times) (AA . . . m times) Ux
= (AAA . . . (m + n) times) u^
= Aw+w%^ ..... (6)
Similarly it may be shown that
16. The equation ^Ux=i{^x+ ^^a: may therefore be written —
^Ux=(l + ^)Ux
whence E = l + A   (7)
and, transposing, A=E — 1 (8)
This relation forms the base of nearly all the formulas to be hereafter
demonstrated.
17. The algebraical demonstrations of the three laws are of general
importance, and the results may be verbally stated as follows :
(1) The difference of a sum of several quantities is equal to the
sum of their differences.
The increased value of a sum of several quantities is equal to
the sum of their increased values.
(2) A constant coefficient of a function, remains sl constant
coefficient of its difference. * ^
(3) The nth difference of the mih. difference of a function is the
{7i + m)th difference of the function.
To these we would add the following :
(4) A constant term disappears in the operation of differencing,
for —
^(ux^ c) — {ux.vh\ c) — (uxV c)=^Ux^hUx^^itx . . (9)
THE ELEMENTS OF FINITE DIFFEBENCES. 7
18. On page 4 we gave a schedule showing successive values of the
function a:^, and pointed out the importance of the fact that the third
order of differences were all equal or constant. We shall now prove
that if Ux be a rational integral function of x of the nth degree, then the
nth. difference is constant.
19. As defined in Hall and Knight's Higher Algebra, p. 94, a
rational and integral function of a; of the nth degree is an expression in
the form —
where a^h, c . . . s are quantities independent x, and n is a positive
integer.
Let i(,x denote this expression, so that —
ic^z=ax^\'bx^~'^\cx^^~^+ . . . \rx{s
then
= {a(a?h^)*'+J(^ + ^)*'^ + c(a? + A)«2f . . . [r{x+h)\s}
— {«^*^fi^*^' + ca;»*2+ . . . ^rx + s}
Now, if we expand the quantities on the right of this equation by
the Binomial Theorem, we obtain —
/^Ux=[a(x''^nhx''^\ ^^'^~ h^x^^+ &c. j
+ bfx^'\nlhx'''^+ ^ ^'^ ^ h^x''^\&Q?j
a:n2 ^ n2hx^^ + h^x^'' + &c. J
+ &C.
— {ax^'Ybx^'^^cx'^^^ . . . +r.i' + s}
=^a(nhx'''^\ ^'^"~ Ji^x'^^ + kcA
+ bCn—Xhx'^^ + ' ~~^^^ A V^3 + &c^
■fc(w— 2^07*^3 H&c.)f . . . ^rh
8 THE ELEMENTS OF FINITE DIFFERENCES.
where the coefficient of
x^~^ is anh
n.n—1
a?»2 is ah^ ^T^ — + bhn— 1 = Ji (say)
^ n.n — 1. n— 2 ^^n — l.n—2 , , .
x^^ IS ah^ — +6^2 +c/i;i— 2 =c^ (say)
&c. is &c,
^« is «A*^ + 5^»*^+ . . . +rA =Si (say)
where hi, Ci , </i . . . ^i are all independent of x.
And it will be noted that Aw.^ is of degree (^—1), i.e.,
^Ux=anhx'*^^\hiX^^\CiX'^^\ . . . +riX\Sx
Again,
= {an7i(x^hy'^ + bi(x+h)^^ + Ci(x + h)^^.+ . . . \ ri(x + h) \ Si]
^{an7ix'''^ + hix'^^+Cix''^+ . . . +riX\Si}
or, expanding —
AHx=\anJi(x''^ + n—Vix''^+ ^ — AV^3 + &c.J
■^bifx''^ + n—2hx''^+ ^^~ \^~ h^a^^'^+Socj
+ Ci(x'^^+n—Shx''* + &c.) + &c.\ri{x + h) + Si\
and, after subtraction.
= anhfnlhx^^+ ^J_J^__^^2^n3_^&c.)
+ &C. ^rilt.
or, rearranging,
= an n^ . h'^x^''' + (an ^~\^' ^ ^*^ + bxn^2h\ x'*''^ \ &c.
THE ELEMENTS OF FINITE DIFFEEENCES
where the coefficient of x^''~'^ is ann—\1i^
n—l.n—2
f
x^~'^ „ ( an
h^ + bin—2hj=:b2 (say)
&c.
&c.
&c.
C2 (say)
S2 (say)
and b2, C2 . . . and 52 are all independent of ic, and we see that
A^Ux is of degree (jn—2), i.e.,
Repeating the process, we can obtain
Summarizing these results, we have
A?Ux
is a function
of X of the
wth degree
(^^l)th „
(^~2)th „
(7i3)th „
(^j— 7^)th or 0th
degree
and the
coefficient of
the highest
power of
X is
a
anh
ann—lh^
The highest power of x in A^^w.v being x'>^'\ or a:^ we see that
A^Ux is independent of x, (since x^=zl) and consists of one term only,
namely, a\nli'^.
That is to say, the nih. difference is constant.
20. When A=l, A'>^Ux=a\n.
Again, when «=1, and h, e, d, &c., all equal zero, A^Ua;=\n.
From the foregoing, we see that all higher orders of differences
vanish, for
= {}n—\n)ah'"
=0
10 THE ELEMENTS OF FINITE DIFFERENCES.
21. Hitherto we have dealt with functions of which the mathematical
law is known, or a sufficient number of terms have been given to enable
us to obtain an order of differences which is constant for all values of
the variable.
We have just demonstrated that it is always possible to obtain a
constant order of differences when dealing with rational integral
functions of a finite degree; generally, however, the law of the series is
unknown, or too involved to be readily discovered, but we shall find that
in practice we often have to deal with series in which, although we
cannot obtain constant differences, yet some low order (say third or
fourth differences) may be so small that we can assume it to be
constant, and so obtain results which are sufficiently correct for all
practical purposes.
For example : —
If ^^=161845, Wa;+ 1 = 154150, 2^0^+2= 14*6980, and w,;+3=140292,
we can construct the schedule as follows: —
X ttx A% A^Ux A^Ux
a: 161845 7695 +'0525 0043
07 + 1 154150 7170 +;0482
a: + 2 146980 6688 i +0439
a? + 3 140292 ; 6249 +'0396
07 + 4 134043 5853
07 + 5 128190
The values below the line have been found by the continued addition
of A%^, A%a;, and Aw^? (as explained on page 4), the assumption
having been made that A^Ux is constant.
The correct values of u^+i and Ux+5 are 13*4047 _and 128212
respectively, thus the error introduced is very small, and for all practical
purposes may be disregarded. The example has been taken from page 9
of A Short Collection of Actuarial Tables, the different quantities
representing the values of an annuity for 21 years when interest is taken
at 2\ percent, 3 percent, 3 percent, &c., respectively.
22. A further example of the construction of tables by the continued
addition of differences is afforded by a consideration of the method of
reconstruction that would be adopted in the very unlikely event of the
complete loss of all existing tables of Logaritlims. It is sufficient here
THE ELEMENTS OF FINITE DIFFERENCES. 11
to remark that the original tables were constructed in a much more
laborious manner.
23. In Articles 225 and 226 of Hall and Knight's Algebra, we have
a formula for . , ^ ,
logeOi + l)logeW, , ,^ ' _
thatis, ^iog,„=2{^3^+g:^^^+*c.f :.:(a)
which, it is stated, is useful in the construction of Logarithmic Tables.
This formula could, of course, be directly applied by continuously
substituting fresh values of n, but the labour would be very great.
24. By the method of Finite Differences, a much easier process is
available, thus : — ^
A log ioW=log 10 (w+ 1) —log 10^
and from formula (12),
A2logioW=logio(w + 2)~21ogioOi + l)logioW,
and, as shown by Solution 231,
log 10 (^ + 2) 21og 10 (wf 1) +log 10 w = 2/x— — ^ + «fcc.
= Anogio^^ . . . . (y8)
We thus have convenient formulas for Alogio^j and A^logio^.
25. It will be recognized that the greater the value of n the smaller
will be the value of the differences — in fact, if 7i be equal to 10,000,
the second difference will be (000,000,004,342,076), hence, if we find,
by means of formula (a), the value of Alogio 10,000, we can, by.
constant additions of the value just found, form a column of
Alogio^i which will be correct to the seventh place of decimals. Then,
again, by continued addition of our values of Alogio^j, we can form
values of logio^?.
26. In applying formiila (a), it will be seen that when n is 10,000
it is sufficient to calculate two terms only, the second term ~ —
^' 3 {2n{iy
is only 000,000,000,000,036, and the following term has 22 cyphers
between the decimal point and the first significant figure. In
formula (^) the first term is quite sufficient, for the second term
mounts only to 000,000,000,000,000,021,7.
12 THE ELEMENTS OF FINITE DIFFERENCES.
CHAPTER II.
Demonsteation of Elementary Formulas.
27. To develop Ux+m7i in a series consisting of u^ and its successive
differences, z.c, to prove the formula
Ux+mh=Ua;hmAUcc\ ,. A^Ux\&c. . . . (10)
I
where ^Ux^Ux+ji—Ux
This can be demonstrated in two ways : — 1st, by Separation of
Symbols ; 2nd, by Induction.
28. First method : —
By formula (3) we have
l^a:+h= E«^a;=(HA)Wa;
&c, = &c. = &c.
and finally t^x+mh='^'^Ux={l + ^y^Ux
whence, expanding the righthand side by the Binomial Theorem,
m(m—l) ^„
Ux+ mh—t^x + m^Ux \ 7^ A2wa, 4 &c.
•_
29. Second method: —
By formula (3) we have
Ux+h=Ux + ^Ua; . (a)
Similarly, Ux+2fi=^Ux+h + ^Ux+h
= Ux + ^Ux + ^ (Ux + ^Ux)
=Ux + 2^Ux + ^''Ux (13)
THE ELEMENTS OF FINITE DIFFEEENCES. 13
In the same way, we may show that
Ux+3h=Ux + ^^Ua; + ^^^ll.v + ^^Ux . . . . (y)
Here we see that the coefficients of the successive orders of
differences in equations (a), (/?), and (y) are those of the successive
powers of .v in the expansion of (l + a?), (i\xy, and (l + a^y
respectively.
These special results suggest the general theorem —
m(m—l) „ 
Ux+mh=Ux + m/\Ux + ^ A%a; + &C.
Let us assume this to be true for a particular value of m, then for
the next greater value of m we have —
/ ^ m(m—l) ,„ „ \
= lux + m^Ux ^ ^ —  C^^Ux + . . . &c. J
^l^[ux\ml^Ux\ ^2 ^^•^'+ • • • ^^j
= Ux^{m\ 1) Awo; + \ ^hix + &c.
r
the form of which shows that the theorem remains true for the next
greater value of m.
But we have shown that it is true when ??i=3, therefore it is true
when w=4j and so on, that is, it is true for all positive integral values
of m.
30. Now suppose mh=:n, so that m= j and for x write a, we then
h
have
n h \h )
Ua+n — Ua+T^Ua+ — r^ A.^Ua + &C. . . . (11)
It has been demonstrated that this formula is true when  is an
h
integer. It will only furnish an approximate value when  is
fractional.
31. We have taken Ua as our leading term in order to emphasize the
fact that it is not always necessary to commence with the first term of
the given series.
14 THE ELEMENTS OF FINITE DIFFERENCES.
For example :
If we wish to find the 19th term in the series u^, ti^i, 1C2, U3, W4, &c.,
we may commence with the first term Uq, when m will have the value
18, or with the 6th term u^ when m will have the value 13. Similarly,
any value may be given to a so long as « + m=18.
32. To develop ^^Ux in a series consisting of the successive values
of the function, that is, to prove the formula —
Ttl 71 ^^ 1 1
A'^U:c=Ua;+7iJi—nux+^;;::rih\ ^^ ^^a;+¥^2;^— &c. . . (12)
33. First method (separation of symbols) —
By formula (8) —
A=E1
we have A»*=(E — 1)^
whence ^'^Ucc= (E—l)^Ux
= 'il'X+nh—nUx+nlhi jx 2^a;+7j  27) — &«•
34. Second method (induction) —
By formula (1) —
^Ux=tta;+h—Ux
also A^Ux=^Ux+7i—^Ux
= (j^x+2h — Ux+7i) — (ttx+h — Ux)
Similarly, we may show that —
A%a;= Ux+3h — ^Ux+2h + ^Kx+Ji — Ux
Proceeding on the line of argument adopted in paragraph 29,
we assume this form to hold for a particular value of n. Then
by an application of formula (2) we can show that a similar form is
reproduced for the next higher value of n. The universal application
of the formula is therefore proved.
35. When each term of a series is the continued product of a given
number of factors which increase or decrease by a constant difference,
the first factors of each term also having the same relation to each other,
the series is said to be a series of "factorial terms," and each factor is
called a " factorial.*'
THE ELEMENTS OP FINITE DIFFERENCES. 15
36. The series here dealt with will be restricted to those having the
following general terms, viz. :
(ax + b) (a^Th + b) (ax+2h+ b) . . . {ax 4 (w  1)1+ b) . {a)
{ax^h) (ax^ + b) (ax2h +b) . . . (ax—(j)i—l)h + b) . (b)
1
{ax + h)(ax + h^b) . . . (ax\ (m—l)h + b) ' * * ^ >'
and
1
(ax+b)(ax—h + b) . . . {ax^m — l)h + b) ' ' ' ^ ^
when h=l, a=l, and &=0, these become —
x(x+l)ix + 2) . . . (x + m1) (a)
x(xl)(x2) . . . (xm + 1) (P)
^ ^ (,)
x(x\l){x\2) . . . (x + m1) ^^^
x(xl)(x2) . . . (xm + 1) ^^^
37. Expressions fS and y are usually denoted by x^"^^ and a;^»»'
respective^, no special notation being employed to denote expressions
a and 8 ; this point is, however, not of great importance, since writing
expressions a and S in reverse order we see that they may be denoted by
{x + m—iy*^^ and (x—m + iy^^ respectively. We will therefore
confine our attention to expressions /S and y and their corresponding
general forms.
38. In the following examples we have assumed that Aa?=l: —
Let tCx=ic(xl){x2) . . . ixm + l)=a;^^')
then Attx={oi:\l)x(x—l) . . . (x^m\2)—x{x—l). . .(x—m^\)
= {(x + l)(xm\l)}x{xl)(x2) . . . {xm + 2)
=wa?<'»i> (13)
Similarly,
A^Ux=m{(x + l)x(x—l){x—2). . .(x—m + S)—x{x—l). . .{x^\2)}
=m{(x + l)(xm + 2)}x(xl)(x2) . . . (xm + 3) V
=m(m—l)x^'''^
«
Proceeding in this way, we see that
/^'^Hi^.=m(m—l)(m—2) . . . 3.2.1.a7<"»"»'
=z\m since x^'^~'^^=x^^^ = l
16 THE ELEMENTS OF FINITE DIFFERENCES.
This result is obviously correct, since a?'*"' is a function of x of
the mth degree.
39. (la) Let Ux=(ax\h){ax^l^b) . . . (ax—{m—l)\h)
then ^Ux={cix{l\h)(ax\h){ax^l\h) . . . {ax—{m—2)\h')
{ax + h){ax—l^h) . . . {ax—im^l)\h)
= {(ax + l + h) — {ax—(m—l)\h)}{{ax\l)(axl + h) . . .
(ax—(m—2)+b)}
=am(^ax + b)(ax—l + b)...(ax—(m—2) + h) . . (14)
and finally,
40. (2) Let ua;=
then Aux~~
x(x + l)(x + 2) . . . (^ + ^1)
1 1
{x + l){x + 2) . .. (x\7n) x(x+l)(xi2) .. . (x + m1)
x—(x + m)
07(0?+ 1) (07+ 2) . . . (x+m)
o7(o7 + l)(o7 + 2) . . . (x + m)
= wo7<'»+"^> (15)
Similarly,
A2i^a?=wj(m + l)o7<"^2)
From which we see that the successive orders of differences are
alternately negative and positive, and never vanish.
This is as it should be, for, if we perform the operation of division,
we see that x^~^^ represents an infinite series.
41. {2a) Let Ua;=
then Au,v'=
(ax\b){ax\l + b) . . . (ax + m — l + b)
1
(ax+l + b){ax + 2\b) . . . {ax + m + b)
1
(ax+b)(ax + l + b) . . . (ax+m—l\b)
—am
(ax\b)(ax+l\b) . . . {ax\m\b)
. . {15a)
«C,. may
be written ^j
»P,.
1")
^ «('•)
n
>»
n^n)
_ n^2)
,j(3) ^
THE ELEMENTS OF FINITE DIFFERENCES. 17
42. Before leaving this subject, it is well to point out that
also
' and
. Similarly,
43. The function '^Cr is a function of oi and r, if n varies while r
is constant, we have
{£ ,r r— 1
44. We have shown, in formulas (4) and (5), that
and that ^{Ux:kVx:kWa') = ^2lx:h^V,x:h^Wx
It now remains to obtain expressions for A(?^.^iv) and A( ~ ) .
^(UxVa) = Ux+nVx+h — UxVx
= (Ux + ^Ux) (Vx + ^Vx) — UxVx
=Wa?AVa.+ i7ipA?^.rHAwj;AiJa' (16)
Again, A O^ajt'.r) = Ux+hVx+h—l^xVx
or A {UxVx) = EWa?E't'aj — UxVx
where E has reference to ttx alone, and E' to Vx alone, then, by
separation of symbols,
WL ^{UxVx)=^iJ£^^'r}UxVx
whence
^^'^(UxVx) = (WE^'iyUxVa
expanding by Binomial Theorem,
= [(EEyHEE^)^^f ''^''~^^ (EE^)^2&c.}?^,.t^^
^Ux^nhVy^^nli—nUx^nlhVxirl^^ni^&O. (17)
c
18
THE ELEMENTS OF FINITE DIFFERENCES.
45. If we substitute (1 + A^) for E\
\^^(u^.v^,) = [E(l + AO l]^i^..r.,
= [/^ + EA']^u^Vj, (since E — 1 = A)
= rA>^ + ?jA^iEA^+ ^^^^~^^ A^^WA'^ + S^c^ua^va;
and, remembering that A and E apply only to Ux, while A' applies
only to Van
2
46.
o=
(18)
VxVx+h
— (^^a7 + A2^a?)^a; — ^rg(?^a'HAya;) _ Vx^Ux — Ux^Vx
VxVx+h VxVx+Ji
(19)
DiFFEEENCES OF 0.
47. The leading differences of any power of the natural numbers,
0, 1, 2, 3 ... w are known as the differences of ; thus we have the
differences of the series 0^, 1^, 2^, 3^, 4^, or generally of the series
0"S 1"*, 2"*, 3*^, 4'^* ... w^. The expression " Differences of " is used
because the leading term of the series is always zero. It will, of course,
be understood that the first powers of the natural numbers will have the
first differences constant, the second powers will have the second
differences constant, and so on.
48. A table of the differences of is conveniently set out as
follows : —
Wo
A
A2
A«
A4
A^
A6
01=0
...
,,,
...
...
...
02=0
2
...
...
...
03=0
6
6
...
...
...
04=0
14^
36
24/
...
...
0^=0
30
150
240
120
1
06=0
62
^1
1,560 ,
1,800
720
THE ELEMENTS OF FINITE DIFFERENCES. 19
It must be distinctly understood that this table does not represent
an ordinary table of differences of the function O*', but is a Table of
Reference from which the leading term and leading differences of x^^
can be obtained by inspection.
49. A table of the differences of will be found to be of considerable
value, since the powers of natural numbers are constantly involved in
practical calculations, and it is a great saving of time to be able to
readily obtain them.
50. In order to construct a table such as above, it is not necessary
to set out each series and then obtain their leading differences, we can
obtain a simple working formula thus :
By formula (12),
nin—Y)
when ?^^=.r"^ we have —
A»»ir'»=(a7 + w)^w(ar 4^21)'"+ Z (cc\n2y"&c.
and when cc=0 this becomes
71(71 ~~~ 1^
. \i
= 7i{{7iy''(nl){oiiy''\ (^^)(^^) (^^_2)mi_&c.}
2
=w{(l + 7JlV»i(^l)(Hw2)^i
(l + w3)^«^&c.}
=wA^~^l^*
whence A'^0^=w{A'*iO^» + A»«0^'^i} (20)
The final step is based on the relation — Aw^/
A»^«^;c=AwiW;i.+iA»^%^
transposing,
51. By means of formula (20) we have — ,/(.  /
A»O2=l{A0Oi + AiO»}
= 1(0 + 1}
= 1
c 2
20 THE ELEMENTS OF EINITE DTrFEBENCES.
Again— A^^=2{A^0'\A^0^}
=2(1 + 0}
= 2
Similarly,
A203=2{Ai02 + A202}
=2{H2}
= 6
A303=3{A202+A302}
=3(2 + 0}
=6
and so on.
We see that the expression for A^O^ involves A®0^ .
52. The symbol Aht^ can be used as an alternative for u^, that is
to say, A^ signifies that no operation has been performed.
This notation is analogous to that already familiar, e.g., a^, ^Cq,
10, &c.
THE ELEMENTS OE FINITE DIEFEBENCES. 21
CHAPTER III.
Intehpolation.
53. The word "interpolate" means "to insert", and therefore
refers to the insertion of missing terms in a series. It is, however,
used in a much wider sense, and has been extended to include the
determination of terms beyond those given, as well as the insertion of
intermediate terms.
54. For example : If we have given Wo> ^hi ^2? % and 2/6, we may be
required to interpolate U3, U4 or Uy, although Uj lies outside the series,
and therefore, strictly speaking, it cannot be inserted. So far, with one
exception, we have dealt solely with functions such as .r*^ a^, log^a?,
which can be readily calculated when a: has any particular value; when,
however, we are dealing with a function, the form of which is unknown,
we have already shown (on page 10) that under certain circumstances
we may obtain results which will be very nearly correct. It is to this
process that the term " interpolation " is applied in its widest sense.
55. It will be noticed that we are chiefly concerned with various
series, and that the position of any given term in a series is of primary
importance, this point being emphasised in the notation employed.
Thus, suppose we are dealing with the series formed by the fifth powers
of the natural numbers, i.e., 0^, P, 2*, 3% , we see that the
index does not vary, but that the quantity to be raised to the fifth
power is one less than the number representing its position in the series,
i.e., it is variable, and depends on the position of the term, hence the
lotation Uq, Ui, u^, &c.
56. In tlie case of almost all tabulated functions it has been found
]hat the successive orders of differences rapidly diminish ; it is therefore
3onvenient to assume that some given order of differences, say the wth,
is constant, and that the said function may in consequence be expressed
as a rational integral function of the variable x, thus,
Ux^=^a f 5a? 4 ex'^ 4 dx"^ + . . . f hx'^
22
THE ELEMENTS OF FINITE DIFFERENCES.
To determine the values of the coefficients «, 5, c, &c., we may adopt
the most suitable of the following methods :
v/57. 1st Method. — When n particular values of a function are given,
and it is required to find or interpolate other terms, it is usual to assume
that the function will be of degree (ii—l). This is because we can
only obtain (n—1) orders of differences when we have n equidistant
terms, and A'^~i will be constant only when the function is of degree
Given n equidistant values, tCa, Ua+hi if'a+^ii', &c. . . . ■Wce+7T^;i of a
function, we have the relation —
m(w— 1) _ n(n—l)(n—2)^^
or writing w for nh whence n=
w
Ua+x= Ua ^l^'^a^ ^J^ ^^^a + ^ ^hla + &C.
Example. — The 3 percent annuities by the H^^ Table on lives aged
21, 25, 29, 33, and 37 are respectively 21 .857, 21'025, 20132, 19*145,
and 18'057. Find the annuity for age 30.
Here A;r=4, e^a=21"857, 2fa+4=21'025, &c., and we require to
find 2^a+9.
Differencing, we have —
X
Mx
A%
A'«x
A%x
A^«x
a
21857
 832
 061
033
+ 026
« + 4
21025
 893
094
007
...
a + 8
20132
 987
101
...
...
« + 12
19145
1088
...
...
...
« + 16
18057
...
...
...
By formula (11) —
«««=«a+ ?A»„+ ^^'^''+ ^^^'"'' + *"
THE ELEMENTS OF FINITE DIFFEEENCES. 23
that is —
«„„=21857c832)^(061)^^(033)«^^026)
«, o.« (7488 2745 1485 3510)
6,144)
= 21857  {1872 + 086 + '004 + 000}
=21857 1962
= 19895
>^8. 2nd Methpd. — (a) When the values given and the value sought
constitute a series of equidistant terms, then, whatever the position of
the value sought, the following simple method may be employed :
Let the series, including the missing term, consist of n terms, so
that n—1 terms are given, and denote the series by u^, Ui, e^, U3 . . . Unh
Employing formula (12),
we may assume that the function is of degree (n^2) when A*^"~%o will
have the value zero. U u t ^ ''
i.e., A»'wo=«^wi(w— l)«^>t2f . • • +( — 1)**"'%=0 . . («)
In this last equation there is only one unknown quantity, and its
value can therefore be readily ascertained.
Example :
Givoii log 350= 5441^ log351 = 5453 "^
log 3 52 = 5465^ log 3 53= 5478 J[
log 354= 5490 rf ""
find any one of the given values, Bay log 353, which may be
supposed missing.
Let log 3*50= '5441=^0? and so on.
We have four terms given exclusive of the missing term, and
theref dre assume t he function to be of the third degree.
Whence A%o=^\
i.e., by formula («},
«^4 — 4t^ ^ 6^2— 4wi 4 «^o=0
.'. transposing; ;og 353=1^3= —
I =5477
24 / THE ELEMENTS OF FINITE DIFFERENCES.
19. (J) Suppose two terms to be missing out of ii equidistant
terms, then in order to apply the above method we must assume the
function to be of degree (?i— 3) whence
Aw%o=0, also A'^hci=0,
and we have two equations from which to determine the values of the
two unknowns.
i.e., A^%o=«*n2(>^2)/^«_3+ . . . f (l)^%o=
60. It must be carefully noted that the order of differences to be
equated to zero is the same as the number of term« given, not the
number of terms in the completed series. Also, if more than one term
is missing, we form our equations by equating to zero the correct order
of differences of tco, wi, etc.
61. (<?) The following course may be adopted wheu the terms given
consist of two or three equidistant terms, together with one isolated
term, say, Uq, Wi, u.2 and tCg, then assuming A%o to be constant, A'^Uq
and all higher differences vanish.
xCo^ — l)
By formula tt^ = Uq f ^r A % H — —f, —  A%o + &f ' • '
I
9 8 9 8 7
or «9= Uo + 9 Az^o + ir A^^^o + ^r^ A^^o
r I?
!9 8 ^
% + 9Ae^oHn^A2Mo
transposing zx'Wo= —
84
A^^o and ^'^Uq can of course be obtained by differencing %, ?^i, «2, and
having obtained A%o from above equation," the series can now be
completed by addition in the ordinary way.
62. 3rd Method. — If n values of the function, not necessarily
equidistant, are given, assume that
and suppose the given functions to be denoted by
7fa, Ubi 2tc ' ' . 2fnr' /
we then have n equations by means of which to determine the n
coefficients A, B, C . . . . K,
THE ELEMENTS OE FINITE DIFFERENCES.
25
namely, Ua=A{Ba + Ca^ + T>a^{ . . . +Ka^»
&c. = &c.
63. This method is obviously not well adapted to the interpolation
of one or two terms only, since considerable labour is iuvo]yed iii^the
solution of the equations. If, however, a table is being formed in which
the successive values follow a definite law, and do not change rapidly, it
will frequently be sufficiently accurate to calculate every fifth or tenth
value, and to supply the intermediate values by interpolation.
Example. — Given,
/3o= 89,685, 735=86,137, /4o= 82,277, 745=77,918,
Denote, /30, I35, <fec. by Uq, %, &c.,
then assume that, 2^^= A + Bx + Coc"^ + D^r^
so that
differencing,
(*)W
3,548
(o)(i).
3,860
(d)(.e)
4,359
again,
(/)(«)
 312
(ff){f)
 499
again,
(i)W
 187
from (Z)
w
(«)
(«)
Therefore,
89,685=1^0 =A
86,137=% =A+ 5BH 25C+ 125D
82,277 = ^f ,0= A h lOB + lOOC + 1,000D
77,918= «,5= A + 15B + 225C + 3,375D
(a)
(&)v
= 5B+ 25C+ 125D (e) v"
= 5B+ 75C+ 875D (/)
= 5B + 125C + 2,375D (y)
= 50C+ 750D (h)
= 50C + 1,500D (i)
= 750D (/)
D= 2493
C= 25
B= 69086
A= 89,685
2^0= A + OB + 02c + 03D = 89,685
2^1 = A + IB + PC 41^0= 88,991
«2=A + 2Bf22C + 23D=88,291
&c.=&c.
26 THE ELEMENTS OF FINITE DIFFERENCES.
The correct values are given in the column headed Ix on page 14 of
A Short Collection of Actuarial Tables.
64. It will be noticed that the coefficients of B, C and D are the 1st,
2nd and 3rd powers of the natural numbers. The work can therefore
be simplified if tables of these powers are formed by the constant
addition of the leading differences of 0* , 0^ and 0^ to be found in the
table of the "Differences of 0" on page (18).
65. Again, since B, C and D are constant, and "a constant coefficient
of a function remains a constant coefficient of its differences*', we need
only take the product of these quantities into the leading differences of
their respective coefficients, and can then obtain the successive terms by
a single operation.
66. The process of interpolating several consecutive values is
known as Subdivision of Intervals, and can be more readily effected
as in the following example.
Griving Uq, Us, Uio, &c., the same values as in the preceding
example, it is required to complete the series by the insertion of the
intermediate values.
If we represent u^—'Uo by Ae^o
and U10—U5 by A%
so that Ao? = 5
we may also represent Ui — uq by Suq
and U2—U1 by Sui
when, of course, 8x = 1
then we have —
89,685 = 2^0 =Uo
86,137=% =2^0+ 5^Uo+ 1082^^0+ 10S%o
82,277 =Uio=Uo\10Suo\ 4582e^o+1208%o
77,918 = ^15 = Wo + 158% h 10582^^0 f 45583%
Now successively differencing both sides of the equation, we have —
— 3,548 =A%= 58%+ 1082%+ 108^%
3,860= Aw5= 58%+ 3582^^0+ 11083%
4,359= A2^io= 58% + 6082%+ 3358^%
 312 =A2%= 2582% + 1008%o
 499=A2%=2582%+2258%o
 187=A3%=12583%
THE ELEMENTS OE FINITE DIEFERENCES. 27
whence h^Uo= —1*496
8^Uo= 6496
8wo= 693616
2^0=89,685
From which the remaining values of the function can be easily
obtained.
Lagran&e's Theorem.
67. 4th Method. — lo be employed when the given values are not
equidistant.
As before, we assume the function to be rational, integral and of
degree (w— 1), but instead of adopting the form with which we are
familiar, viz., Ux=A\Ba;\Ga:^+ . . . + Ka;**~^ the following expression
must be employed :
+ B{x—a)(^£C—c)(a:—d) . . . (x—n)
}C(a>—a){a:—b)(a: — d) . . . (x—n)
\ D(x—a) ({v—h) (x—c) . . . (x—n)
+ &C.
\'N(x—a){a:—b){x—c) . . . (a;—m)
where there are n terms, ^.e., as many terms as the given values of the
function, and where a, h, c . . . n denote the numerical values of the
suffixes. J^ote. — Ua, Ub, Ua &c., need not he equidistant.
68. It is important to note that each term is of n dimensions, and
that (a?— «), (07—5), (d7— c), &c., are replaced by A, B, C, &c.,
respectively. Only (ji—V) terms involve a?, and the function is therefore
of degree {n—V) as required.
69. To determine the values of A, B, C, &c., first, let x=a
.'. Ux—Ua, and all terms on the righthand side involving {x—a)
vanish, whence
Ua=Ps.{a—h){a—c){a — d) . . . {a—oi)
and A= ^— ^ , "^ — , r^
{a — o){a — c){a — d) . . . {a—n)
Similarly, lei x=h
(21)
then B= ''^
ila){h~c){hd) . . . (J)n)
28
THE ELEMENTS OF FINITE DIFFERENCES.
and so on ; substituting the values in the original equation
(a:—b)(x—c)({r—d) . . . (a;—n) >^
Ux^Ua
\Ub
(a — b){a—c)(a—d) . . . {a—n)
(j)c—d){x—c){x—d) . . . {x—ri)
ijba){hc){bd) . .. (bn)
+ &C.
or, dividing, both sides by {x—a)(a:—b)(j;—c){a:—d)
have finally
Ux Ua
(22)
{x—7i), we
{x—a){x—h){x—c) . . {x—n) (x—a){a—b){a—c)(a—d) . . (a—n)
■^ {bd){xb){bc){bd) . . (bn)
(23)
{c—d){c—b)(x—c){c — d) . . {c—n)
+ &C.
70. This formula will be readily remembered when it is noticed that
the factors of the successive denominators on both sides are formed by
taking the difference between the suffix of the numerators and the
various suffixes of the given values of the function, replacing all factors
in the form {a— a), {b — h), &c., by {x—a), (x — b), &c., respectively.
Example :
Given log 654=28156 log 658=28182
log 659=28189 log 661=28202
Find log 656.
. Let ua=2'8166 and a=0
«f6=28182 so that b=4i
Wc=28189 „ c=5
2^^=28202 „ d=7
and ^=2
then
also
(xa)(xb)(xc)(xd)=::(2)(2)(3)(6) =  60
(ah)(ac)(^ad)(xa) = (4>)(5)(7)(2) = 280
(ha){bc){bd)(xb) = (4<){l)(S){2) =  24
^ca)(cb){cd)(xc) = (5)(l)(2){S) =+30
(da)(db){dc)(xd) = (7)(S){2)('5) =210
THE ELEMENTS OF FI?TITE DIFFERENCES. 29
Therefore —
1 _ _ Ua^ ll'b Uc U£
■" 60 ^^•'~ ~ 280 " 24 "^ 30 ~ 210
L.C.M. of denominators =840
14iUa;=SUa + S5ub — 2SlCc + 4iUd
= 844681 986370
789292 + 112808
= 1183646 789292
=394354
whence 2^^;= 2 '8168.
True value of log 656 = 28169.
71. It may be pointed out that the sum of the coefficients pn the
righthand side of the equation must equal the coefficient of u^ . Also
the terms in the neighbourhood of the term to be interpolated will
always have the larger coefficients.
Functions of Two Vaeiables.
72. Functions of this kind are most frequently met with in practice
in the form of premiums to assure a sum of money on the death of the
first of two lives, or annuities payable during the joint existence of two
lives.
73. If we require the value of an annuity payable during the joint
lives of two persons aged 43 and 57 respectively, using the data given
on page 27 of the Actuarial Tables, denote this value by «43:57 in
conformity with the notation at the top of each column, and take as the
given values «40:50? ^40:55> ^40:60} ^45:50) ^45:55? ^45:605 ^50:50) ^50:55) ^50:60»
For the purpose of calculation, represent these by «o:Oj ?^o:5> ^0:10?
W5:0j «^5:5, %:105 ^lOcO, ^10:5, 2^10:10, ^ud Ict t^x:y rCprCSeut a43:57, SO
that a?=3 and ^ = 7. Assume second differences to be constant.
74. We may now proceed as follows : — Separate the given values into
groups of three, according to the values of the first variable, and obtain
the successive values ?/o:7> %:;? «^io:7 by the use of the formula —
tKen by the same formula obtain the value u^ 7, using as data the three
values just obtained.
30 THE ELEMENTS OF FINITE DIFFERENCES.
75. A double application of Lagrange's Theorem will enable us to
arrive at the same result.
Thus by formula (22),
„ _„ (75)(710) (70)(710) (70)(75)
(0_5)(010) ' "••^(5~0)(510) "'"(lOOXlOS)
3 21 7
—^^OiOok" "^^OtSoK +^0:1025
_„ (75)(710) (70) (710) (70(75)
 — W 5:0 77;^ ^T77^ — TTvT ~r^5:5 7^ 7^^^^ 777T ' "^ 5:10
(05)(010) ■ ^"(50)(510) ■ '•^"(100)(105)
3^ 21 7
(75)(7 10) ^ (70) (7 10) ^ (70)(75)
''^"=^'''"=\o5)(oio) +'^^«=^(5o)(5^io) +^^«=^«(lo:ro)(io=5)
_ 3 21 7
— '^10:0rtF +^10:5^ +^10:10rtK
Now, similarly employing 2^0:7 > «^5:7> «^io:7, we can write down the
value of %:7, thus:
_ (35)(3>10) (30) (3 10) (30)(35)
^3:7^0:7 ^^_g>j^Q_,^Q) +''^=^(50) (510) +''^«=^ (100) (105)
or
7 ^ 21 ^ 3
«^3:7 = «^0:725 +^5:725 +^^10 = 725"
Substituting numerical values for Uq.q^ ih.^^ &c.
^07= ^{(3) X 10894 + 21 x.9796h7 x 8553}
25 I
=9316 >
1 ^
«^5^7= — {(3) X 10591 + 21 X 9583 + 7 x 8*417}
=9136
^^lO:7=^^ {(3) X 10059+ 21x9181 + 7x8132}
=8782
and, finally,
u^.^= ^ {7 X 9316 + 21 X 9136+ (3) x 8782}
= 9229 = «43:57
True value 9190.
THE ELEMENTS OF FINITE DIFFERENCES.
31
76. A closer approximation would have been obtained had third
dilterences been employed.
77. The numerical values of the coefficients of Uut Uh, ih in the
formula (22) when a^ h and c are in arithmetical progression with a
common difference 5, are of such frequent occurrence that it would
appear useful to tabulate them for central values of x.^
Tables of Coefficients of
•8 2
'6 '4 \ When 1st Differences are assumed constant.
•4 6
•2 8
Uq
+ •28
+ 12
•08
12
+ •84
+ •96
100
+ •96
+ •84
12
•08
+ •12
+ •28 j
When 2nd Differences are assumed
constant.
X
8
9
10
11
12
«o
6 048 +864 +216 032
7 _.064 +672 +448 056 )
8 056 +448 +672 '064
9 •032 +216 +864 '048 J
When 3rd Differences are
assumed constant.
«5
WiO Mi5
0336 +3584 +8064 ^1536
•0176 +1584 +'9504 1056
10000
_.0144 1056 +9504 +1584
+ •0224 1536 +8064 +^3584
+ •0224
+ 0144
0176
•0336
When
4th Differences
are assumed
constant.
* The values of x can be restricted to these limits, as explained on page 33 in
Chapter IV,
32 THE ELEMENTS OP FINITE DIFFEBENCES.
[It is interesting to notice that the coefficients in the table of second
differences are those of the successive terms in Mr. W. S. B. Woolhouse's
graduation formula, which formula may be directly obtained by Lagrange
without the introduction of Central Differences. See Jotirnal of the
Institute of Actuaries J vol. xiii, p. 95.]
THE ELEMENTS OF FINITE DIFFEKENCES. 33
CHAPTER IV.
Centeal Differences.
78. It will probably have been noticed that the examples illustrating
Interpolation were all of one form, that is to say, a few values only of
a function were given, and it was required to furnish one or more
additional values, making use of all the data at our disposal. In
actual practice, the problem is seldom presented in this form ; so far as
the actuary is concerned, he is generally in possession of fairly complete
tables, in which every value, or at least every fifth value, of the function
is tabulated: — thus, in A Short Collection of Actimrial Tables, it is
not until we come to page 27 that we find the first table which affords
scope for practical interpolation. Before we can apply any interpolation
formula, it is necessary to determine two points, namely : (1) what
order of differences may be assumed to be constant, ^(2) which values
of the function are to be employed. ^Formula (10) is best adapted for ^^'
exhibiting the first point, while the importance of the latter will be
more clearly illustrated if we consider Lagrange's theorem, where the
interpolated term is ihown to consist in the algebraical sum of certain
proportions of each of the terms employed, the proportion being
greatest in the case of those terms nearest to the interpolated term.
That is to say, we assume the successive terms in the series to be
mutually dependent on each other, and to be influenced to a greater
extent by those in their immediate neighbourhood.
79. If, therefore, the term to be found occupies as nearly as possible
a central position among the terms employed, we may anticipate more
accurate results than could be obtained if it were otherwise situated,
provided we give full weight to both of the points just discussed and
employ the data selected without any arbitrary adjustments.
80. Our object will be attained by taking values of the function
Uq, Ui\ ii2, ?^3) «^4? so that itx lies in close proximity to u^^ and employing
formula (10), whence
«^2+.T=«^o+(2 + ^)Az^o+ ^?±^^i:t?)A%o + &c.
34 THE ELEMENTS OP FINITE DIFFEEENCES.
but the following notation was introcliiced by Mr. W. S. B. Woolhouse,
and is generally employed in Central Difference Formulas : — Instead of
denoting the given terms by Wo> ih, th, ^hi «^4, replace the suffixes by
— 2, —1, 0, 1, 2, respectively. Thus we have
Old Notation New Notatioii
^U2 ^2 ,
Aui i^H2 ^1 ^1
where Az^_2, ^^U2, &e., are denoted by «_2, 6_i, &c.,
and «o=2(«i + «i)
Co=i(c_i + Ci)
Rewriting the formula for U2^x given above, we have —
(2+ar)(l + ar)a;.,
!£
and it is required to convert this into a formula involving the central
differences «o? ^o? <^0) ^o> together with the central term Uq\\^^
We have ^o=J(^i + ^i) whence c_i = 2(?o— ^i
also (^0=^1 — ^1 ,» c_i = Ci — r7o ^
adding and dividing by 2 ^ » ^_^_^^_i^^
again, ' J_i = io— c_i ^
again,
finally, «/_2=2^i — «2
= (i/o— «i) — («o— Po + ^o— l^o)
= («^o— [«o— Po]) — (^/o— io + <?o— l^o)
=7^0— 2^^0 + 2 Jo— ^o+i^o
THE ELEMENTS OF FINITE DIFFEREXCES.
35
collecting, Kx = tfo —2ao\2bo—Co+lck
(2 + ^)(l + a,)
+
(h—Co + ^do)
(2+^X1 + ^)^,
+ j3 {co—ido)
, (2 + a?)(Har)a:(a7l)
H Ti dn
rearranging,
itx=th\oe\(^o+ T^hoj
^ i.\ I ^(^^~1)
13
(.o+l^o)
(24)
81. If in equation (24)
we substitute «o=«i— 2^0 and Co=Ci — <?o
we have —
=2;o + ^'^( «i — ioo+ 12*0)"^ (3 ( ^1— 2ao4 ^doj
^(071) , (^ + l).r(^l) {.v+l)x(xl){a^2)
=i(o + xai +
ho +
Ci +
Bl . a?'!) iv^^\ (a7 + l)<3) (a^ + iy^^
Ci4
1 ^ ' 12' 13 " • 4
82. These two formulas must be learnt.
Formula (25) can be easily remembered as follows :
, /"(^o Co\ , rho do\ ^ (Cq e^ (d^ fo\ 1 ,
do .(26)'
• (27) I
or «.r=«/o+.Wo+ 12 *o+ jg <'o+ j7 <?o+ g eo+&c.
/y» /y»2 /y»3
Again, formula (27) may be extended:
.(1) ^(2)
^^a;=1lo+7Y(fl+ ^2^0+ g
a7<2) (.r + l)(3) (.r + l)<^> (.i7 + 2)<^
^^+T4^^«+ 5
^1
/o+&c.
D 2
86 THE ELEMENTS OF FINITE DIFFERENCES.
■A
83. These formulas have been derived from the original formula —
2^.^=w_2+ (2 f^) A«^_2+ i^±fKi±f} A%_2 + &c.
by the substitution of identities. They will therefore give exactly
identical results when employed in full. It is only when we neglect the
higher orders of central differences, either &o or ^o> that we obtain
different results, for it will be noticed that Co=c_if ^t^o? so that,
although we appear to discard cIq, we still retain a portion of it.
84. If, therefore, the central difference formula is taken as far as
first, third, or any odd order of central differences, we may obtain truer
results than would be possible by the use of the corresponding order of
ordinary differences, but it must be remembered that third central
differences require five given terms which would enable us to obtain
fourth ordinary differences, and so on. If, however, the central
difference formula is taken as far as an even order of central differences,
the result obtained is precisely the same as when the ordinary differences
are used. Central differences, therefore, do not appear to possess any
considerable value, since with the same data we can quite as easily
obtain the same results with ordinary differences. If any of the
adjustments in central differences are employed, there is always the risk
of obtaining a worse instead of better result, as is necessarily the case
in any arbitrary adjustment.
THE ELEMENTS OF FINITE DIFFERENCES. 37
CHAPTER V.
Summation or Integration.
85. Summation has been defined as the act or process of forming
a sum or total amount. Integration has been defined as the act or
process of making whole or entire.
Summation is met with by the actuary under two aspects, that is to
say, (1) it is either definite or (2) approximate.
86. (1) Definite summation is possible (a) when all the terms to
be summed are given, or (6) when the algebraic form of the terms is
known.
87. (a) It is not always convenient to adopt the simple expedient of
ordinary addition, and even where this is possible an independent method
l^_obtaining the same result is always advisable.
I^P Let the series to be summed be represented by ita, tia+u ^(a+2, «^a+3,
&c. . . . iia+nii then, since
we have na=Ua
L
a common ratio of (ljA),
&C. = &C.
i.e., Ua\tla+\+ • . . \Ua+ni
__ (lh A)^l
"^^(1 + A)1
38
THE ELEMENTS OF FINITE DIFFEEENCES.
1 + 71A +
= Ua
\2~
A2 + &C.1
f . n(n—l) ^ 7i(n—l)(n — 2) „ „ )
(28)
88. The result obtained by this formula will be identical with thait
obtained by ordinary addition, provided (n) is an integer and Ua+x is a
rational integral function of a; of a finite degree.
89. (b) When the algebraic form of the function is known, it is
frequently unnecessary either to find the value of each term and sum
in the ordinary manner, or even to make use of the formula (28) just
given, which necessitates the finding of the leading term and leading
differences ; we may instead employ the method of Finite Integration,
which method will be more clearly understood by consideration of the
following extension of the scheme given in paragraph 8 : —
Ai A« Ai A2
^0
Uo
Auo
Vi
Ul
Awi
^
th
^3
Vn2
Vn^\
Un2
tlUn.
In the same way as Ux precedes and is the function from which ^Ux
is derived, so we see that it is possible for another function, Vx, to bear
the same relation to Ux .
In other words, as Ux is the integral function of which t^Ux is one
of the elements, so Vx is the integral of Ux»
90. Integration is here shown to be the inverse of the operation
of differencing, and is denoted by the symbol A~^, or '%. 2^^^♦ is
referred to as the indefinite integral as distinct from the definite
integral, '^^^Uxt which latter represents the definite summation between
the limits n and n + m, as hereafter explained.
THE ELEMENTS OF FINITE DIFFEBENCES. 39
91. Now applying formula (9), we have ^W
/^(Va; + c) = Ua;
which becomes, after integration,
or V;c^c='^Ux (29)
92. It will be noticed that, in the j^rocess of integration, a constant
c is introduced, the reason for which is readily understood when it is
remembered that a constant term always disappears in the process of
differencing.
93. When dealing with indefinite integrals, the nature of this
constant requires more careful consideration, but in the case of definite
integrals, the constant introduced by integration will disappear.
Thus, if ^Ux —Vx\c
as shown in paragraph 95.
94. Applying formula (l)',^^nd giving h the value unity, we have,
successively,
'i^l —Vq = Uq
V2 —Vi = Ui
Va —Vai = Uai
Va+i — Va = Ua
I^ni — ^w2 = 'ihi+2 ««
Vn —Vni = Uni
Summing both sides, we obtain
'Vn — V(i=tio + Ui + U2\ . . . \Uni
also 'i)a — Vo=tCo + Ui+ . . . +Ua1
whence 'Vn—Va=UaiUa+i+ . . . +Unm (30)
95. In all these cases we have assumed that summation is to be
commenced and terminated at two definite points, which are known as
the inferior and superior limits of summation.
Thus ^«V^,=t'^— t'o
= Vn — Va (31)
40 THE ELEMENTS OF FINITE DIFFERENCES.
96. Vx is the indefinite integral. Hence we see, from the above
formula, that to integrate between limits, say a and (;i— 1), we have
merely to give x in the indefinite integral the value n and subtract from
this the corresponding integral when x has the value a.
07. We have stated that an integral may exist in the case of
any function, but although in certain functions the integral is known or
may be obtained by a consideration of the law connecting the function
and its successive leading dilFerences, we possess no formula by means of
which it may be universally derived, as we have, for example, with the
formula (1). We shall, therefore, later on, give a list of the more
generally required integrals.
98. We have already obtained the first differences of x'^^ «^,
n(n — 1^
I^x'^z^nx'^^ + ^ ^ x^^\&c.
A«^=(«l)a^
99. In the first case we have a complicated expression, which does
not lend itself to integration, but can be dealt with by means of
formula (28). In the other three cases we have simple expressions,
which, after transposition, become —
A«^
a^=.
^ini—\)—
m
100. Therefore, by integration, we have —
S«^=2^ =^+c (32)
a—1 a—1
2^(m.i=2^f^ =— +<; («)
m m
2^<.»i'=S=^ = ^— +c (i)
— 7W —m
where c is the constant introduced by integration.
THE ELEMENTS OF FINITE DIFFEEENCES. ^ 41
In equation («) write m for {m — V) .', (w + 1) for m,
then %x^^'^= — —\c (33)
y
—m—1 becomes
m + 1
In equation (h) write m—1 for
— (w— 1) — 1 or —m,
then
^^im)— _ _ __c
(34)
■(m1)
101. (2) Approximate Summation.
This method is due to Sir J. Lubbock, and was given by him in
1829. The formula is therefore known as " Lubbock's Formula," and
is applicable when the data consists of certain equidistant terms and it
is desired to sum the series including the intermediate terms.
102. Let Uo, Uf, ^t2t, • • • u^H^^n, Unt, u^it, • . • ^^tJ^^^i^, be the
successive values of any function u^ where t is fractional and n integral,
and where nt=l, so that t=  and Untt ^hnt, 'Usnt, • • • Umnt ^^Y be
n
written Ui, U2, zht ^w
Then employing the formula — ,
cc( X ~~~ 1"^
ttx — tfO + X^tfo + " 7^ ^^Uq + &C.
Uq = Uo
t(i—l)
If
lC2t = Uo + 2t^Uo+ —^ — ^ A22^0 + &C.
we have
&c. = &c.
. — ^,, nlt(nltl) „
Hence, by addition.
\+ut+ . . . +w,T3i<=?iWo + ^(1 + 2 + 3+ . . . +n — l)Auo
+ ^ {(^1) +2(22^1) +&c.}A2wo+&c.
=nuo+t
— l.n
G*'
^^0+ 7^(1+22 + 32+. . . + W12)
  (1 + 2 + 3+ . . . +?il)1 A2«^o + &c.
42 THE ELEMENTS OF FINITE DIFFEEENCES.
n—1.7i . rt^n—l.n.2n—l
n ^ n
t 91 — 1.71
12 2
A%o+&c.
Similarly,
+ &C.
n—1^ n'^—l^„
:=nUo+——Auo r— AX + &C.
2 12^
w— 1 . ^2—1
Therefore,
n1 ^ n^1 ^^ , ,
= nuo\ — — Auq— —t— A%o+&c.
2 li^i
^_1 ^2—1
1ww, + "2 ^Ui j2^ A%i + &C.
+ ^^^^2+ 7, ^^2— jir— A22^2+ &C.
2 12?i
+ &C.
= n{uo+Ui + U2+ . . . 42^mi}
w— 1
12n { )
+ &c.
But Al^o+ A2^i+ . . . + ^Umi = Um—Uo
and A2wo + A2wi+ . . . + A2e^,„_i = Aw,„— Awo
So, finally :
n—1 %2— 1
4 "2" (^'« — ^0)— j^2,2 {^U,n^l(o)
+ &C (35)
THE ELEMENTS OF FINITE DIFFERENCES.
43
103. I£ the series Uq, ut, e%, &c., be an infinite and convergent
series, formula (35) may be written
Uo + Ui\U2t\ . . . ad inf.=n{uo\Ui.+ . . .}
where the series 2^o+^'i + ?^2 + &c. is continued until the numerical value
of any term is negligible.
104. If the interval of differencing be n, so that ^Ux=Uj;+n—Ua:;
and if ^=1, then formulas (35) and (36) become
and
n—1 n^—1
+ g (^mn—Uo) j^— (Aw,„,i— Ai^o) +&C. . (35«).
^0 + 2^1+ • • . ad inf.=n{uo + thi + ^i2}i+  • .ad inf.}
n — 1 n^—1 , „ , ,
^ Wo4 j^— Ai^o— &c. . . . (36a)
Indefinite Integrals.
Function
Indefinite Integral
a^
xim)
ar(  '»)
x{x — \)
2
a;(w+l)
+ c
+ c
■+ b){ax — 1 + b)... {ax — m + l + b)
{ax + b)(ax + 1 t b) . . . {ax + ml+b)
m ■
{aT + b){ax + l + b) . . . (ax + m — l + b)
{ua! + b){axl + b) . . . (axm + l + b)
(ax + b
){ax
1 + b) . . . (ax — mi
^Kc
+ b){
a{m + l)
■^■"'^ 1 .
(ax1
%x + b) . . . {ax + m
a{m + 1)
1
— a{m
l){ax + b){ax + l+b) . .
1
. {ax + m
2 +
+ c
h)
^.7=: +C
i
a{m — V){ax — l + b) . . . {ax — m + 1 + b)
QUESTIONS.
CHAPTER I.
1. Divide
1887, Oct.— 1st Paper, No. 6.
2. Find the numerical value of the following expression :
rhere rt=16, 2>=2, ^=3, ^=4.
1887, Ofit.— 1st Paper, No. 5.
T„ h — c c — a a — h
3. If + + =0,
y—z z—x x—y
prove that (^h — c){y—zy\{c—a){z—xy^{a — l){x—yy=.^.
11900.— 1st Paper, No. 2.
4. Simplify the expressions —
x^\{a^}))x'^^{ah^V)xVh
(1)
(2)
hx'^V («i + 1).t2+ {a + l)x^ 1 '
^^12 + 6v/3
1901,— 1st Paper, No. 2.
4G QUESTIONS.
5. State and prove the rule for finding tbe Greatest Common
Measure of two or more algebraical expressions, explaining what factors
may be introduced or rejected in the process.
Prove that if a and h be any two integers greater than unity,
a% — ah^ is always divisible by 3.
1892, Oct.— 1st Paper, No. 3.
6. Prove that every common multiple of two algebraical
expressions is a multiple of their Least Common Multiple.
Find the Greatest Common Measure of —
6a;54^4_ii^^3_3^2_3^_l and ^x^\'lx^—\^x'\'^x^
1887, Oct.— 1st Paper, No. 7.
7. Find the L.C.M. and H.C.F. of ^«+l and x^\x''^\.
Simplify the expression —
^3_^_6 072 + 2^+6
X
^2 + 2^7 + 3 x^^2x—Vl
8. Find the H.C.F. of
a7416^'3+93^2_234^.216 and
2^'3_24^2 + 93^_117.
Obtain the square root of — 2\/6^
1899.— 1st Paper, No. 2.
1902.— 1st Paper, No. 3.
9. Extract the square root of 7 + V^13, and the cube root of
10+yi08.
1893.— 1st Paper, No. 2.
10. Simplify the expression : —
«2 52 g2
{a^r00){ay)(ac) (b + x)(bc)(ha) (c\x)(ca)(cb)'
and prove that, if n be an odd integer, then (a—h)^ + {b—c)^^(c—a)^
is always divisible by (a — b){b — c){c—a).
1895.— 1st Paper, No. 2.
QUESTIONS. 47
11. If I take any number such as 84,796,205, and subtract from it
a number formed of the same digits, such as 60,594,287, I am sure to
find a remainder consisting of a multiple of nine, and having the sum
of its digits a multiple of nine. Prove this to be true for all numbers,
and explain the reason.
1898.— 2nd Paper, No. 4.
12. Show that the product of any number of positive quantities,
whose sum is given, is greatest when the quantities are all equal.
1902.— 2nd Paper, No. 21.
13. Prove that the product of any n successive integers is always
divisible by ]^. ^>/ca6
If n is an integer, prove that
120 "24 + 30''*^'°*° '"*"§*'■•
1899.— 2nd Paper, No. 9.
14. Resolve  . \ ^ .  r into partial fractions.
{x—a){x—o){(c—c)
Show that the coefficient of x'*^ in the expansion of
[iXx) (l..r) (l«^.r)]  i. \C7ilrP ■
1900.— 1st Paper, No. 8.
I
15. Explain what is meant by a Continued Fraction. By means of a
Continued Fraction find a series of convergent fractions giving successive
approximations to the probability of drawing a white ball at the first
jal from an urn containing 1,000 balls, of which 785 are white.
1890, Oct.— 2nd Paper, No. 18.
16. Find the value of
^*l
1887, Oct.— 1st Paper, No. 1.
48 QUESTIONS.
CHAPTER II.
17. Solve the equations —
(1) y/6—a:+ v/5 + ^= \/Sx+4i ;
(2) x^+f=S6, 07+^=5.
18. Solve the equations —
a^^+f + z^^Sa^
19. Solve the equations —
(2) \/x^\2x4< + 2cc^=lS4iX.
20. Solve the equations —
(1) cc—ay\a^z^a^,
and x—cy^d^z=c^\
(2) \/2^Tl + v/7a727= \/3ar+4.
21. Solve the equations —
(1) lx—ay=l\
»
(2) a?2a7+\/2ar23a?+5=15+.
1901.— 1st Paper, No. 3.
1899.— 2ncl Paper, No. 4.
1899.— 1st Paper, No. 3.
1897.— 1st Paper, No. 3.
1902,— 1st Paper, No. 4.
QUESTIONS. 40
22. Solve the equations —
^ + 4 07—2 _ 4
(2) (^ + l)(^+5)(^ + 9)(^ + 13) = 105;
^ ^ x^—xy+y''= 19./
1900.— 2nd Paper, No. 15.
23. Solve the equations —
^^^ 3^2~2 3""^^'
and (2) Vxia + h—co) + \/»(6— a4^) + \^b{a + x—h) =0.
1896.— 1st Paper, No. 3.
24. Solve the equations —
10^+17 1207+2 5o7— 4
(1)
18 13a?16 9
a;2 + a?y+y2=io9
^ ^ xy= 2).
1898.— 1st Paper, No. 3.
25. Solve —
(1) y^M^ + v^^^z:^=?±^
a:23y=l
and
(2) ^^^^=n
1895.— 1st Paper, No. 3.
26. Solve the following equations —
1 1_ . _! \ L.j._J: 1 1 _
^^07 1 + 07 2 + 0; 3 + 07 4 + 0; 5 + 07 6 + 07 7 + 07~ '
(2) ^(1 + 0^)2 ^(1^)2= 4/1^:^2.
1892, Oct.— 2nd Paper, No. 2.
57. Solve the equations —
(1) \/o72 + 2o7l + \/o;2 + o7 + l = \/2 + \/.3 ;
^ 07 7
(2)
072 — 3 ^2_3 ^3_^3 •
1893.— 1st Paper, No. 3,
50 QUESTIONS.
28. Solve the equations
(a) (74iy3)a;2(2~v/3)a?=2;
(&) 12a7456a73+89^2_56^^12=o.
1888, Oct.— 1st Paper, No. 5.
29. Solve the equations —
(i) J 072+^2 + ^2 = 45^
I yz=20;
and in positive integers the equation
(ii) 19^2307=7.
1894.— 2nd Paper, No. 4.
30. Prove that integral values can always be found to satisfy the
equation ax—hy=:.c provided that a and h be prime to one another.
Solve in positive integers the equation 3o7— 4y=ll.
1896.— 1st Paper, No. 10.
CHAPTER III.
31. Show that a quadratic equation cannot have more than
two roots.
Show that the roots of the equation  A 1 =0 are
real if a^— ah + 62 fg positive.
1901.— 2nd Paper, No. 15.
32. Find the condition that the roots of the equation ax^+hx + c=0
may be real and different.
If x be real, find the limits between which the expression
a726o? + 25
^^:pg^— musthe.
1000.— l8t Paper, No. 3.
QUESTIONS. , 51
33. Prove that there are two values of 'p for which the equation
«a?2+2A^ + 6f/?(a'a;22^'a7l&')=0 has equal roots.
1897.— 2nd Paper, No. 4.
^ 34. If a and ^ be the roots of the equation 3a?2+6a?f2=0, show
that the equation whose roots are — — and — — will be 3a?2— 18a7+2=0.
p a
1889, April.— 1st Paper, No. 10.
35. Find the condition that the roots of the equation
flrar2+5a? + c=0 may be (1) both positive, and (2) opposite in sign, but
the greater of them negative.
1889, Oct.— 2nd Paper, No. 6.
36. Find the condition that the expression ax^^hx^c may retain
a constant sign for all values of x.
If a be positive and less than 2, show that the equation —
(a?+3)(l«.r)(« + l)(ar+3)=2
has no real roots.
1902.— 2nd Paper, No. 14,
CHAPTER IV.
37. State the rules for the position of the decimal point in the
product and quotient of multiplication and division of decimals
respectively. By the use of algebraical symbols show the rationale of
the rules.
1889, April.— 1st Paper, No. 3.
m
^■88. Prove the rule for the conversion of a recurring decimal into a
^Hilgar fraction.
^K 1891, Oct.— 2na Paper, No. 13.
52 QUESTIONS.
39. state and prove the rule for contracted multiplication of
decimals.
What is the value to the nearest farthing of
328794 X £8. 17s. 4i^. ?
888, April.— 1st Paper, No. 4,
40. Divide 73*64 by '432 by the method of contracted division so
that the quotient may have four places of decimals.
, ,• 1887, Oct.— 1st Paper, No. 2.
41. Express as a decimal fraction the quantity
9^1i^^l.x.3xl74i.006x ''
3.3 9 '" 4207
1894.— 2nd Paper, No. 2.
42. State the rule for the decimalization of a sum of money, correct
to three places of decimals — i.e., for writing down at sight any sum of
money in terms of pounds and decimals of a pound. Illustrate by
expressing £487. 17s. lO^d. as a decimal, and also by writing down the
value of £33*693 in pounds, shillings, pence, and farthings.
1896.— 1st Paper, No. 1.
43. 12,000 sums of money, consisting of (besides pounds and
shillings) equal numbers of all possible sums of pence from 0^. to lid.
inclusive, are each expressed in pounds and decimals of a pound, three
places of decimals being retained, and the usual rule being observed by
which the figure in the last place is increased by unity if the first of the
figures rejected be not less than 5. What will be the error in the
amount obtained by adding these decimals together ?
1890, April.— 1st Paper, No. 2.
44. Find the least sum of money which can be paid in pence,
shillings, fiorins, halfcrowns, crowns, sevenshilling pieces, halfguineas,
sovereigns, and guineas.
1888, Oct.— 1st Paper, No. 1.
45. A cube contains 1,953,125 cubic inches. Find, to four places
of decimals, the difference between the lengths of its edge and its
diagonal.
1890, Oct.— 2nd Paper, No, 14.
QUESTION'S. 53
46. Express 23 lbs. 9 ozs. 12 drams avoirdupois decimally in terms of
a lb. troy, assuming 7,000 grains troy equal to a lb. avoirdupois.
1888, April,— 1st Paper, No. 2.
47. The cost of levelling and turfing a square cricket field, at
£175. 9s. 4^d. per acre, is £987 ; find tbe cost of surrounding it with a
railing at 3s. 2d. per yard.
1900.— 2nd Paper, No. 13.
CHAPTER V.
48. If a? vary directly as the square of y, and inversely as the cube
root of z; and if a^=2 when y=4 and z=:8, find y when .r=3 and
2;=27.
1888, Oct.— 1st Paper, No. 7.
49. If m sovereigns in a row stretch as far as n pennies, and
p sovereigns in a heap are as high as q pennies, compare the value of
equal bulks of gold and copper, it being assumed that the area of a
circle varies as the square of its radius, and that the value of the copper
in 240 pennies is equal to that of the gold in a sovereign.
1895.— 2nd Paper, No. 7.
50. Two vessels contain mixtures of wine and water ; in one there
is thrice as much wine as water, in the other five times as much water
as wine. Find how much must be drawn from each to fill a third
vessel which holds seven gallons, in order that its contents may be half
wine and half water.
1892, Oct.— ^nd Paper, No. 5,
I
51. A and B start at the same instant and each walks at a uniform
rate, the former in x hours from P to Q, and the latter in y hours from
Q to P. They meet on the road a hours before A arrives at Q, and
I purs before B arrives at P ; show that —
1901.— 1st Paper, No. 4.
54 QUESTIONS.
52. If each of two quantities varies as a third when the other is
constant, show that when both vary the third varies as their product.
Assuming that the quantity of work done varies as the number
of days when the number of working hours per day is constant, and that
the quantity of work done per hour varies as the excess of 12 over the
number of working hours per day, find what the number of working
hours per day must be in order that the work of 27 days may be
equivalent to the work of 20 days of 9 hours eact.
1892, April.— 1st Paper, No. 4.
53. A body is composed partly of copper and partly of tin. If the
copper had been tin, and the tin copper, the weight of the body would
have been greater by 9 per cent, than what it actually is. The weights
of equal volumes of copper and tin are as 8'96 to 7'29. Find how much
of the body is copper and how much tin.
1893.— 1st Paper, No. 6.
54. If the speed on a railway be 20 miles an hour, it is found that
the expenses are just paid. If the speed be more than 20 miles
an hour, the increase of the receipts is found to vary as the increase
of the velocity, and the increase of the cost of working to vary as the
square of the increase of the velocity. At the rate of 40 miles an hour
the expenses are just paid. Find the velocity at which the profits are
greatest.
1899.— 1st Paper, No. 9.
55. The value of a pound of gold is 30 times that of a pound of
silver, and the weights of equal quantities of gold and silver are as
19 to 10. Find the value of a bar of silver equal in bulk to a bar of
gold worth £2,500.
1S97.— 2n(i Paper, No. 2.
56. If A vary as B when C is constant, and if it vary as C when
B is constant, show that it varies as BC when both B and C vary.
Three circular coins of radii , i, and ^ of an inch, and of
thickness ^^, ^, and ^^ of an inch respectively, are melted together
and formed into a circular coin ^th of an inch in thickness. Assuming
that the areas of circles vary as the squares of their radii, find, to
three places of decimals of an inch, the radius of this single coin.
1902.— 2ud Paper, No. 15.
QUESTIONS. 55
CHAPTER VI.
57. One of two clocks, which both indicate the true time today at
noon, gains a second an hour, and the other loses three seconds in two
hours. When next will they both indicate the same time, and when
next will they both indicate the true time ?
1892, Oct.— 2nd Paper, No. 4.
58. Two persons, A and B, entered into a speculation to which B
subscribed £15 more than A. After four months C was admitted, who
added £50 to the stock, and at the end of twelve months from C's
admission they found that the total gain from the commencement was
£159. A then withdrew, and received for principal and gain £88.
What did A originally subscribe ?
1891, Oct.— 2nd Paper, No. 19.
59. A ship started on a voyage with a sufficient supply of water
for all on board, numbering 175 souls. After 30 days there was a daily
loss of 3 lives from scurvy. A storm delayed the ship 3 weeks, and it
got to port just as the supply of water ran out. What was the length
of the voyage ?
1891, Oct.— 1st Paper, No. 7.
60. Two casks containing, the one a, and the other a' gallons, are
filled with a mixture of wine and water, such that the first contains /8,
and the second P' gallons of wine. By means of two smaller casks a
gallon is drawn from each of the larger casks simultaneously, and
transferred to the other. This operation having been repeated n times,
how much wine will there be in each of the two casks ?
1891, April.— 2nd Paper, No. 16.
61. A man buys a certain number of eggs at 11 a shilling, and
three times as many at 15 a shilling ; he mixes them and sells them at
13 a shilling ; how much does he gain or lose per cent. ?
1901.— 2nd Paper, No. 13.
56 QUESTIONS.
62. What is the price of meat per lb. if, on a reduction of 25 per
cent, in the price, 8 lbs. more than before are obtained for a sovereign ?
1897.— 2nd Paper, No. 3»
63. A tradesman sells his goods at a price which gives a profit of
X per cent, on the selling price and x plus 50 per cent, on the cost price.
If his profit for a single day be £20, find his gross receipts for that day.'
1889, Oct.— 1st Paper, No. 8.
64. A retail dealer, who reckons his average cash sale to be Is. 8<?.,
adds to the wholesale price 25 per cent, for profit, and further additions
to provide for discount of 10 per cent, for cash, and to meet expenses at
the rate of 2t?. per average cash sale. He buys one set of articles
wholesale at 81s. per hundred. What should be his nominal retail
selling price ?
1892, April.— Is Paper, No. 1.
65. A man invested £1,874. 8s. in 2f per cent. Consols at 98;
how much money must he invest in a 3J per cent. Stock at 104 J, so
that the rate of interest on his whole investment may be 3 per cent. ?
[Brokerage \ per cent, in each case.]
1900.— 1st Paper, No. 1.
66. A landlord has an estate which brings him in £4,000 a year,
but the repairs, &c., cost him 15 per cent, of the gross rental. He sells
the estate at 30 years' purchase on the gross income and invests the
price in 2f per cent. Consols at 112 (brokerage i). Find to the
nearest penny the resulting difference in his net income.
1898.— 1st Paper, No. 1.
67. A man sold out from a Stock, paying 2f per cent, per annum,
at 96i, and invested the proceeds in another Stock paying 4 per cent.
per annum. If thereby he raises his income 5 per cent., find the price
of the second Stock.
1899. 1st Paper, No. 1.
68. The gross income of a man was £30 more in the second of two
particular years than in the first, but in consequence of the incometax
rising from hd. in the pound in the first year to ^d. in the pound in the
second, his net income was unaltered. Find the income.
1890, April.— 2nd Paper, No. 15.
QUESTIONS. 57
69. A man holds certain amounts of a 5 per cent, stock and of a
6 per cent, stock, and he sells out from both when the former stands at
102 and the latter at lOli, and invests in a 5^ per cent, stock at 103.
He then finds that his income is unaltered. Compare the amounts of
the stocks that he held, brokerage being reckoned in each case at
ith per cent.
1S97.— 1st Paper, No. 5.
70. If a man buys equal amounts of a 3^ per cent, stock and a
4^ per cent, stock, his income is £1,880. If he invest so that his
income is the same from each kind of stock, his income is £1,872.
Find what his income will be if he invests half his capital in each of the
two stocks.
1896.— 1st Paper, No. 9.
71. How soon after halfpast one o'clock will the second hand of a
clock
(«) Overtake the hour hand ?
(h) Bisect the angle between the other two hands ?
(c) Overtake the minute hand ?
1891, April.— 1st Paper, No. 5.
72. It is found that it takes 8 minutes to fill a certain cistern by
ineans of one tap, 24 by means of another, and only 5^ minutes when
both taps are used. Show that this indicates that the cistern has a
leak which would empty it in 66 minutes.
1892, Oct. 1st Paper, No. 6.
73. During the time that the shadow on a sundial, which shows
true time, moves from 1 o'clock to 5, a clock, which is too fast by
m hours and n minutes, strikes a total number of strokes equal to m + n,
and it is observed that the number of minutes is less by 41 than the
square of the number which the clock strikes at the last time of
striking. The clock does not strike 12 o'clock during the time. How
much is it too fast ?
1891, April.— 2nd Paper, No. 17.
74. The beam of a defective balance is horizontal when the weight
in one scale is onefourteenth more than the weight in the other. A
tradesman placing a pound weight alternately in the two scales of the
balance, sells what he thinks to be two pounds of a certain commodity.
What weight does he actually give his customer ?
1896. 2nd Paper, No. 2.
58 QUESTIONS.
75. A cricketer, on getting out in the first innings of a match, finds
that he has in that innings increased his average for the season by 2 ; on
getting out in the second innings he finds that he has still further
increased his average by 2. If in the whole match he made 80 runs,
find how many runs he made in each innings of the match.
1895.— 1st Paper, No. 9.
76. Write down any sum of money consisting of any number of
pounds, shillings, and pence, the number of pounds being greater than
the number of pence, and the number of pounds less than twelve.
Subtract from it the sum obtained by interchanging the numbers of
pounds and pence. To this result add the sum obtained by
interchanging in it the numbers of pounds and pence. Show that the
answer is always £12. 18s. lid.
1901.— 1st Paper, No. 14.
77. A cyclist travels the first half of the distance from P to Q at
the rate of 10 miles an hour ; he then rests 15 minutes and travels the
remainder of the distance at the rate of 15 miles an hour. Had he
travelled the whole distance without stopping at the rate of 11 miles per
hour he would have taken exactly the same time. Find the distance
from P to Q.
1898.— 2nd Paper, No. 3.
78. A train travelling at the rate of 30 miles per hour starts from a
town A to a town B ; 12 minutes later a second train travelling at
40 miles per hour starts from B to A, and goes half a mile beyond the
middle point between A and B before it meets the first train ; find the
distance between A and B.
1900.— 2nd Paper, No. 16.
79. A train passes two men walking beside a railway in 3i seconds
and 3f seconds respectively; a second train passes the men in 4f and 4^
seconds respectively. Prove that the latter train will overtake the
former, and, if they be on difierent rails, will pass it in 36 seconds.
1894.— 1st Paper, No. 6.
80. The men in a regiment can be formed in a solid square and also
in a hollow square four deep. The number of men in the front of the
latter formation exceeds twice the number in the front of the former
formation by four. Find the number of men in the regiment.
1898.— 1st Paper, No. 5.
QUESTIONS. 59
81. Find four equidifferent numbers such that their product shall
be 384, and the sum of their squares 120.
1893.— 2nd Paper, No. 7.
82. There are two numbers, each consisting of two digits, and the
digits of the second are those of the first but reversed ; the product of
these numbers is 8,722 ; if the first number be divided by the second,
the quotient will be 1 with a remainder consisting of one figure only.
Find the numbers.
1896. 1st Paper, No. 5.
83. Find that number of six digits such that when the extreme
lefthand digit is transposed to the right hand, the rest being
unaltered, the number is increased threefold. Show that there are two
solutions, the greater being double the other.
If the smaller solution be taken and the same process be
repeated on it 2, 3, 4, and 5 times successively, then the numbers
obtained are respectively 2, 6, 4, and 5 times the original number.
1894.— 2nd Paper, No. 7.
84. A father's age is equal to those of his three children together.
In 9 years it will amount to those of the two eldest ; in 3 years after
that, to those of the eldest and youngest ; and in 3 years after that, to
those of the two youngest. Find their present ages.
1894.— 2nd Paper, No. 3.
85. In a race of a quarter of a mile A allows B 15 yards and C
40 yards start, and they all reach the goal together. How much start
can B allow C when B has to run the whole distance ?
1902.— 1st Paper, No. 1.
86. The proprietor of a boardingschool, having already 30 pupils,
finds that an addition of 5 increases his gross yearly expenditure by
£300, but diminishes the average cost per head by £1. What did
his expenses originally amount to ?
1902.— 1st Paper, No. 2.
87. Three men, A, B, and C, throw dice upon the condition that he
who makes the lowest throw shall give each of the others the sum he
has already. Each loses in turn in the order named, and at the end of
the third game all have the same sum, namely, £n. How much had
each at first ?
1890, April.— 1st Paper, No. 7.
60 QUESTIONS.
88. A man pays insurance on his life amounting to onetenth o£
his income; after paying incometax at the rate of Is. 2d. in the £ on
the remainder of his income he has £1,066. 8s. Qd. left; what was his
gross income ?
1902, 2nd Paper, No. 13.
89. A is three times as old as B was when A was as old as B is
now ; and in 24 years' time B will be twice as old as A was when B was
half as old as A is now. Find the ages of A and B.
1898.— 2nd Paper, No. 7.
90. Given one solution in positive integers of the equation
ax{hy=.c, find the general solution.
How can I divide 5 halfcrowns, 6 florins, and 29 sixpences
between two persons so that each person shall have the same amount of
money and also the same number of coins ?
1901.— 2nd Paper, No. 20.
91. Given one solution in integers of the equation acc\hi/=c, find
the general solution.
In how many ways can £100 be paid by using halfcrowns and
sovereigns only ?
1898.— 1st Paper, No. 11.
CHAPTER VII.
92. Define Permutations and Comhinations, and obtain general
formulas for the number of each kind which can be made out of
n things taken ^ at a time.
1888, April.— 1st Paper, No. 9.
93. Find the number of combinations of n things taken r together
in each of which p of the given n things always occur.
How many different arrangements can there be made of
7 persons at a round table ?
1901.— 1st Paper, No. 6.
t
QUESTIONS. 61
94. Find the number of permutations of n things taken altogether,
when^ are of one kind, q o£ a, second, and r of a third.
A committee of eight is to be elected from 14 men, of whom
six are British, and eight are Canadians. In how many ways can the
committee be selected so that the British may not be outnumbered ?
1898.— 1st Paper, No. 6.
95. Find the number of permutations of n things taken r at a
time, and deduce the number of combinations.
A closed chain is to be formed of twelve links of different
metals ; how many different arrangements can be made ?
1894.— 1st ^aper, No. 5.
96. Find the number of combinations of n things, taken r at a
time, without assuming the formula for permutations.
How many words can be made by using all the letters in the
word AbracadajbraQa ?
1900,— 1st Paper, No. 4.
97. Find the number of permutations of n things taken altogether,
there being p things of one kind, q of another kind, and the rest being
all different.
Find the number of different words that can be formed from
the letters in the word repetition, there being always a consonant both
in the first and last place.
1895.— 2nd Paper, No. 5.
98. A bag contains 9 counters marked 1 to 9. How many
different numbers, each of nine digits, can be formed from 5 counters
taken from the bag, and four other counters each marked ?
1893.— 1st Paper, No. 5.
99. Find the number of permutations of n things taken r at a time,
and hence deduce the number of combinations.
In a railway carriage, which seats 5 on each side, there are
10 people. Of these, 3 wish to travel with their backs to the engine,
and 2 refuse to do this. In how many ways can they be arranged ?
1899.— 1st Paper, No. 6.
62
QUESTIONS.
100. There are a sopranos, h contraltos, c tenors, and d basses.
How many choirs of eight voices, two of each kind, can be formed ?
1890, April.— 1st Paper, No. 3.
101. Seven prizes are to be distributed amongst five boys, each boy
being eligible for all the prizes. In how many ways can they be given
away?
1888, Oct.— 1st Paper, No. 11.
102. Find the number of combinations of n things taken r
at a time.
How many games of lawn tennis, in which each side consists of
a lady and gentleman, can be arranged at a party consisting of 7 ladies
and 4 gentlemen ?
1897.— 1st Paper, No. 6.
103. Define Permutations and Combinations, and, given n things,
find the number of each when the n things are taken ;• at a time.
Twelve balls can be arranged in three separate parcels, of
which the first parcel always contains three balls, the second always four
balls, the third always five. In how many ways can this be done ?
1887, April.— 1st Paper, No. 9.
104. Show that the number of ways in which mn things can be
divided among m persons, so that each shall have n of them, is ,~^^ .
1888, April.— 1st Paper, No. 10.
105. Show how to find the number of permutations of n things
taken all together, which are not all different.
Find the sum of the diff'erent numbers which can be formed
with three 6's and four 2's, all these seven digits being employed in the
formation of each number.
1892, April.— 2nd Paper, No. 18.
106. Find the number of permutations of n things taken r at
a time.
10 similar presents are to be distributed amongst 6 children so
that each child receives at least one present ; in how many ways can the
distribution be made ?
1902,— 1st Paper, No. 6,
QUESTIONS. 63
CHAPTER VIII.
107. Assuming the truth of the Binomial Theorem for a positive
integral exponent, prove its truth for a negative or fractional exponent,
stating the necessary limitations in the value of any symbol employed.
If X be so small that its square may be neglected, prove that
(1^^ = 2+32^
1900.— 1st Paper, No. 5.
108. Assuming the truth of the Binomial Theorem for a positive
integral exponent, prove its truth, with a certain condition, for a negative
or fractional index.
Write down the first five terms of (1 + a?)^, and hence show
that A/l6i= 100498756
1898.— 2nd Paper, No. 5.
109. Assuming the Binomial Theorem to be true when the
exponent is a positive integer, show that it is true when the exponent is
a positive fraction.
If N be a large number and a the integer next greater than
its fourth root, show that
'S^^H^ approximately.
1893.— 2nd Paper, No. 5.
110. Find the numerically greatest term in the expansion of
{a\xY'^ where w is a positive integer.
What will be the numerically greatest term of (3«+7&)7 ?
1887, April.— 1st Paper, No. 10.
64 QUESTIONS.
111. Assuming the truth of the Binomial Theorem for a positive
integral index, prove its truth, with a certain restriction, for any index.
Show that the sums of the two infinite series —
3 3^ 3.9.15 3.9.15.21
"^8 8.16 8.16.24 8.16.24.32 "*"***
and lfi + i4.i + _i_+. ..
are equal.
1902.— 2nd Paper, No. 18.
112. Expand, to four terms, the quantities (1 — £p)^ and (1 + 2;^)^,
stating under what restrictions the expansions are respectively valid.
Prove that the two infinite series
11^ J^^ . 1.4.7.10
"^4 4.8 4.8.12 "^4.8.12716 "*■ * ' *
1 L ? 4 — ^ ^'^'^ L 2.5.8.11
and ^+6 ^6. 12 "^6. 12. 18 "^6. 12. 18724"^
are equal to one another.
1894.— 2nd Paper, No. 5.
113. In the expansion of (1 + ^)^ by the Binomial Theorem,
n being integral, prove that the sum of the even coefficients and the sum
of the odd coefficients are each equal to 2^~^ .
Eesolve ~ ——^ —z — into partial fractions, and find
(a?— l)(a?— 2)(ir— 3)
the coefficient of a?** in the expansion by the Binomial Theorem, x being
less than unity.
1896.— 1st Paper, No. 6.
114. Find the greatest term in the expansion of (^x + a)^, and
[pand ( J in a series of ascending powers of a:.
\a—xj
1890, April.— 2nd Paper, No. 14.
115. Write down the general term \i.e.^ the (r + l)th] in the
expansion of (a + ^r)** by the Binomial Theorem.
Obtain, in their simplest forms, the coefficients of x^ in
(3 \*7 / iv
2 +  a: J and of ir^r+i j^ the expansion of f x J
1902.— 1st Paper, No. 7.
QUESTIONS. 65
116. Find the first four terms in the expansion of —
(l + 4a:):^X (1307)*
in i^ovvers of x.
Find the coefficient of cc^ in the expansion of — j — w'\~ •
yL ~~~ ^Ou )
1901.— 2nd Paper, No. 17.
117. Prove the truth of the Binomial Theorem for a positive
integral exponent.
Show that the sum of the squares of the coefficients in the
\2n
1S95.— 1st Paper, No. 6.
expansion of (lH.r)»*, where n is an integer, is
118. Prove the truth of the Binomial Theorem for a positive
integral exponent.
If {1\xY=zcq\CiX{C2X^\ . . . f . . . Vcnx'^ show that—
\2n
CqC2{CiC^\C2C^\ . . . +Cw2<?W =
n—2\n\2
and that ^»_^J + f _ . . . +(_i)» ^» 1
1 2 ' 3 ' V / ^^__j^ ^^^j^
1901.— 1st Paper, No. 7.
119. If {1 + oc:)''=Cq{Cicc + c^^+ . . .^CnX^, find the value of
Find the coefficient of x'*' in the expansion of ) ~ , where
n and r are positive integers and r>n.
1900.— 2nd Paper, No. 18.
120. Find the sum of the products of the coefficients of the powers
of X taken two together in the expansion of (Jl\xY.
1892, Oct.— 2nd Paper, No. 8.
66 QUESTIO?fS.
OHAPTEll IX.
121. What is meant by the probability of an event happening,
and how is this denoted mathematically ?
If ^ denote the probability of an event happening, what
inter^Dretation is to be assigned to the cases where ^=0, ^=1, and^ = ?
1887, April.— 2nd Paper, No. 8.
122. Explain how the probability of the happening of an event is
estimated mathematically.
Two persons, A and B, play a match at a certain game, the
winner being the one who first wins two consecutive games, no game
being drawn. Their chances of winning any particular game are as
a : h. Show that A's initial chance of winning is = • — 7— 77: . If
a\b a^ + ab\b^
A wins the first game, prove that his chance of winning is then increased
to
(fi \ai\b^' 1899.— 1st Paper, No. 11.
123. A general orders 2 men drawn by lot out of 150 mutineers to
be shot. If there are 12 ringleaders of the mutiny, find the chances
that (a) only one, (b) two, (c) none of the ringleaders are chosen.
Prove your answers.
1800, Oct.— 2nd Paper, No. 21.
124. Twenty persons are arranged at random in a straight line ;
show that the chance that four given persons, out of the twenty, shall be
consecutive is —  .
^oD 1901.— 1st Paper, No. 10.
125. There are three events. A, B, and C, of which one must
happen. The odds are 3 to 7 on A, and 3 to 6 on B. Find the
odds on C.
1888, April.— 2nd Paper, No. 6.
126. Five balls are drawn from a bag of 12 balls, of which 3 are
known to be marked ones. What is the probability that all the three
marked ones are among the 5 drawn ?
1887, April.— 2nd Paper, No. 9.
QUESTIONS. 67
127. In a certain competition the chances are 7 to 3 against one
competitor A, 11 to 5 against B, and 13 to 7 against C. What is the
chance that one of these three wins ?
1897.— 1st Paper, No. 10.
128. Five men speak at a meeting; what is the chance that A
speaks immediately before B ?
What is the chance, if B agrees not to speak till A has
spoken ?
1890, Oct.— 1st Paper, No. 7.
129. The figures 1, 2, 4, 5, 7, 8 are arranged at random as the
period of a circulating decimal, which is then reduced to a vulgar
fraction in its lowest terms. What are the odds against the denominator
being 7 ?
1890, April.— 2nd Paper, No. 20.
130. The sum of two positive integers is 80 ; find the chance that
their product is greater than GOO.
1896.— 2nd Paper, No. 10.
131. What is the most probable number of aces that will be thrown
in 35 throws with 1 die ? Give an algebraical proof of the correctness
of your answer.
1890, April.— 1st Paper, No. 5,
132. Define Probability, and show how it is mathematically
measured.
A pack of 52 playing cards is dealt out at random (with the
cards face downwards) into 4 sets of 13 cards each. The top card of
one of these sets is lifted and found to be the ace of hearts. Show that
the chance that the king, queen, and knave of hearts are in this same
set is about ^^.
1902.— 1st Paper, No. 9.
133. What is the probability that the first card dealt out from
a full pack will be a spade ; and that being so, that the second will be
a heart ? And a spade and a heart being already out, what probability
remains that the third will be a spade ?
^^^ 1889, April,— 2nd Paper, No. 8.
68 QUESTIONS.
134. Two cards are drawn from a complete pack. What is the
chance that the sum of their values will be
(a) Exactly 20;
(b) At least 20;
assuming that each ace and court card is reckoned as 11.
1892, April.— 2nd Paper, No. 21.
136. Find the chance that a player, taken at random from a whist
party, has —
(1) All the trumps in his hand ;
(2) No trumps in his hand ;
and (3) No cards of any particular suit. "I
1894.— Ist Paper, No. 11.
136. The probability of the happening of an event in one trial
being known, find (I) the probability of its happening o times exactly
in n trials, and (2) the probability of its failing exactly r times
in n trials.
1888, April.— 2nd Paper, No. 7.
137. The chance of an event happening in any trial being known,
find the chance of its happening at least 7' times in n trials.
From a pack of fifty two j)laying cards four are drawn at
random. Show that the odds against there being one of each suit
lies between 8 to 1 and 9 to 1.
1901.— 1st Paper, No. 11.
138. The chance of a given event happening in one trial is 1 in 7i
where w is a large number ; prove that it will be an even chance that
it happens at least once in "693 . , . x n trials.
1894.— 2nd Paper, No. 10.
139. The probability of the happening of an event in one trial
being known, find the probability of its happening 1, 2, 3,
times exactly in n trials.
Find the chance that in 7 throws with a pair of dice (each
having six faces marked 1, 2, 3, 4, 5 and 6 respectively) the sum of
the readings of the faces turned u^^ will be 9 in exactly 3 of these
trials.
1897.— 2nd Paper, No. 10.
QUESTIONS. by
140. The chances of an event happening in one trial beings, find
the chance o£ its happening exactly r times in n trials.
Three dice with faces marked 1 to 6 were thrown, and the sum
of the numbers cast was 12. Find the chance that the number shown
by each die was 4.
1899. 2nd Paper, No. 10.
141. Three black balls and five white ones are placed in a bag ;
three men draw in succession (the drawn ball being continually
replaced before the next drawing) until a black ball is drawn ; prove
that their respective chances of success are as 64 : 40 : 25. If the
drawn ball be not replaced, show that their respective chances are as
27 : 18 : 11.
1900.— 2n(l Paper, No. 22.
142. Show that the probability that two independent events should
both happen is the product of the separate probabilities of their
happening.
A throws twice with a cubical die whose faces are numbered
from 1 to 6, and B throws four times with a counter marked 1 on one
side and 6 on the other ; find the chance that the sum of the numbers
thrown by B shall be double the sum of those thrown by A.
1900.— 2na Paper, No. 21.
143. Four men, A, B, C, D, throw with two dice whose faces are
numbered 1, 2, 3, 4, 5, 6 respectively; the man who first throws
7 wins the stake ; if they throw in the order A, B, C, D, and so on
continually, what are their respective chances of winning ?
1901.— 2iul Paper, No. 21.
^» 144. A die has 2 sides marked with the number 3, and 4 sides
marked with the number 5 ; it is thrown 100 times. What is the
probability that the sum of the numbers thrown will be exactly 400 ?
1891, April.— 2na Paper, No. 22.
145. In two throws with a pair of dice, what are the probabilities
of the followina: events :
I
(ft) 8 the first throw, followed by 9 the second ;
(i) 9 the first throw, or, if not^ then 8 the second ?
1891, Oct.— 2nd Paper, No. li
O
70 QUESTIONS.
146. A and B play a match at a certain game, the winner being the
one who first wins three games, no games being drawn. A*s chance ol
Q
winning any particular game is  . Find the chance of his winning thi
o
match after B has won the first game, and prove that the odds against
him are nearly 11 to 10.
1898.— 1st Paper, No. 10.
147. A man writes three letters and directs three envelopes into
which he puts them at random ; find the chance that all the letters will
go wrong.
1889, Oct.— 2nd Paper, No. 7.
148. A cylindrical saucepan, 8 inches in diameter, has a circular
hole in the bottom, centrally placed, 1 inch in diameter. A coin of
2 inches diameter is thrown in. Show that the chance of its falling
clear of the hole is 3 to 1.
1891, April.— 1st Paper, No. 9.
149. A ball, 1 inch in diameter, is thrown against a wire netting,
the apertures of which are squares of 3 inches a side. Find the chance
that it goes through without hitting a wire. The thickness of the wire
may be neglected.
1889, Oct.— 2ncl Paper, No. 8.
150. Define and illustrate the term Expectation as used in the
Doctrine of Probabilities.
1889, Oct.— 1st Paper, No. 6.
151. A bag contains 10 tickets numbered from 1 to 10. A ticket
is drawn and replaced four times. What is the chance that the sum of
the numbers drawn is 33 ?
If in that case the drawer receives £1,000, what is the value
of his expectation ?
1892, Oct.— 1st Paper, No. 7.
152. A and B throw two dice once for a pool of £1,000. If the
throw is 7, A takes the whole ; if 8, B takes the whole ; in any other
case, each takes onehalf of the pool. What should A pay B to equalise
their chances ?
1891, April.— 1st Paper, No. 8.
QUESTIONS. 71
153. Two players of equal skill, A and B, play a set of games.
They leave off when A wants 3 points and B 2. If the stake is £16,
what share ought each to take ?
1888, Oct.— 2ikI Paper, No. 7.
154. A has a box containing 4 dice, with the faces numbered
1 to 6, and B has a dicebox containing 3 dice similarly numbered.
They throw in succession, A commencing, and the one that first throws
10 receives the stake of £1. Find the values of their respective
expectations.
1902.— 1st Paper, No. 10.
155. Explain what is meant by Inverse Probability.
A bag contains 8 balls, each of which is equally likely to be
white or black ; three balls are drawn and turn out to be two white
and one black; these three balls are then removed, and another ball
is drawn ; show that it is an even chance that this is a white ball.
1902.— 2nd Paper, No. 22.
156. An event must have happened from one of n mutually
exclusive causes, the antecedent probabilities of which are Pi, P2, . . .
Vn respectively ; the probabilities that when these causes exist the
event happens are respectively ^1, ^2? • • •  Pn Show that on any
occasion on which the event happens, the probability that it was
due to the rth cause is r =r ^^^^ t^t .
Pi.^l + P2.^2 + . . . . + rn.pn
A purse contains five coins, each of which is equally likely to
be a sovereign or a shilling and cannot be anything else ; two of
these coins are drawn at random and found to be sovereigns. Show
that the fair price to offer for the purse is £3. lis. Gd.
1000.— 1st Paper, No. 10.
157. If there are two independent events, the respective
probabilities of which are known, find the probability that both
will occur.
A makes a correct statement 3 times out of 4, and B makes
a correct statement 9 times out of 10, and thoy both assert that a
ball drawn from a bag known to contain 10 balls, all of different
K lours, was white ; find the chance that it really was white.
1901.— 2ad Paper, No. 22.
72 QUESTIONS.
158. The a 'priori odds against a certain event having happened are
10^2 to 1 ; if 13 independent witnesses, each of whom makes a correct
assertion nine times out of ten, assert that it happened, prove that the
probability that it did happen is now about f .
1898.— 2nd Paper, No. 11.
159. A dynamically perfect coin having been tossed 10 times in
succession has always come up heads. What is the probability that the
11th toss will give a head {a) when the coin is known to have an obverse
and a reverse, {V) when the probability of, a coin being made with two
sides alike is one in a million ?
1890, Oct.— 1st Paper, No. 8.
160. A bag contains 10 balls of which 6 are known to be red, and
of the rest each is equally likely to be red or white. Three balls
are drawn, found to be red, and replaced, and then a white ball is drawn :
what is the chance that there are at least two more white balls in
the bag ?
1899.— 2nd Paper, No. 11.
161. There are {m\n) coins in a bag which are either sixpences or
shillings, and n of them are known to be sixpences ; of the remaining
m coins any number of sixpences is equally probable : prove that the
, p, . . . 2n^rm
chance or drawing a sixpence is —  •
189G.— 1st Paper, No. 11.
162. From a heap of playing cards, which originally consisted of
three of each suit, one card has been lost. From the remaining
eleven cards five are drawn at random, and are found to be two
hearts, two clubs, and a spade. If from the remaining six cards two
be now drawn at random, prove that the chance of at least one of
them being a diamond is  .
/ 1897.— 2nd Paper, No. 11.
QUESTIONS. 73
CHAPTER X.
y 163. Explain the nature of the Calculus of Finite Differences, and
furnish reasons why the value usually given to the difference of the
independent variable is unity.
Prove that if Ux be a rational and integral function of x of
the wth degree, its nih. differences are constant.
1892, April.— 1st Paper, No. 9.
^ 164. Define the symbols A«^^., A%a?, A%a;.
Obtain a formula for expressing Ux+n in terms of Ux and
its successive finite differences.
1887, Oct.— 2nd Paper, No. 10.
2a? + 7
165. If Wa.=: —^ — — ~, find A%^, expressing your result in
its simplest form.
1892, Oct.— 1st Paper, No. 10.
166. Find the value of A^^e^^., where Ux is equal to
x{x\l){x\2) . . . {pc^m—l),
and show that the ;ith difference of a rational integral algebraic function
of the nih. degree in x is constant.
1902.— 1st Paper, No. 11.
167. Find the value of ^^Hix^ when Ux is equal to —
(1) xn
(2) lr[_x{x+l){x + 2) . . . (xVml)].
1901.— 2nd Paper, No. 23.
I
168. Prove the formula —
^ . n(n—l) ^„ n(n — l)(n—2) ^ •
Given 2^0=89,685, 2^1=88,994, 2^=88,294, and e^3=87,585,
hd the value of i^y.
1902.— 2nd Paper, No. 28.
74 QUESTIONS.
169. If A^=«, and ^ttjc=^Ux+a—Ua;, prove that : —
n , n(n—a) ^„ 7i(n—a)(n—2a) ,„
Ua,+n=Ua:+  • Al^^+ ' , ' A%;, + ^ ^ ^^ ^A3«^^ + &c.
1891, Oct.— 1st Paper, No. 9.
170. Prove that ^^Ux=Ux^.nh—p\'Ux+ii:=\h\p2Ux+^^2h—&>G'i where
/?! , ^2 . . . are the numerical values of the coefficients in the expansion of
1888, Oct.— 2nd Paper, No. 10. :
171. Prove the formula —
n(n—V) .
If all the terms, except %, of the series Ui, u^,    th be
given, show that the value of % is —
70
1901.— 1st Paper, No. 12.
172. Show that
^'"Ux^^Ux+n— T^x+ni H .. ^ 2^a;+M2 — &C.
Hence show that, second differences being constant
A^+22A;,,+i + Aa;=0. ^
1887, April— 2nd Paper, No. 12.
173. When m consecutive equidistant values of a function Ux are
given, find an approximate general expression for it.
Find a rational integral function of a^ which for the values
a=2, 3, 4, 5 shall have the respective values 2, 14, 40, 86.
1902.— 1st Paper, No. 12.
174. («) Given ^^o= 2,844,
2^1=2,705,
W2= 2,501,
2^3=2,236,
find u^.
{h) Find by finite differences the first 10 terms of the series
1, 3, 5, 13, 33, &c.
1889, April.— 2nd Paper, No. 11.
QUESTIONS. 75
175. Let Ux l>e a function whose differences, when the increment of
X is unity, are denoted by hx, S . . ., and by A^?, A^ . . . when the
increment of x is n. Then, if 8'J, 8^+1 . . . are in Geometric Progression
(common ratio q), show that
(2._1)_^(^_1)(^_1)2
1888, April.— 2na Paper, No. 12.
176. Explain what is meant by Interpolation, as applied to the
calculation of tables, and show how to find a missing term in a series
of equidistant terms.
Find «^2, given that «^o=98,203, i^i = 97,843, ^^3=97,034,
and 2^4=96,569.
1900.— l.st Paper, No. 12.
177. Given log 350=2'54407,\Au
„ 351=254531, '^t
„ 352=254654,^'^
„ 354=254900, l/>j
find log 353. (/ ,
^ 1889, Oct.— 2nd Paper, No. 11.
178. Given log 422= 6253125.
„ 423 =6263404.
„ 424= 6273659.
„ 425 = 6283889.
Find an approximate value of log 421684.
1888, April.— 2nd Paper, No. 11»
179. The amount of £1 in 50 years —
at 2i per cent. =34371090.
3 „ =43839061.
3> „ , =55849264.
4 „ =71066845.
Find the amount at 3 per cent.
1888, Oct.— 2nd Paper, No. 9.
76 QUESTIONS.
180. Given that the annual contributions payable by members
joining a friendly society at the ages of 20, 25, 30, and 35 are as
follows : —
Age at Entry. Annual Contribution.
£ s. d.
20 . . 2
25 . . 2
30 . . 3
35 . . 4
9 : 7
18 : 5
10 : 3
6 : 3
find by interpolation the corresponding contributions for the inter
mediate ages at entry (A^ to be constant) .
1887, Oct.— 2nd Paper, No. 11.
181. Having given the following values of annuities, find by
interpolation the value of ^43 at 3^ per cent.
3 per cent.
3
I per cent.
4 per cent.
^40 17176
16103
15135
«42 16566
15568
14664
«44 15924
15001
14162"
1892, April.2nd Paper, No. 23
Having given
2^0=
: 9545
U^ =
: 14018
U2=
:l8322
U3 =
=22474
Find X when 2^j,= 1*5000
.
1891, April.— 1st Paper, No. 11.
183. Show how to interpolate an unknown value of a function, by
means of known values which are not consecutive.
By means of the formula, having given
log 280=24472
„ 281=24487
„ 283=24518
„ 286=24564
deduce log 282.
1890, April.— 1st Paper, No. 10.
QUESTIONS. 77
184. Demonstrate Lagrange's Theorem of Interpolation ; and, by
means of the formula, find the probability that a person aged 53 will live
a year, having given —
Probability that a person aged 50 will live a year= '98428.
„ 51 „ =98335.
„ 54 „ =98008.
„ 55 „ =97877.
1890, Oct.— 1st Paper, No. 0.
' 185. Demonstrate Lagrange's Formula of Interpolation, and
employ it to find the number of which the log is 2^, having given —
log 200=230103
t . 210=232222
] 220=234242
230=236173
1891, Oct. 1st Paper, No. 10.
H
186. In thfe series Uq, Ui, U2, &c., in which the successive differences
Az(x, A^Ux, &c., converge slowly, it is observed that the ratio — —  is
nearly equal to  —  . What expedient would you adopt to obtain the
Ux
best results upon interpolation ?
1891, April.— 2nd Paper, No. 23.
187. Show how "central differences" may be derived from
ordinary differences.
1890, Oct. 2nd Paper, No. 20.
188. («) Explain, in few words, the principles and advantages of
the method of " central differences."
(J) What is meant by the " differences of nothing " ? How
may they be calculated ?
1890, April— 2nd Paper, No. 22.
189. Deduce a formula for the sum of the series —
Hence find the sum of the fourth powers of the first n natural numbers.
1892, April.— 1st Paper, No. 10,
78 QUESTIONS. I >
i
190. Find the sum of the squares of the terms of the series
1, 4, 7, . . . 'Sn2,
1902.— 2ud Paper, No. 17.
191. 2 railway trucks travelled for 3 hours at 4 miles an hour,
^ ») ")•) ?J 5) *^ )> 5> t) ,, ,,
&c. &c. &c.
and 28 „ „ „ „ 29 „ „ 30 „
What was the mileage run in the aggregate ?
1890, Oct.— 1st Paper, No. 9.
192. Find by the method of finite differences the sum of n terms
of the series —
1 + 6 + 15+28 + &C.,
and of the series —
I2+22 + 32+42+&C.
1892, Oct.— 2nd Paper, No. 11.
193. If Ux=cix\h, prove that —
^ =c
UxUx\\'U'x+2 • • • y^x+ni a{n — l)UxUx+\ . • • 1(x+n'i
where C is a constant.
Find the sum of the series jg + ^^ + ^^^ + ...
to n terms.
1900.— 2na Paper, No. 24.
194. Find the integral of an expression of the form —
Ux'1f'x\\''i^x+2 ' ' • 'ifx+mli
•where Ux=^(iic\h, a and b being constants.
Find by this method the sum of n terms of the series
1.4.7 + 4.7.1047.10.13+ . . .
1901.— 2nd Paper, No. 24.
195. Find the integral of an expression of the form —
1^
'^X''i('x+i1^x+2' • • ^a?+wi
where iix=^ax + h.
Sum to n terms the series ;r— r + —  + — r + . . .
2.4 3.5 4.6
1902.— 2nd Paper, No. 24.
QUESTIONS. 79
19'6. Sum to 11 terms the series
(1) 12 + 22 + 32+... ^n^,
and (2) 1.3.7 + 2.4.9 + 3.5.11+
1896.— 2nd Paper, No. 6.
197. Find the sum of n terms of the series
(1) whose Tith term is (37^—1) (^ + 2) ;
^^^ 1.2.4"^2.3.5'^3.4i.6"^ * * *
1898.— 2nd Paper, No. 6.
198. Sum to n terms the series
(1) 1.2.4 + 2.3.5 + 3.4.6+
^"' (2)ll4 + 2i5 + 3X6+
State the rule for writing down the sum to n terms of the
series whose ^th term is
l^{«w + &}{«(w + l) + Z>}{«(^ + 2) + 5} .... to r factors.
1894.— 2nd Paper, No. 6.
\
199. Find the sum of the infinite series —
and of the series —
8X8 + lO^l + 12^4 + ra +*'=• "^ '■"•^
1893.— 1st Paper, No. 4.
200. Sum to n terms the series —
(1) 1 + 13 + 61 + 253+
^^^ (^)li3 + 2X4 + 3X5 + 
1897.— 1st Paper, No. 11.
201. Find the sum to n terms of the series —
(1) I3+23 + 33 + 43 + &C.
(2) whose wth term is
{ji—x){ii—\—x)
1899.— 2nd Paper, No. 6.
80 QUESTIONS.
202. Find a convenient expression for the sum of the series —
ihoc\U2^'^\U2X^\it4!X^+ , , . ad. inf.,
when the differences of u are rapidly convergent.
1S91, April.— 1st Paper, No. 10.
CHAPTER XI.
203. Sum the following series —
(1) P423+ . . . to ^ terms;
(2) 320a?30^2_290:r3 ... to infinity, where cc<\,
the series being a recurring one.
1901.— 2ncl Paper, No. 19.
204. By the method of differences, or otherwise, find the wth term,
and the sum of n terms, of the series 4 + 5 + 8 + 15 + 30+61+ . . .
1900.— 1st Paper, No. 9.
205. Find the wth term, and the sum of n terms, of the recurring
series —
6 + 31^ + 145.^2 + 6430^3 + &c.
1899.— 1st Paper, No. 10.
206. Find the wth term in the recurring series —
2 + 17ar+95a72 + 461a?3+ . . . ,
and sum the series to infinity, x being less than unity.
1895.— 2ncl Paper, No. 6.
207. Find the generating function, and the general term, of the
recurring series 1 — 5^— 23a?2—77^3— . . .
Sum the series 2.5 + 3.6 + 4.7+ . . . tow terms.
1900.— 2nd Paper, No. 20.
208. Sum the series —
(1) 1.2 + 3.4^ + 5.6072+ ... to infinity, x being <1,
and (2) 2. 5. 8 + 5. 8. 11 + 8. 11. 14+. . .tow terms.
JS95,— 1st Paper, No. 10.
QUESTIONS. 81
CHAPTER XII.
209. Define Arithmetic Progression, Geometric Progression, and
Harmonic Progression.
Find formulas for the sum of n terms of series in Arithmetic
and Geometric Progression respectively.
Sum the series —
4 5 4 5 4 5, , . ^
1S87, Oct.— 1st Paper, No. 10,
210. Find the sum of n terms of an Arithmetic Progression whose
first term is a and whose common difference is d.
Insert 10 arithmetic means between 8 and —5.
1902.— 1st Paper, No. 5.
211. Having given the sum, the first term, and the common
difference of a series in A.P., show how to find the number of terms ;
and explain how you would interpret negative and fractional results.
Prove that if any term n of an Arithmetical Progression be the
( —^ Jth of the series, the sum of the first m positive terms =m^.
1892, Oct.— 1st Paper, No. 5.
212. S is the sum of n terms of a series in Arithmetical Progression
of which the first term is unity and the common difference h. Find a
value of h such that Sf3 is divisible by 2k for all odd values of n.
1890, Oct.— 1st Paper, No. 4.
213. If three quantities are in A. P., find under what circumstances
the second will have to the first a greater ratio than the third has to the
second.
1887, April.— 1st Paper, No. 7.
82 QUESTIONS.
214. Prove that the sum of the (m{n)th. and the (m—n)ih terms
of au Arithmetical Progression is equal to twice the mth. term.
Sum the series —
lfll fill f nil + 11111+ . . .ton terms.
1894.— 1st Paper, No, 4.
215. When are series said to be in arithmetic progression, geometric
progression, and harmonic progression respectively ?
2 j_ 2
If a, h, c be positive integers, and «&, Joe, cl be in G.P.,
2 _1_ 2^
show that «6», 5a»c», cb» are also in G.P.
18S8, April.— 1st Paper No. S.
216. If «, 5, c, d are in G.P., prove that —
Sum to n terms the series —
ar + 2a72+3ar3 + 4^i+ . , .
1902. 211(1 Paper, No. 10.
217. Sum to n terms the series—^*
11 1 o
1+a/2 ' 3+2a/2 ' 7 + 5^/2
1891, April.— 2nd Paper, No. 19.
218. The sum of 2n terms of a Geometric Progression whose first
term is a and common ratio r, is equal to the sum oin terms of a
Geometric Progression whose first term is h and common ratio r^.
Prove that h is equal to the sum of the first two terms of the
first series.
1888, Oct.— 1st Paper, No. 8.
219. {a) Find the sum of n terms of a G.P. whose 1st term is a,
and whose common ratio is r.
(h) If r be < 1, deduce the sum of an infinite number of
terms of the progression.
(c) If the pih. term of an A.P. be  and the o'th term be ,
prove the sum of pq terms is  (pg' + l)
1895.— 1st Paper, No. 4.
QUESTIONS. 83
220. Find the nth. term of an Harmonieal Progression whose first
two terms are a and b.
If a, bj c, d are four unequal quantities in harmonieal
progression, and if d, «, 6, c are in proportion, prove that 3aHc=0.
1901.— 2nd Paper, No. 16.
221. Compare the values of the arithmetic, geometric, and harmonic
means between two quantities ; and show that if a be the arithmetic
mean between b and c, and b the geometric mean between a and c,
c will be the harmonic mean between a and b.
1892, Oct.— 2nd Paper, No. 3.
222. Prove that the arithmetic mean of any number of positive
quantities is greater than their geometric mean.
Show that the sum of the harmonic series —
111 . .
 ^ J \ 7 + .... to w terms
a a + 6 a\2b^
is greater than
272
2a+{n—l)b
1894.— 1st Paper, No. 10.
223. Given P and Q, the ^th and qih. terms, find the series
(1) in arithmetical progression,
(2) in geometrical progression,
(3) in harmonic progression.
1893.— 2nd Paper, No. 3.
224. Insert 5 harmonic means between 5 and 10.
If «, 5, and c be three positive quantities in harmonieal
progression, prove that a^\c'^> 2b^.
1896.— 1st Paper, No. 4.
225. Find the sum of the squares of the first n natural numbers.
Prove that the difference between the square of the
arithmetical mean of the n quantities a^x, b\x, c\cc, . . . and the
arithmetical mean of their squares is the same whatever be the
value of ir.
1900. 2nd Paper, No. 17.
^ita n 9.
84 QUESTIONS.
226. Prove that the arithmetic mean of any number of positive
quantities is greater than the geometric mean.
Find the greatest and least values of the expression —
— — ■ , a: bemg real.
1897.— 2u(l Paper, No. 7.
227. The population of a certain town at the end of any year it
found by subtracting eleven times the population at the end of th^
previous year from ten times the population at the end of the succeedinj
year ; also ten years ago the population was 11,000, and eleven years
ago it was 10,000. Show that the population increases in geometrical
progression.
1901.— 1st Taper, No. 5.
CHAPTER XIII.
228. State, and prove the truth of, the Exponential Theorem.
Show that —
e1
e\ 1. \^ M I** I"
1900.— 1st Paper, No. 7.
229. Assuming the truth of the Exponential Theorem, prove that
Find the value of log^^^ to 9 places of decimals, and show
how the value of logio^^ could be obtained.
yy 1902.— 2nd Paper, No. 20.
230. Write down the expansions of w^ and loge(l+^) in ascending
powers of sc, stating the limitations, if any, on the values of x, and show
how to calculate the values of logarithms to base 10 by means of the
expansion of log e(L + .r).
Find the value of log^S to three places of decimals.
1901.— 1st Paper, No. 9.
QUESTIONS. 85
231. Assuming the truth of the expansion of a^, deduce the
Logarithmic series :
log(Ha7)=a7— ^a:2 + ^a;3 — ^07^*+ . . . ad inf.
Are there any limitations to the value oi x?
Prove that —
logo;— ilog(arf 1) ilog (^1)
1 1/ 1 V 1/ 1 V o
1899.— 1st Paper, No. 8.
232. Show that—
log.^=^ 1  l(a.^ 1) + l(^ i) &c.
1888, April.— 2n(l Paper, No. 3.
233. When is the expansion of jloge(Ha:) suitable for practical
calculations?
Why is it unsuitable in most cases ?
Deduce from it a more convenient formula.
Show that log.4=l+ J_^ + ^^ + ^^ + . . .
1888, Oct.— 2ncl Paper, No. 6.
234. Write down the series for loge(l— o^ + ^r^^ as far as oa^, and
show that —
^ r 3 1 / 3 \3 1 / 3 \5 ^
+^253 + 3(253) +5(253) +•••}•
1892, Oct.— 2nd Paper, No. 10.
235. Write down the expansion of a^ in ascending powers of a: and
deduce therefrom the expansion of loge(l+y) in ascending powers of y.
Show that the logarithm to the base e of the product of —
l+ax 1ax
(l\aa:) 20 and (1—ax) 2a
ax^ a^x"^ a^x^ _ . „
1.2 3.4^ 5.6 ^
1893.— 2nd Paper, No.
86 QUESTIONS.
236. If ^ < I, prove that—
logf(l + ^)=^— 2^^+Q^^~4^^+ • • • (^dinf.,
and show how the value of logarithms to base 10 may be calculated.
Prove that —
5 7 17
logf(l + ^— 2^2)_^_^2_j_^3_ ^4__ , ad inf.,
^ *J TC
and find the general term of the series.
1895.— 2nd Paper, No. 8.
237. Fmd a? from the equation 2^' + 2^i = 10.
1887, Oct.— 2nd Paper, No, 6.
238. Define a logarithm, and show that logaW**=wlogam, and that
logaW=log6m X logai.
Solve the equation 2^=35, correct to five places of decimals,
given that log 28 = 14471580 and log 49 ='6901960.
1900.— 1st Paper, No. G.
CHAPTER XIV.
239. Define the characteristic and mantissa of a logarithm, and
show how the value of the former is determined for any number by
inspection.
Making use of the tables, find the value, to two places of
decimals, of —
(8845)3 X Vr S9
(233)^=v^46^ '
1901.— 2nd Paper, No. 18,
240. Prove that loga'ni'^=nx\ogam.
A wine seller has 40 gallons of Vine ; as soon as he has sold
half a gallon he mixes with the remainder half a gallon of water ; how
often can he repeat this process before the amount of wine in the
mixture is less than half of the whole ?
1902.— 2ud Paper, No. 19.
QUESTIONS. 87
241. Within what limits are the approximations obtained by the
use of a table of proportional parts in a book of logarithms accurate ?
1889, Oct.— 1st Paper, No. 2.
242. The assumed value of a quantity exceeds its real value,
12, by 01. Find the amount of the error in the common logarithm
of this assumed value, expressing your result to 7 places of decimals,
having given that the modulus of common logarithms= '43429
approximately.
1890, April.— 1st Paper, No. 4.
243. Define a logarithm, and prove that logaa'=logja; x loga^.
With the help of the tables, find the value of —
^^ (7839)'*f v/IiF 5
1235 (00037) i *
1902.— 1st Paper, No. 8.
^
244. Given log 2= 30103 and log 3= 4771213, find after how many
years the number of trees in a plantation will be less than onethird of
the original number if, in each year, there be cut down onesixteenth of
the number standing at the commencement of that year.
1898.— 1st Paper, No. 9.
245. Find the value of (34782)i x (289)i, given log 3478=
35413296, log 3479 = 35414544, log 17 = 1*2304489, and
log 35389=4548869.
1900.— 2nd Paper, No. 19.
CHAPTEE, XV.
246. Assuming the expansion of log e(i \ iv) , show how to calculate
the values of logarithms, and find the value of loge2.
For all ordinary rates of interest prove that the time in which
a sum of money will double itself at compound interest is nearly
70
— + "35 years, where r is the rate per cent, per annum.
1898.— 1st Paper, No. 8.
88 QUESTIONS
247. Find the present value of £P due at the end of n years,
compound interest being reckoned at the rate of £r per unit per annum.
What does this formula become if the interest be convertible halfyearly?
A man is entitled in 20 years' time to receive £10,000 ;
allowing interest at 4 per cent., what is the present value of this right to
the nearest penny? Given log 1 '04 ='0170333, log4'5638 = '6593266,
and log4'5639= 6593361.
1899.— 1st Paper, No. 7.
248. Find the present value of £P due n years hence, compound
interest being reckoned at £r per unit per annum.
At 4 per cent, compound interest find what sum should be paid
down now to receive a freehold estate, worth £400 per annum, in
12 years' time, given log 2 = '30103, log 13=1'1139434, and log 6246
=37956.
1897.— 2nd Paper, No. 8.
249. Find, correct to a farthing, the bankers' discount on
£345. Qs. 7d. due 57 days hence, at 4 per cent, per annum.
1897.— 1st Paper, No. 1.
250. Prove that the true discount on a given sum of money for a
given time and at a given rate of interest is half the harmonic mean
between the sum and the simple interest on the sum for the same time
and rate.
1894. 1st Paper, No. 8.
SOLUTIONS,
CHAPTER I.
1. It is first necessary to rearrange the dividend in descending
powers of y, thus : —
a722;%Hy*^22y^2;2_^2^4+y2^4_^_2;r42;2_a7V(y<+2yVy%2^ar4 + ^V
I
2y4^2_ y4^2_ ^2^4
— fz^\ y'^ci^ + 2y'^xH^+yH^
yVz^ — ccf^z'^—xH^
y'^x'^z^ — x^z^—x'^z^
Answer.— y'^V2y'^x'^—yH^Voi^\xH^
The accuracy of this answer may be easily tested, thus : —
[y4+2/a:2— /;22^^4_.^2^2] ^2_^2_2;2]
= [(y2+a?2)2_2;2(^2_^2)] [ ( ^2 _ ^2) _ 2^2]
= (y2 + a72)2(y2_^2)_^2[(y2_,_^2)2^_(^2_^2)2+2;4(y2__^2)
= (y2 + ^2)(y4_^)_^2(2y4^2a?4)+;S4(^2_^2)
=/ — a?6 + X'^y^ — a?V2 _ 2z'^y\ _ 2;2;2^4 4. ^Y _ 2;4^2
90 SOLUTIONS.
2, After substituting the given values, the expression
4 X 162 /3 X 16 , /.]
4x162 /3xl6_^ /i? . 42 . i.l^'^^^'x^'^^^
162x22 > 4 V 2 36x2
4x16x16 / ,
^ v^3 X 4 + v/8 X 16+16+113 x 8 x 16
= 128 \/l2 + 12 + 11,664
= 128x51,664
= 6401,664
= 1,024
Answer. 1,024.
3. This sohition will be very lengthy, if the expressions are taken
as they stand. If, however, we write p for (x—y) and g[ for (y—z),
so that (072;)=^ + ^, Ave have
h — G . c—a a—h
+— ^ + =0
Multiplying up and changing signs,
Eearranging, p^{b — c)—2pq{c—a){q^{a—h) =
Add and subtract {p'^\ g^) {c — a),
whence 'g^{h—c) — {p'^{2pg\q^){G—a)—'p'^{a — h)=Q
Multiply by ( — 1) and replace or, y, and z,
whence {b—c){y—zy\{c—a){—cc—zy+{a—b){x—yy=0
Q.E.D.
4. (1) Factorizing both numerator and denominator —
^3+ {a\h) x'^+{ah^^l)x + h _ x^ + a^+^ + i^ + abx\b
bx^+(ah + l)x^^{a\b)a;+l "" ha^\ahx^ + hx\x'^\ax^l
_x{x'^\ax+l)\h{x^Yaa:\l)
~ hx{x^+ax{l)\{x'^\ax^l)
_ {x\h){x'^+ax\l)
~ (J)x\l){x^+ax\l)
_ x\h
Answer. — , , .
ba:\l
SOLUTIONS. 91
^12 + 6^3 'v/3a/4 + 2'/3
^ ^ VS + l ^/3^l
^^3^3 + 2^/3 + 1
73 + 1
_ ^3^(^3 + 1)2
~ V^+1
Answer. — a/3.
5. (1) Arts. (146) and (147) Elementary Algebra.
(2) a^ab^=ah(^[i^b^)
=ab(a + b)(a—b)
Any integral number greater than unity can be expressed in the form
of (3m— 1), Sm, or (3m + 1), where m may have any integral value from
unity upwards.
It follows, therefore, that a and b must be capable of expression in
one of these forms.
There are, therefore, nine possible cases to be considered.
1st,
where
«=3m
and
b=Sn
2nd,
)?
a=Sm
jj
b=Snl
3rd,
57
a=Sm
55
b=Sn\l
4th,
5?
a=Sml
55
b=Sn
5th,
)5
a=Sm — l
55
b=Snl
6th,
55
a=Sm—l
55
5=3^ + 1
7th,
55
«=3m+l
55
b=Sn
8th,
57
a=Sm + l
55
b=:Snl
9th,
55
«=3m + l
55
5=371 + 1
^B It is evident that in cases 1, 2, 3, 4, and 7 the expression must be
divisible by 3, since the factor ab must contain either Sm or Sn. Cases
5 and 9 are provided for by the factor (a — b), which reduces to
(Sm — 3w), and is therefore divisible by 3.
I^ft Similarly, the factor {a\b) reduces to (3m + 3^) under cases 6 and 8.
92 SOLUTIONS.
6. Todhunter's Algebra, (Art. 126) .
407^+ 2x^ 18^2.^ 307 5
7
28a:H14^'126o72 + 21o:35
28o?*^64ar3+ 30ar218o;+ 4
39)78o73156a72 + 39a;39
6a;5_ 4^4_ii^3_ 3^2_ 3^_
1
2
12a:5 8ar422a73 Qx'' Qx
2
12075+ 6o;*54o73+ 9o7215o7
1407H 32073 15072+ 9^_
2
— 14074 + 28073 7o72 7^
4073 8072+ 207
2
4073 8072+ 207
2
3o?
707
Therefore the H.C.F. is 2o73— 4072 + 07— 1.
Answer. — 2o73— 4072+07— 1.
7. («) 076+1)078+074 +l(o?2
 078 +072
074—072+1)076 +1(^2^1
076—074+072
074—072+1
074—072+1
H.C.F. =074072+1.
Since the H.C.F. is 074— 072 + I, we may rewrite the two expressions
thus :
(0^— 072 + l)(o72+l) and (074— 072 + 1) (074 + 072+ 1).
It is evident, therefore, that the L.C.M. is —
(074072+l)(074 + 072 + l)(072+l).
f(o74O72+l)(o74 + 072+l)(072+l).
Answer. — \ \ „ ,
[074— 072+1.
(h)
073—07—6 072 + 207+6 _ (072)(072+207 + 3) 072 + 2o7 + 6
072 + 207 + 3 073 + 20712 072+2o7 + 3 (072) (o72 + 2o7+ 6)
= 1.
Answer. — 1.
SOLUTIONS.
93
8.
2x
2a7324a;2 + 93ar117
^4_i6a;3+ 93a;2 234^7 + 216
2ar314a72+24^
2
10
10a?2 + 69a?117
2^7432073 + 186^2_468a7 + 432
10072 + 70^120
2a74240;3+ 9307211707
>
x^ 3
 8073+ 93072 35107 + 432
 8073+ 96o72372o7+468
3)3o72+ 21o7 36
072 707+12
072— 307
 4o7+ 12
 4o7+ 12
Therefore, the H.C.F. is (o73).
Assume
then
Equating the rational parts
Answer. — (07— 3).
a/— —2 ^6= V.r v^y
— — 2 V6=o7— 2^/^+^
11
„ irrational „
Since
2^/6=2 \/o7y . . .
(07y)2=(a7 + y)2_4^y
12196
(1)
(2)
that is, X
From (1) and (3)
That is, the required root is 2— /W.
25
4"
5
2
(3)
07=4 and v=
Answer, — 2
4
94 SOLUTIONS.
9. (a) Assume ^7+ ^^13= '/^+ v'y
then 7i V 13= x{ 2^x^ + 1/
whence a: + i/=7 (1)
2V^=Vis (2)
Again, since (^— y)2=(a?+y)^— 4ir,y
=4913
=36
a^2/=6 (3)
13 1
From (1) and (3) ^—~o ^"^ ^— o *
That is, the required root is a/— + a/ = ^^ — ,
(b) Assume ^^10+ A/l08=a?+ Vy (1)
then Art. (88) v^lO ^108=^? \/y (2)
(1) X (2) v^lOO108=a72y
whence — 2=07^— y (3)
Also from (1) cubed 10+ VlOS=a^^ + Sa:'^y^\Sw^+^\^^, '
equating rational parts 10=£c^ + Sa^i/ (4)
substituting y in terms oi ic ^ 10=ic^\Sa:(x^ + 2)^
from (3) J =4ar3 + 6^,
transposing 4^ + 6a?— 10=0,
factorizing 2 (a? — 1) (2a?2 + 2a: + 5) =
whence 071=0 (5)
or 2ar22a7 + 5=0 (6)
From (5) x=l.
_2±a/4— 40 2±6a/^
„ (6) 07= = .
_1±3a/^
 3 '
SOLUTIONS. 95
Neglecting tlie imaginary roots and substituting a:=l in equation (3),
^=3.
That is, the required root is 1 + VS.
^^^ V2(A/i3+Vl)
Answers.
2
XV) 1+^3.
10. {a)
«2 52 C2
{a\sc){a—h){a—c) "^ (& + ar)(5 — c)(5— ^) (c + a?)(c— «)(c— i)
changing the sign of one factor in each denominator, so as to preserve
cyclic order, we have Art. (224) Hall & Knight's Elementary Algehra —
— a2 —12 _c2
+
{a\cc){a — h){c—d) (h ■\ x) {h — c) {a — h) {c\x){c—a){h—c)
_ — «2(& + a?)(c + ^)(5— c)— ^>2(fl5 4a?)(c + iF)(c— ^)— c2(« + ;r)(5 + a?)(«— 5)
"~ {a + x)(J) + x'){c\(xi){a—l))ih—c){c—a)
— [a^hc{hc)\h'^ac{c—a)^c'^ah{a — h)']
[a^(hc)\h^(ca)\c^(ab)']x^
{a + a?) (5 + x) (c \x){a — h) {h — c) (c—a)
Eemembering the identities given in the Article just referred to, we see
that the numerator reduces to —
_[_(«_5)(&c)(c«)]a72,
and the whole expression reduces to —
{a\x){h\x)(c\iic) '
(b) In Art. (514) it is shown that a "rational integral function"
of X vanishes when x=aj if it is divisible by (x—a). The expression
(« — &)»*+ (^»—c)^+(c—fl^)^ is a function of a, b, or c— Art. (122). It
will, therefore, be divisible by (a—b) if it vanishes when a is replaced
by b ; effecting this change, the expression becomes —
I
{b—by^{b — cY^ {c—by^= (5 — c)«f ( — 1)«(J— c)«
= (&c)^(l+[l]^).
96 SOLUTIONS.
that is, it vanishes when n is an odd integer. It is, therefore, divisible
by (a — h) when n is an odd integer.
Similarly, it may be shewn that it is divisible by (h — c) and (c—d)
when n is an odd integer.
That is, (a — h)^\ (b — c)^\ (c—a)^ is always divisible by —
(a — b)(b—c)(c—a)
when n is an odd integer.
11.— 84,796,205
60,594,287
9) 24,201,918
2,689,102
oT digits in remainder=2 + 4 + 2 + l + 9 + l48=27=3 x9
=a multiple of nine.
Numbers are expressed by means of multiples of powers of 10,
Art. (76).
Suppose a number to be represented by —
^10'* + Z>10^J + clO*^2+ . . . +ml0'^ + nl0\2)
where a^h, c, . . . &c., are integers, all less than 10, of which any one or
more may be zero.
From this subtract a number formed of the same digits, say : —
wa0^ + cl0»*»+^10»2 . . . +J102 + «10 + w.
Then the remainder may be written —
«(10»^10)4^»(10»i102) + c(10^210»0+ . . . +w(10210»)
lw(lOl) +^(110^2)
=«10(10«il) + &102(10^3l)cl0^^2(10l)+ . . .
ml02(10^2l)+w(10l)^(10»»2l).
Now every term in this expression is divisible by (10—1) Art. (55)
Elementary Algelra; that is to say, the remainder is divisible by
9 whatever values (less than 10) », J, c, &c., may have. It therefore
follows that the sum of the digits in the remainder will also be divisible
by 9, Art. (82).
SOLUTIONS. 97
In any arithmetical example negative terms are avoided by the
artifice, commonly known as "borrowing and carrying"; but it is,
of course, immaterial how the subtraction is effected so long as we give
the digits their correct significance.
12. See Art. (252).
13. (a) SeeKvi. (4L8).
A proof of the statement is also afforded by the formula for the
number of combinations of n dissimilar things taken r at a time, viz. : —
^(^l )(^2 ).. .(i ir^1) ,,,.,.,. .,
«U^= j . the value or which is necessarily
integral.
^ ^ i20 ~ 24 "^ 30 ~ 120
"" 120 ~
"" 120
_ (n2)(nl)n{n + V){n + 2)
That is, the expression has been shewn to be the product of five
successive integers divided by 5, and is, therefore, an integer. Art. (418).
14. (a) Assume that —
■
clear
{a;—a){a;—b){x—c) x—a x—h x—c ^ ^
clearing of fractions —
x'^\px\g[= A.{x—h){x—c)\^{x^a){x—c)
j +Q{x—a){x—'b).
JjQi x= a, ihevi a^ + 'p a \q=. K{a — h) (a— c)
whei
nee A= ^ ^ .
(a — h){a—c)
98 . SOLUTIONS.
Let .'r=J, then h'^\ph + q=B{h—a){h^c)
h^Yph + q
whence B=:
Similarly, C =
(ha){hc)
c'^Ypc + q
(c—a)(c—h)
Therefore ^^^±Pf^±I aH^pa + q__ h^+ph + q
xnereiore, (.^_^>J^^_^)(^_^^^ (^^_a)(ab){acy{a;h){ba){bc)
c^ + pc+q
{x—c){c—d){c—h)
(h) It is first necessary to resolve the expression into its partial
fractions, thus :
Assume that —
{1 — X){1 — GX){\ — C'^X) 1—X 1 — CX 1 — C^X
clearing o£ fractions l=zK{l — cx){l — c^x)\B{l—x)(l — c'^x)
^Q{Xx){lcx).
l=K(lc){lc^) (2)
Let x=
:1,
then
Let x=
1
then
Let 07=
1
, then
from (2)
» (3)
1=b(^1J)(1c) (3)
A= 1
(l«)(lc')
— c
B=:
(!_,)(!_,)
" (^^ ^~(lc2)(lc)
Inserting these identities in (1) —
1 1
{1x){\cx){1g''x) ilx){lc){lc'')
c
+
(lcx){lcy^ {lc''x)(lc){lc')
i.e., [{lx){lcx){lc''x)']' =
1
(l_,)(l_c^)
[{lx)^c{l + c){lcx)^ + c\lc''x)'}
SOLUTIONS. • 99
I£ we expand both members of this identity by the Multinomial
and Binomial Theorems respectively, we shall obtain two convergent
infinite series in ascending powers of x. Therefore the coefficients of like
powers of x in the two series are equal. Art. (314).
It is, therefore, only necessary to obtain the coefficient of x'"' in the
right hand member.
The required coefficient is —
lc^^((?)(l + c)+cV^^ Art. (186).
(l_C^ + 2)(l_c« + l)
(l_,)(l_e')
Q.U.JD.
15. A Continued Fraction is one which has a denominator
consisting of an integer and a fraction whose denominator again consists
of an integer and a fraction, and so on.
Thus, a Continued Fraction is an expression of the form
b
« +
d
c +
e\&c.
which, for convenience, is sometimes written a\ . . . ., where
c\ e +
a, h, G, &c., may denote any quantities whatever. Art. (331).
The fractions formed by neglecting the fractional part of one of the
denominators is called a convergent fraction, because each successive
convergent is a nearer approximation to the true value of the continued
fraction than any of the preceding convergents. Arts. (334)
and (339).
The probability of drawing a white ball at the first trial, from an
785 157
urn containing 1,000 balls, of which 785 are white, is = or>n'
Art. (449).
Expressing this fraction as a Continued Fraction — Art. (333),
157_ _1 1 1 1 1 1_1
200 ~ "^ IT "3T"l^^ TT TT 6+2*
H 2
100 SOLUTIONS.
The successive convergents formed according to the law established
in Art. (336) are
3 4 7 11 73 157
^' ' 4' 5' 9' 14' 93' 200*
The first convergent is here given as zero, since the inclusion of this
convergent emphasizes the statement contained in Art. (335), that
"the convergents are alternately less and greater than the Continued
Fraction."
16. This being a terminating Continued Fraction, the value may be
found as shown 'in Art. (332) or Art. (438).
Adopting the latter course, we find the successive convergents to be
5 29 233 2,329
' 3' l9' 151' i;5n'
the last convergent being the value of the Continued Fraction.
2,329
Answer. —
1,511
CHAPTER II.
17. (1) A/5a7+V5 + a7=\/3a7 + 4.
Squaring, (5a^) +2A/25a^'^+ (5 + ^) = 3^ + 4, i
transposing, 2V25—£c^=Sa:—G. j
Squaring, 4i(25cc'^)=da;^S6a; + S6, '\
transposing, 13^^ _ 3 g^ _ 54 — q, j
(13^ + 16) (^4) =0, i
whence ^=4 or — — :. ]
13 1
1 r»
But a?=— — does not satisfy the equation, therefore the root is ^=4.
±0
See Art. (281) Elementary Algebra.
Anstver. — ir=4.
SOLUTIONS. 101
(2) 073+^3=35 (1)
^+y=5 ....... (2)
(l)(2) ^2_^^+^2=7 (3)
(2) squared .^2 + 2^^+^2—25 (4)
In equations (3) and (4) put y=mx, Art. (206) Elementary Algebra)^
then x^—mx^\m'^x'^=1 (5)
and a72 + 2???^2_.^i2^2=:25 ....... (6)
. ,,, l + 2m + m2 25
7 + 14«z + 7w2=2525w + 25w2j
or 18m^39»t+18=0,
whence (6?;^9)(3m2)=0,
whence ^=0 ^^ o*
2 o
substituting ^;i=  in (5), ^^^(1—  +  j = 7,
0?= ±3, whence y= i2.
Also, substituting w=  in (5),
4
^2^7 X =4
a; = ± 2, whence y := ± i) .
The negative roots do not satisfy the equations.
ra;=3 or 2.
Answer. — \
(y=2 or 3.
18. x\y—z=a (1)
a;^ + ^2 4.2;2^3^2 (2)
xy^aP—^ ....... (3)
transposing (1) x\y=a{z.,
whence x''\2xy\y'^—o}^1az^z^ . . . . (i)
102
SOLUTIONS.
From (3) xy——a^,
substituting in (4) cr — 2a^ +f=a^j2az + z^
(2) (5) 2z'^=2az,
whence z=^0 ov —a .
Substitute z=0 in (1) a:+y=« . . .
„ „ (2) x^ + if=Sa^ . . .
deduct 2xy=^—2ci^ from (7), then —
from (6) and (8) 2x=a{\^ \/5)
Also
2y=a(l=F ^5)
y=(l=F^/5)
Again, substituting z^=^—a in (1), then —
» in (2),
^2__y2_2^^2 ,
deduct 2xy= — 2a^ from (10)—
V
(if— y')=4«^,
iK— y=±2a
from (9) and (11) x=±a .
(5)
(6)
(7)
(8)
(9)
(10)
(11)
Answer. —
x=l{l^V^)
y=(l=F^/5)
=
or
x=±a.
y=^a.
z=—a.
19. (1)
SOLUTIONS. 103
^ + 4 07 + 5 ^ + 6 ^ + 7
Simplifying each side separately. Art. (186) Elementary Algebra.
(^ + 3)(a7 + 5)(^ + 4)2 (^^ + 5)(;r + 7)(^ + 6)2
whence
(a7 + 4)(^ + 5) (^ + 6)(^ + 7)
1 1
(^ + 4)(^ + 5) (a7 + 6)(^ + 7)
Hence, since the numerators are equal, the denominators must be equal ;
i.e., (^ + 4)(^ + 5) = (^ + 6)(^ + 7),
or ;r24.9^_l_20=^^2 + i3^_j_42;
4^= 22,
11
^=¥
Answer. — x=— —
(2) v/a72 + 2^— 4 + 2;r2=184ar.
transposing, 2^2 ^ 4.r — 8 + Vaf^ + 2ar— 4 — 10=0,
or 2(^2_.2^_4)+ V^2 + 2a74— 10=0,
put ^/x^ + 2x—'k=y Art. (130) and the equation becomes —
2y2+y_10=0,
(2y45)(y2) = 0,
whence
Squaring
or
and
y=
Vx^{2a!'
4=
2'
x^+2a:
4=
25
4:^2 + 8a?
41 =
0,
8±n/64 + 16x41
8
8±Vl6x45
8
8±12>v/5 2±3'v/5
104 soLniONs.
Also, ify=2, A/^24. 207—4=2.
Squaring, . 072 + 2^:— 4=4,
^2^2^8=0,
(^ + 4)(ar2)=0, '
x=—4i or 2.
The roots ^= satisfy the equation —
'/^2__2^_4_2;r2=184^,
but not the given equation.
Answer. — ar=— 4 or 2.
20. (1)— a;ai/ + a''z=a^ (1)
x—ly^Vz^l^ (2)
x—cy\c^z=c'^ (3)
(2)(l) («_J)^(,,2_5^)^=_(a3_i3) (4)
(3) (2) (5_c)y(J2_c2);2=_(j3_c3) (5)
divide (4) and (5) bj {a—h) and (h—c) respectively, '
whence ^(^a + h)z= — (a^ + ah + b^) .... (6)
and ^(^h + c)z=(h' + hc\c'^) .... (7)
(6) (7) (ca)z=c^ + hca^ab,
i.e.f z=c{a\hj
substituting in (7) and] ^=z(^l^c){a + l + c) — (h'^ + hc + c^),
transposing J =a(h + c)^hc,
substituting in (1) —
{V—a^(b + c)—aic + a^(a + h + c) = a^,
i.e., ic=abc.
fa=ahc
Answer. — ! y=a'b\ac\hc.
yz=^a\'b\c.
SOLUTIONS.
105
(2)
N/2ar+l+\/7;r27= \/3ar + 4,
transposing
x^3x+4<'>/2x+l= V7^27,
squaring both sides —
3^+4 + 207+1
2 ^/ (3^+4) (2^7+1)= 7^727,
transposing
2V(3a7 + 4)(2a7 + l)=2^32,
or
~A/(3a7 + 4)(2^ + l)=^16.
Again, squaring
(3^ + 4) (2^ + 1)= (^16)S
or
6^+ll^+4= 07232^7+256
507^4307252=0
43±v/(43)^ + 20x252
10
43±Vl849 + 5040
10
43±A^6889
10
~43±83 ,
= in =4or12?.
By inspection the required value of 07 is evidently 4, as, by
substituting the other value, imaginary quantities are involved.
Answer. — 07=4.
21. (1) hxay=l^.
transposing io7 — oy — 5^ = 0.
(a— 5)o7 + %— ^2=0.
Applying the Rule of Cross Multiplication. Art. (16).
—a —52 ^5 _^
+ J a" +(«5) +5
cc y 1
whence
or
and
a3 + 53 a^bb\ab)~ b^^a(ab)'
: «3+53
07 =
= (« + J)
«2_^J^52
Answer. —
ccz=(a + b),
y=b.
106 SOLUTIONS.
(2) x\x+ A/2^23a: + 5=15+ 2,
transposing and clearing of fractions —
2^2
3a: + 2V2a723^ + 5=30,
add 5 to both si
ides
3^

2a;^
• + 5 + 2V2a723a; + 5 = 35,
put
V2^2— 3^7 + 5 =y,
then
(y + 7)(y5)=0,
whence
y=— 7 or 5.
Ify=7,
V2^23a7 + 5=7,
squaring
or
2^23^^ + 5=49,
2^23^44=0,
i.e..
(2^ll)(a; + 4)=0,
whence
11
07= — , or 4.
If ^ = 5, a/2^2_3^_^5 = 5^
squaring 2x^—Hx\5=25,
or 2;r23^20=0,
i.e., (2^ + 5)(^4) = 0,
5
whence x= — , or 4.
The root ^=—4 satisfies the equation x'^—x—\^2x^—Sx\5
X
=15+ , but does not satisfy the given equation.
Answer. — x=. — ,— , or 4.
ir + 4 _x—2 _ 4
^^ ^^^ ^+3 ^^~l5'
i.e.,
1 1 _ 4
iT + o ^ — 1 15
SOLUTIONS. 107
clearing of fractions, 15{^— l + a7 + 3} = 4{^^ + 2a7— 3} ,
or 407^22^42=0,
or 2^'^lla?21 =
(2^ + 3)(a77)=0,
whence ^ = — , or 7.
Answer. — ;r= —  or 7.
2
(2) (^ + l)(^ + 5)(^ + 9)(a7 + 13) = 105.
Let ^ + 7=y, and the equation becomes
(y6)(y2)(y + 2)(y + 6) = 105
or (/36)(/4) = 105
or /40/ + 144=105
/40/+ 39=0
(/l)(/39)=0,
whence ^=±1 or ± a/39,
and x= —6 or —8, or — 7± ^^39.
Answer. — a? =—6, —8, or —(7 ±^39).
(3) a;H^y+/= 133 (1)
x^— xy +/= 19 (2)
(l)(2) ^2^^^_^^2^ 7 . . . (3)
(3) (2) 2;ry=12
or xy=— 6 . (4)
(2) (4) (^y)^= 25
(3) + (4) (^+y)^= 1
whence a? y=± 5
and ^y=± 1
by addition 2^= ± 6 or ±4, whence ^= d=3 or ±2,
by subtraction 2y==F 4 or =f6, „ y==F2or=F3.
(^=±3 or ±2,
Answer. — {
(y==F2 or =f3.
108 SOLUTIONS.
2^ « 1 + 1 = 1 w
2 ~ 3 ~ 2
(2)
Multiplying up andl 207 + 3^15=0 . . . . . . . . (3)
transposing, J 3^2y15 = (4)
By Cross Multiplication — Art. (16) —
^ _ y _ 1
whence ^r = — ,
Id
15
13*
__75 _15
and
(2)
'\/x{a + h—x)+ \/a{h — a + x)\ Vb{ci^x — h)=0.
Transposing, ^ a{x + h'a)\ ^b(x—b—a)=:—^x{a + h—x).
Squaring,
a(x + b — a)+h(x—b—a) + 2'/ailx^—(b—ay]=x(a + h—x),
or {a{'b)x(J?ay + 2yablx^^(bay]=x(a\bx).
Cancelling (a + b)x and transposing,
^^2_ (j_«)2= ^2Vab[x'~(baf]
Squaring, [x^— (Z> — «)^]^=4«5[ir— (5— ft)"].
Dividing by 07^— (5 — ft) ^, x^—(b — ay=0
or x^—(b — ay=4iabf
whence ir=±(i— «)
or x=^(b + a)
Answer. — ^=±(&=tft).
[Note. — None of these roots satisfy the equation, although it may
be pointed out that they would do so if the signs connecting the three
expressions were varied.]
SOLUTIONS. 109
10^+ 17 12^ + 2 _ 5^74
^^' ^ ^ 18 "13^16 9 '
10^+17 5^—4 12^ + 2
transposing, ^^ — = — ^ ,
. ,., . 10^ + 1710^ + 8 12^+2
simphf ymg, = j^^ ,
25 12.T + 2
18" 1:^^16'
i.e., 25(13.1716) =18(12^ + 2),
109^=436,
Answer. — ^=4.
(2) ^2+^^_.^2=io9 ..... (1)
^^=2 (2)
From (2) y=a72.
Substituting in (1), ^2^^(^2) + (^2)2=109,
3^2__6^ + 4=109,
or 072—2^735=0,
(^7)(^ + 5)=0,
whence a?=7 or —5,
and from (2) y=5 or —7.
ra7=7 or —5.
Answer. — s
ly=5 or —7.
25. (1) Vw''+aa;+ Vx'^lx^ ^~ (1)
we have identically^ Art. (132),
{x~\ax) — {x'^—lco)=x{a^h) (2)
Divide (2) by (1) then—
\/x'^\ax— ^ x'^—lx=.2x (3)
add (1) to (3) 2 V'^H^= !^^±^ ,
2
110 SOLUTIONS.
clearing of fractions and squaring —
or S(a — h)a:=(a\hy,
I. (a + hy
^^^"^^ "=8(^4) W
Deduct (3) from (1) 2 V^^:::j^= (^ + ^)^^ ^
whence \^{x'^—lx) = {a^iy—'^{a^l)x^\Qx^,
or 8(« — 5)^=(a + 5)2,
and, as before, , x^ —, ^r .
8(<2— o)
Answer. — ^= ^^ — .
8(« — 6)
(2) ^2_3^=1 (1)
^y^=l . (2)
from (2) 07=/ + !.
Substituting in (1) y * + 2y2 + 1  3y = 1 ,
y4 + 2y2^3y=0.
Therefore y(y3 + 2y3)=0,
y3 + 2y3=0 (3)
Also y=0 (4)
Equation (3) is obviously satisfied by the root y=l, and is therefore
divisible by (y— 1). Art. (514).
and 5^= ^ • • • v^;
_i_t.A/z:ii
But y= ^ does not satisfy the equations.
Substituting the roots ^=0 and 1 in (2)
we have ^=1 and 2 respectively.
[x—\ or 2.
Answer, — \
(y=0 or 1.
SOLUTIONS. Ill
(
26. (1)
1 1
or
or
a; 1 + x 2 + cc ^\x 4 + ^ 5+.^ Q\x l + x
Transposing and simplifying each side separately —
ri jvr— ^vr— ^^(~ ^a
\x 1 + xJ \ij + x 7 + xJ \4 + ^ 5\xJ \2 + x 3 + xJ'
1 1 _ 1 1
x(l + x) (6 + ^)(7 + ^)~(4 + a?)(5 + ^) (2 + ^)(3 + ^)'
(x'' + ldx + 4<2){x' + x) _ (^ + 5^7 + 6) (^H 9^ + 20)
x{l + x)(Q\x)(7 + x) ~ (4i + x){6 + x)(2 + x){S + x) '
12;rH42 — (4ar + 14)
or '— zzz
x(l + x){6 + x)(7\x) (2 + ^)(3H^)(4 + a^)(5 + ^)
—7
Dividing both sides by (4^ + 14), whence x= —— (a)
and multiplying up —
3(2 + ^) (3 + ^) (4 + ^) (5 + ;r) = 07(1 + 07) (6 + 07) (7 + ar),
or 3o74 + 42o73 + 213o72 + 462o7 + 360=(o74 + 14o;3^55^2__42^^,
or 407^ + 56073 + 268072 + 50407 + 360=0,
or o74 + 14o73 + 67o72 + 126o7 + 90=0.
Divide by 07^, then^ —
^2 + 14^+67+^ + ^=0,
07 072
90 , / 9^
or
'+"+K^+3+*^^=°'
9
subtract — from both sides and rearrange —
07
whence 07 H 17=±'^— 1
07 07
112 SOLUTIONS.
Clearing of fractions —
^•^ + 707 + 3 (3 ± V^) = 0,
7±^4912(3±\/l)
whence
2
7j.^^l3dbl2^/l
7 7±v^l3±12yl
Ansiver. — ^= — , or
2' 2
(2) v^(l + ^)'^v^(la7)2=^(l^2)
The solution of this equation may be obtained by the use of the
identity that if
then (u—vy=c^=u^—v^—Suv(u—v)
Denote v^(lH^)^ by w,
v^(l^)2 by V,
and v^CT^) by c.
Then {^^(l + a^y^^{la;yy=lx^
transposing o:^ + 4^ — 1 = 3\/ (1 + ^) ^(1 — a:) ^
= 3(l + a7)(l^),
=3(10^2),
or transposing and dividing by 4 —
a;2 4irl=0,
whence ^= 7^ .
l±^/5
i(;er, — ^=
SOLUTIONS. 113
27. (1) y^2 + 2^l+A/^H^ + l=v/2+^^3 . . . . (1)
Squaring (x'' + 2a;l) \ (x' + a; + l)+2\/{x^ + 2xl){x^ + x+l)
= 5 + 2v/6.
Equating the rational parts on each side, and also the irrational. Art.
(275) Elementary Algebra.
(^2+2071) + (a7H^ + l) = 5 (2)
and 2^{x'\2x1){x'^x + 1) = 2Vq (3)
hence (^H2^l)2v/(^^h2^l)(^'^ + ^ + l) + (^'+^41)
= 52v/6,
or 'yx''{2x—l—^x'^^x{l=V2—V^. ... (4)
adding (1) to (4) and dividing by 2,
^x''\2xl= \/2,
Squaring and transposing ^^ + 2a7 — 3 = 0,
whence • (^— l)(a7 + 3)=0,
and 0^=1 or —3.
But 07=— 3 does not satisfy the equation.
Again, subtracting (4) from (1) and dividing by 2,
I' \/;r2 + ^ + l=v/3,
Squaring and transposing ^^ + ^'—2=0,
whence {x\2){x—\)=0^
and x=l or —2.
But x=.—2 does not satisfy the equation.
Other solutions might be given, where the roots obtained include
imaginary quantities, but one of the roots will always be 07=1, and
this is the only root which satisfies the equation.
Answer. — x=l.
(2) _y_=__£_ = _JL_
I
y Art. (12),
y + ^  ^ (1)
(«23)(/3) x^f
I
114 SOLUTIONS.
1 __ 7 _ y _ —X
x—y x^—y^ x^—^ y^
"—y ^3_,« „2 O .2 O V /
Taking =
x—y x^—y^
we have x^\xy+y^=i7 (3)
^^^^"' J^y = ^^
whence x^— xy+y^=:S (4)
put x=zmy, Art. (136), Ex. 4,
then, from (3) and (4), ^^^f+^'+l'
or Sm^+Sm+S=7m^—7m + 7,
transposing 4<m^—10m + 4i=0,
or (4w— 2)(m— 2)=0,
whence m=^ or 2 (5)
Substitute x=2y in original equation, whence —
y __ —2y
or ^3_3^=6^_8^3,
i.e., 9y^9y=0,
whence y=Oor ±landar=Oor ±2 (6)
1/ ^ X
Substitute x=^y in fs = ~T~^^
whence ^^ = ^ , '^ ,
^2_i2 2(y^3)'
8y3_24y = 12yy3
%3_36y=0
or y=0 or ±2 and a:=0 or ±1.
The root a?=y=0 does not satisfy the equation. Art. (271).
r ;r=±2 or ±1
Answer. — \
y y=±l or ±2
SOLUTIONS. 115
28. (a) (74^/3)^2+ (2^^3)^=2.
After adding I to both sides, the equation is transformed into an equation
of two complete squares, viz. : —
{(2^3).+ ^}^ = ?,
— 1 3
whence (2— ^3)a7+  = ±,
and ^=: —7ZZ or 7= .
2^/3 2^/3
Eationalising these roots ^=(2+ ^^3) or —2(2+ ^3).
Answer.— x={2^ ^W) or —2(2+^3).
{h) See Art. (133), Example 2.
Answer. — x=2, ,  or .
29. (i) x^y^z^W (1)
^'+y2_^;22=:45 (2)
y^=20 (3)
Multiplying equation (3) by 2, and adding the result to equation (2),
we have —
put u for (y + 2^), then this equation becomes —
x'^^u^z=.%^ (4)
also from (1) (:r + #)2=121,
or 072 + 22^^7 h?^2_i21 (5)
(5) (4) 2ux=m (6)
(4) (6) (xuy=4:9,
or x — u=zk7 (7)
(l) + (7) • 2^=18 or 4,
whence x=9 or 2 (8)
and u=2 or 9=(y\z) .... (9)
I 2
116
SOLUTIONS.
Substituting the value of x in equation (2), and combining with
equation (3) multiplied by 2, we have —
(^yzy=1Q or 1.
The negative value (—76) must be discarded, since it will involve an
imaginary value for y—z^ and, consequently, the corresponding values
for X and u^ viz., x=Q and u=2 must also be discarded; we
therefore obtain —
y^=±l (10)
(9) + (10) y = 5 or 4,
(9) (10) z=^ or 5.
'x=2.
Answer 
=4 or 5.
j2;=5 or 4.
(ii)
Divide by 19
lQy2Zx=1
(1)
since y and x are to be integers, we must have
407+7
and, therefore,
or
whence
^ + 1 +
19
20a: + 35
19
;r + 16
19
^+16
19
= an integer,
= an integer,
= an integer,
= an integer =^, suppose,
^=19p — 16
Substitute this value in equation (1) —
J/ — 19p + 164p + 3=0,
or y=23p19
Here p may have any integral value except zero.
(2)
(3)
(a:=S, 22, 41,
ly=4, 27, 50,
&c.
&c.
SOLUTIONS. 117
That is, i? = l, 2, 3, &c.
a?=3, 22, 41, &c.
y=4, 27, 50, &c.
Answer.
30. (a) See Art. (347).
(5) 3;r4y=ll.
y . 2
Divide by 3 ^y=3+^,
o o
but 0? and y are to be integers, therefore
•^^—  =an integer =^, suppose,
o
whence y=3^— 2.
Substitute this value in the original equation, then,
307 — 12^ + 8=11,
or x=^p\\.
Here p may have any value, except zero.
That is, if p=l, 2, &c.
ir=5, 9, &c.
y=l, 4, &c.
p=5, 9,
;r=5, 9, &c.
Answer. — \ , . .
&c.
118 SOLUTIONS.
CHAPTER III.
31. (a) See Arts. (112) and (538).
a: x^a x\o
Multiplying up —
{x^a){x\h)\x{2x\a\h)=0,
or 3a:^ + 2ir(a + J) + «&=0,
2(« + i)±v/4(« + ^)'12«6
whence x=^ ~ .
6
In order that the roots may be real, the expression under the radical
sign must be positive ; Art. (113), that is
4i(a + hy—12ah must be positive,
or 4a^— 4«5 + 4&^ „ „ ,,
n J)
Q.E.D.
32. («) See Arts. (Ill) and (113).
then — •
aXyl) + 6^(y + l)+25(yl)=0
whence 6(y + l)± ^/36( y tl)100(yll
2(yl)
_ 6(y + l)=b v"16(l 4y )(y4)
2(yl)
And since x is real, (1— 4y)(y — 4) must be positive, hence the
factors must be both positive or both negative. They cannot be both
positive, but will be both negative when y lies between 4 and  .
Answer. — The value of the expression must lie between 4 and  .
SOLUTIONS. 119
33. The equation may be written in the form —
2(}l^■ph'):^^/^[ {h\ph'Y{a+pa'){h^rph')']
whence iP= tt? , 7^^ •
a; will he real and the roots will be equal if
(U +ph'y {a\pa') {h +ph') =0,
or A2 ^2phh' +pW^  [ah +(ab' + a'h)p ^p'^a^h''] = 0,
or {h'^a'b')p^{(2Wah'a'b)p + h^ah=0.
Art. (113).
This is an equation of the 2nd degree, therefore p has two values.
Art. (538).
Q.U.B.
34. In this question the roots must not be considered singly.
Art. (117).
By problem a + ^=— 2, j
2 [Art. (114).
and ^^~3' J
Whence ^^ = 3,
a/3
— — d,
ap{a?ap + P^)
whence ^^!^' = _3[(a + ^)2_3«^]^
= 3(42),
= 6.
q2 02 TO
or — — = 6 = — = Sum of roots of required equation ;
pa. o
— ^jf j=ay8=  = Product of roots of required
equation.
120 SOLUTIONS.
Therefore the equation whose roots are — r and—— is
pa
Sa?" 18a; +2=0.
Q.MD.
35. See Art. (118)— Example.
36. (1) /SfccArt. (121).
(2) (a;+3)(la^)(fif + l)(a7 + 3) = 2.
or aa;^ + 4iaa; + Sa + 2=0,
whence a;= ^ ,
2a
_ —4ia±2^a'—2a
" 2a '
If a^—2a is negative, then cc has no real roots. Ai't. (113).
But when a is positive a^— 2 a is negative,
so long as 2a > d\
» „ „ 2>a.
If a is negative a'^—2a is always positive.
Therefore, the equation has no real roots if a be positive and less than 2.
CHAPTER IV.
37. See Pendlel)ury''s Arithmetic, Arts. (170) and (175), or
Brook Smith'' s Arithmetic, Arts. (148) and (150).
In the case of multiplication : —
Let there be two numbers, represented algebraically, as in Solution
No. 11, and suppose the lowest order of decimals in each number to be
multiples of 10~"* and 10~^ respectively, so that there are m and n places
of decimals in each number respectively. Then the product of these two
SOLUTIONS. 121
numbers will contain a multiple of IQ^tn+u) ^ and this will be the lowest
order of the powers of 10. Hence the product will contain {m\n)
decimal places.
In the case of division : —
Let the highest order of units in the dividend and divisor be the
mth and nth respectively, so that the numbers may be represented
by alO'^ + 510^1 + clO'^^ + &c. and «'10^ + i'lO^ "^ + clO^2 + &c.,
respectively.
The quotient will therefore be —
alQm + hW^ + cl0"^2 + &c.
a'lO^ + J'lO^i + c'10^2+ &c.
_ «10''*±»*4 JlO"*±^i + clO"*±^2+&c.
where the upper or lower sign is taken according as n is negative or
positive. That is to saj^, the highest order of units in the quotient will
be m:izii or mzhn — 1, according as a is greater or less than a'.
38. /S^ee Article (59).
In this proof we have algebraical symbols employed as though they
were ordinary numerals ; the reason for such a course is not obvious, and
the confusion of two distinct methods of notation is to be deprecated. We
therefore suggest the following adaptation of the first part of the proof.
To find the value of a recurring decimal : —
Let P denote the figures which do not recur, and suppose them
p in number ; let Q denote the recurring period consisting of q figures ;
let D denote the value of the recurring decimal ; then —
10^ 10p+^ 10^^+2^
and D X 10i'+9=Pxl02 + Q+^ + ]^ + • ' ' ^"^ '""f'
by subtraction, I>(10p+^—10p) = (P x lO^ + Q) P,
(Pxl0'?4Q)P
D=
102^(1031)
122 SOLUTIONS.
39. (a) See Pendlehury's Arifhmetic, 14th edition, Art. (199), or
Brook Smith's Arithmetic, Arts. (152) and (153) .
In a note to Art. (200) of Pendlehury' s Arithmetic, reference is
made to an alternative method which has little to recommend it. The
multiplier, being reversed, becomes a mere collection of meaningless
digits, while the position of the decimal point is entirely lost
sight of.
The principal difficulty experienced in the method of Art. (199) is to
determine the digit in the multiplicand with which to commence.
If the multiplier consists solely of a digit in the unit's place, the
number of decimal places in the product and multiplicand will be the
same. If, therefore, we require six decimal places in the product, we
shaU. commence with the sixth decimal place in the multiplicand.
If the multiplier commences with a digit in the ten's or hundred's
place, we shall commence with the seventh or eighth decimal place in the
multiplicand ; proceeding in the opposite direction if the first significant
digit in the multiplier is in the first, second, third . . . decimal place.
(5)— £8. 17s. 4i^.= £886875.
The product is to be correct to the third place of decimals ; we must
therefore work to four places and include the " carrying figure " which
would have been obtained if we had worked to five places ; thus : —
886875
328794
2,6606250
1773750
709500
62081
•7981
•0354
2,9159916
£2,915992= £2,915. 19s. 10^.
Answer.— £2,915. 19s. 10^^.
SOLUTIONS. 123
40. Determining the position of the decimal point in the quotient
by the ordinary rule, Art. (175) Pendlehury's Arithmetic, we see
that the quotient will consist of seven digits. The divisor must
therefore be extended to eight places. Art. (201).
•43232323 ) 73640000 (1703355
43 232323
30 407677
30 262626
145051
129697
15354
12970
2384
2162
222
216
6
Answer.— 1703355.
41. Without the use of brackets it is impossible to say whether the
quotient resulting from the division of the product of the first five
• • 32
expressions by '006 is to be multiplied or divided by . See note
to Art. (138) and Art. (140) Fendlehury.
Taking the former case, we have —
96 14023 ^ , 6 1 ^r.V 1 32
X X V« X 3 X 1741 X ^r X
33 9 006 4207
? x^?^ X30x3xl740x32
y yuu
30 9 6
— X — x29x9x999x— X 4207
y J.U yyy
124
SOLUTIONS.
simplifying numerator and denominator separately —
85 X 4207 X 58 X 32
~3x29x9x6x 4207
85x2x32
"" 3x9x6
2720
~ 81
3022
~ 9
= 33580246913.
Answer.— 33580246913.
42. See JPendlehuri/, Art. (203a) and Appendix IV.
In practice, if only a few amounts are involved, it is sufficient to
consider Sd. as '012, and 9d. as "038 of a pound. If the number
of amounts being dealt with is considerable, Sd. may be taken as
•012 and "013 alternately, and the decimal representing 9d. may be
treated in a similar way.
£487 : 17s. : 10^6?. =£487894.
£33693 =£33 : ISs. : lOi^.
43. There will be 11,000 sums of money, consisting of —
Whose approximate Whose true value
value
will end with
will end Avith
1,000 ending
in Id.
4
416666
5 5'
2d.
8
803333
? 55
Sd.
13
125
> )>
4^.
17
1666666
» ii
5d.
21
2083333
» 55
6d.
25
25
5 55
Id,
29
2916666
> 55
Sd.
33
3333333
5 55
9d.
38
375
5 55
lOd.
42
4166666
5 55
lid.
Total
46
Dotal
4583333
276 G
27470000
T
rue Value . .
]rror in Excess
2747
I
13
Answer.—
£1.
6s. in excess.
SOLUTIONS.
125
44. Obviously the sum of money is intended to be such as can be
paid exactly in any one of the various denominations alone; reducing
each of the amounts to the lowest denomination, viz : — pence, it is
required to find the L.C.M. of —
1, 12, 24, 30, 60, 84, 126, 240, 252.
That is 240x21 pence, or £21.
Answer. — £21.
45. The length of the edge is v^l,953,125=125.
This result may be obtained by extracting the cube root according
to Horner's Method. See Brook Smith's Arithmetic, Art. 178 to 182.
The diagonal of cube is \/3 times the length of one of its sides.
See Euclid, Booh 1, Prop. 47.
Therefore the length of the diagonal is \/3 x 125,
= 17320508x125,
= 21650635 approx.
Hence, the difference between the length of its edge and its diagonal
is 915064 inches.
Answer. — 91 5064 inches.
46. 12 drams = 046875 of a lb. avoirdupois.
9 oz. = 5625 „ „
23 lbs. =23 „ „
Total =23609375 of a lb. avoirdupois.
We are told that 7000 grains troy are contained in a lb. avoirdupois,
while we know that a lb. troy contains 24 x 20 x 12 grains, that is
5760 grains.
Therefore, 1 lb. avoirdupois = ^f^^ pounds troy,
and 23609375 lbs. „ = ^^^q pounds troy,
=28691948 „ „ approx.
Answer. — 28*691948 lbs. troy approx.
126 SOLUTIONS.
987*
47. The lield contains ^.^ ^^. acres, that is 5*625 acres or
17o*4d7
27225 square yards, therefore the length of one side is \/27225=165
yards.
The length of railing is therefore 4x165=660 yards, and the cost
is 660x£1583=£1045.
Answer. — £104. 10s.
CHAPTER V.
48. Since x varies directly as y^^ ^nd inversely as \/z^ therefore
X varies directly as "3^ ■ Art. (34) .
whence
r
By probl
em
whence
I
And it is
required
to find y
from the relation
3_1 f
^43'
or,
y^=36,
.%
y=6.
Answer. — y=6.
49. Let r and t denote the radius and thickness of a sovereign
respectively. And let r' and t' denote the radius and thickness of a
penny respectively.
7)1
Then by problem 2wr=2wr', whence r'= — r ;
n
also pt=.qt\ „ 1^—t.
Again, the contents or bulk of a sovereign=cr2^,
and „ „ „ ,, penny =c(r\ t.
SOLUTIONS. 127
Therefore, since £1 is worth 240 pennies, it follows that {crH) of gold
is worth 240c(—r) ^ of copper, that is each unit of gold is worth
\n J (£
7)1/^ 7)
240 — . units of copper.
Answer. — The value of gold and copper varies as Tt^.^ : 240^^^^
3 1
60. The first vessel consists of  wine and  water.
4 4
„ second „ " 6 " 6 "
Let cc denote the number of gallons to be drawn from the first
vessel, leaving (7—^) gallons to be drawn from the second.
3 1 7
Then by problem I^"'"a(^"~^)~9'
or 9a; + 142a7=42,
7^7=28,
iP=4.
4 gallons from 1st vessel.
Answer.
.'nd ,
51. Since A and B each walks at a uniform rate, the times occupied
in walking equal distances will always be in the ratio oo : y.
After they have met A has a hours in which to travel the distance
which B took (y — V) hours to cover, and B has h hours in which to
travel the distance which A took {x—a) hours to cover.
a X — a X
But
yh
~ h
~ y
(^«) :
= {y
&),
a
yV
yi
~ h
h X
~ y
128 SOLUTIONS.
whence,
compounding
the ratios

a
x^
or
4
X
i.e.,
X : y '.
: ^/»
^h.
Q.E.D.
52. (a) See Article (35).
(h) Let n denote the number of working days.
„ H „ „ „ hours per day.
„ W „ work done per hour.
„ Q „ „ in n days.
Then by problem Qx n when H is constant,
that is Qx nK „ ,, „
Also by problem Qx W when H varies and n is constant ;
but Wx (12 H) ;
therefore Qx nKW when H and n both vary.
That is Q=^;mHW,
=:m?iH(12H).
Thus, when the working days are 9 hours each, that is when H=9,
the total amount of work done in 20 days will be
w20x 9(129),
and when the number of working hours per day is x, the total amount
of work done in 27 days will be
m27xx{\2x).
By problem w27x^(12^)=wl80(129),
that is a^2_j2a7 + 20=0,
(ar2)(^10) = 0,
ir=10or2.
Answer. — 10 or 2.
SOLUTIONS. 129
53. Let X denote the number of parts of copper,
and (1 — ^) V >•> 5 J tin,
so that the weight of the body may be represented by
0? X 896+ (1— 07) X 729.
Then by problem —
{896a;+729(l^)}xl09=896(la7)+729^,
or {l67^+729}xl09=896l67ar,
or {167 x 109 + l67}a7=896729 x 109,
34903;r= 10139,
t 07= 29049.
That is, about 29°/o of the bulk of the body is copper, and about
71 % is tin.
Ansiver.— 29 % and 71 %.
54. Let R denote the increase in the receipts.
„ V „ „ velocity over 20 miles per hour.
„ W „ „ increase in cost of working.
By Problem we have R=wV and W=nYK
Also, we are told that when V=0 or 20, R=W, that is mV=nNK
Therefore, giving V the value 20, we have 20^=400?^,
w=20w.
The Profits are represented by R— W, which by substitution
becomes (20wV— wY^), and it is required to find for what value of V
this expression will be greatest.
Let 2QnYnW^=y
then 7iV(20V)=y
Since n is constant the required value of y will depend only on V,
that is to say, we have to find the value of V, which will result in the
largest product of the two factors, V and 20— V. The sum of these
two factors is known, therefore, by Art. (252) their product is greatest
when they are equal, that is when V=20— V,
or V=10.
130 SOLUTIONS.
Therefore the profits will be greatest when the speed is 30 miles
per hour.
Answer. — 30 miles per hour.
55. If gold and silver were of the same weight, then the bar of
2,500
silver would be worth £
30 •
But the weight of silver is only T^th of the weight of gold;
therefore the actual value of the bar of silver is —
2,500x10
30x19 '
this is £438596,
or £43. 175. 2id.
Answer.— £43. 17s. 2\d.
56. See Art. (35).
The contents of the three coins is —
{^oST+ i^6T+ lro(S)V' '"^^' ^^'^'''
(16 + 8 + 3)w2 27 m ,.. ^
= ^ — = cubic mcnes.
1,280 1,280
Let r denote the radius of the new coin in inches, then —
1 27
or
r2=
'256'
r=
S^3
■ 16 '
=
:^xl73205,
= •325.
Answer. — 325 inches.
SOLUTIONS. 131
CHAPTER VI.
57. The first clock gains 1 second an hour. The second clock loses
lo seconds an hour. In order that they may both indicate the same
time, the time gained by the first clock added to the time lost by the
second clock must equal 12 hours or 43,200 seconds.
Let X denote the number of hours required, then —
3
a; + ^ =43,200,
2t
or 07=17,280 hours, or 720 days.
Answer, — 720 days hence.
They will both indicate true time when they have gained and lost
twelve hours respectively. The first clock will gain 12 hours in
43,200 hours, or in 1,800 days, and the second clock will lose 12 hours
2
in  X 43,200 hours, or in 1,200 days,
o
If by the question it is desired to find when the two clocks will both
indicate the correct time together^ it is evident that this will be in
3, GOO days, since at the end of that time the first clock will have gained
24 hours, and the second clock will have lost 36 hours.
Answer. — 3,600 days.
58. Let ^ denote A's capital in pounds, then a? +15 will denote
B's capital in pounds. C's capital of £50 having been invested for
Hhs of the total period, is equivalent to 7 X 50 invested for the full
4 ■
period. Total capital may therefore be considered as —
, , ..x , 75 407 + 105
0? + (07+ 15) +  = — ,
]5ach person's share of the profits will therefore be in proportion to their
capital, and A's profit is therefore —
^ 2o?
X 159= — X 159.
/ 4o7+105 \ 407+105
132
SOLUTIONS.
2x
By problem x^ X 159=88,
4^+105
or 4a?2+105a?+318^=352a7+9240,
or 4^2^7107—9240=0,
71 ±^152881
whence x:
71±391
8
=40 or 57.
Answer. — £40.
59. Let X denote the original length of the voyage in days. Then
the number of daily allowances of water was 175^. The extended
length of the voyage was (or + 21) days, and had there been no deaths
this would have necessitated 21 x 175 = 3,675 additional daily allowances.
The deaths extended over a period of (a?21— 30) = (ir— 9) days, and
the total amount of water which the deceased would have consumed
if alive during that time would have been sufficient for a daily allowance
to the following number of persons : —
3 + 6 + 9+ ■■■+3(.9)=3 ("'y') .
g
Therefore, by problem, 3,675=  (^—9) (a?— 8),
or ^2_i7^_2,378=0,
17=^^9801
whence x=
2
17d=99
2
=58 or 41.
Answer. — 79 days.
60. Before any transfer has been effected, each gallon of the
mixture in the first and second casks contains
 and — parts of wine, respectively,
a a
.SOLUTIOKS. 133
'i After the first transfer, the first cask contains —
V p—Jf.z=B , gallons of wme,
, a a aa
and the second cask contains —
p H , — gallons or wine.
aa
If we denote the second term by yi, the expressions become yS— yi
and y8' + yi respectively.
After the second transfer, the first cask contains —
^_^^ ^' + ^ aaV, + aX /3yO a(^^ + yO
/^— yi + — J— —p T
a a aa
(a'y8a;8') + (a'y3a;8')(«'/3a/8')'^
=^ W ^
aa I aa J
After the third transfer, the first cask contains —
p— y2 1 J— —p ,
a a aa
a' + a
=/8
(a'^a;8') {2 ^"l + ia'PaP')
\ aa J
( aa ) aa
aa ^. aa \ aa / )
Similarly, we may show that —
em
134 SOLUTIONS.
The expression — is less than unity, when a' and a are both
aa
greater than 2 ; therefore, when this is the case, and n is infinitely great —
l3y^_l3——
a fa
Similarly, y8' + y^=— ^^ (^8+^') .
Answer. —
^ a' + a ^ \ aa' y j
a + a ( \ aa J J
61. Let a: denote the number of eggs bought at 11 a shilling;
then Sec denotes the number of eggs bought at 15 a shilling ; that is,
X 3^ 16^ 4;2?
he spent —  +  = — — shillings, and sold the eggs for — shillings.
11 15 55 io
/ 4 16\ 12^7
Therefore his gain or loss is (^ — ^) ^~ to kk shillings, and his
.^ . 12;r 16ic 3
gam per unit IS 13X5^5^ = 52'
= •0576923 . . .
= £5. 15s. 4i^d. per cent.
Answer. — He gains £5. 15s. 4i^d. per cent.
62. Let the original price of meat be ^ pence per lb. ; therefore the
240
number of pounds obtained for £1 will be
£C
i
3
Then, after the reduction, the price will be  ^ pence per lb., and at
TO
Sec
this price the number of pounds obtained for £1 will be 240 i ^
soLrTiONs. 135
3 240
By problem 240 ^  ^= — + 8 ,
, . 320 240+8^
that IS, = ,
X X
807=80,
a;=10.
Answer. — \^d. per lb.
63. Let the gross receipt for one day be £y ; then, by problem —
i^^^=*20 (1)
^^(y20)=£20 (2)
From (1) we have ir5^= 2,000,
J.X. 2 2'000
thereiore x= (3 )
y
From (2) we have ^y+50y20a?= 3,000 (4)
Substituting — for x in (4) we have
u
2,000+50y^ =3,000.
Multiplying by y we have
50y2_l,000y=40,000,
y2_20y— 800=0,
(y40)(y + 20)=0.
Therefore y=40.
Answer. — £40.
64. Let the wholesale price of an article sold at Is. 8 J. be x pence.
Then by problem —
^(l + i)+2=20,
io ^ 72
5 5
136 SOLUTIONS.
72
That is, — pence is the wholesale price of an article to be sold at
o
the nominal retail price of 20 x — pence.
y
Let the nominal retail selling price of an article purchased for
81 X 12
Jqq pence be y pence,
then by proportion
81x12
100
:y=
72
5
200
• 9 '
y=
81x12
100 ^
200
9
X
5
72'
therefore
= 15.
That is, the nominal retail selling price should be I*. Sd. per article.
Answer. — Is. 3d.
65. Let £x denote the amount invested in 3i per cent. Stock at 104.
Adding i per cent, for brokerage, the effective prices will be 99 and 105.
Byproblem 18744 x ^ +a^— =(18744 + a;) j?^, .^
1704X ^ + ^=56232 4 OSo;,
;rx 003=41653
0^=12496.
Answer.— ^ £1,249. 12s.
66. The net income from the estate is —
£4,000 X ('i^—) = £3,400.
The estate is sold for £4,000 x 30=£120,000.
The income derived from Consols =£120,000 x , , ^ ,
112f + i
=£2,923588,
. =£2,923. Us. dd.
Decrease in net income =£476. 8s. Sd.
Answer.— £476. 8s. Sd.
275
SOLUTIONS. 137
67. As nothing is said in the question in reference to brokerage, we
assume that it is not to be taken into account in the answer.*
Let a: denote the price of the second Stock, then, by problem —
96i ^ ^ X
Answer.— 133^ %.
235
68. Let a: denote the gross income in first year. Therefore o^^rr^z
will denote the net income in first year. Also a? + 30 will denote the
232
gross income in second year. Therefore (a? + 30)—  will denote the
net income in the second year. Then, by problem —
235 , _LQm2B2
240 = (" + '^)2S0
3^=232x30
a;=2,320.
Answer. — Gross income in first year= £2,320.
69. The question does not state what kind of stock is intended, and
the student who happened to be perfectly familiar with brokers' charges
could therefore only guess at the actual rate of commission which would
give the answer required.
It is usual with brokers to charge commission on the nominal value
of all Government stocks ; thus, on Consols the commission is th per
cent., on Colonial stocks i per cent. ; so that in these cases it will be
correct to deduct the brokerage from the selling price. On other stocks,
however, such as railway debentures, &c., the commission (viz., ^ per
cent.) is charged on the consideration ; thus, £100 stock bought at
£160 would be subject to commission of 16s.
* The brokerage on the 4 % Stock would depend on the description of the Stock
purchased, and, as no particulars are given, it is impossible to determine the
probable charge.
138 SOLUTIONS.
Of course in many cases special terms are arranged for, where the
dealings are on a very large scale. There are other expenses also to be
reckoned (that is, in practice), such as stamp and transfer fee, which
are charged to the buyer, but not to the seller.
Assuming the stocks mentioned in the question are such as will be
chargeable with commission on the nominal value, we have —
Let a? = amount of 5 per cent, stock
y= amount of 6 per cent, stock.
Then original income ='05^+'06y.
Amount of cash realised by sale of 5 per cent, stock —
= 07(101^).
Amount of cash realised by sale of 6 per cent, stock —
=y(ioii).
Amount of new stock purchased —
= {^(101)+^(101i))j^^
New income = ja:(101)+y(101i)}j^^(055)
Then, by problem —
•05^+06y=^(10H)(^^^)055+y(10H)(~)055
=<f)(i><?)(s4)
50.+60y=.815Q+y809g)
=.163(i)+y809(l)
13 91
3
so
= 15^
CO
3
91
~y
= 13^
7
~5
15
Answer. — The amounts of 5 per cent, and 6 per cent, stock were in the
proportion of 7 to 5.
i
SOLUTIONS. 139
70. Let 2a? denote the amount of stock held which produces a total
income of £1,880, therefore —
0?= 24,000.
Let ^=the price of the 3 per cent, stock and 2;= the price of the
4 J per cent, stock, therefore the man's total capital will be —
24,000 X
\iooJ'
By problem, if the man invest so that he receives equal incomes from
each stock, his total income will be £1,872, that is, £936 from each,
and therefore the amount invested in 3^ per cent, must be
£936 ^ — , and in 4^ per cent, stock £936 ^ . Therefore—
936xf+936x=24,000(g),
whence
^ A ^
y=~z and z=y.
We thus have —
Total invested capital
=240(y+«^)
=240(y+^y)=240xyy
=240^2; 4 ^2^) =240 x^^.
Therefore, half the invested capital =120x — y
= 120x^^.
The income from the 3 per cent, stock will evidently be —
1 K
120xy
x3i=900.
140 SOLUTIONS.
Similarly, the income from the 4^ per cent, stock will be — ,
15
120x— 2;
X 4^=975.
Hence, the total income will be £1,875.
Answer.— £1,875.
71. {a) The dial of the clock is divided into 60 spaces, and the
second hand travels over 720 such spaces whilst the hour hand is
travelling over one space.
At half past one o'clock the hour hand is 7^ spaces in advance of the ^
second hand.
Let a;=the number of seconds required for the second hand to ,
overtake the hour hand, therefore —
• ^='^^"^720
719 _
""720^^
"^^^^719 =^^^ seconds.
Answer. — 7 seconds.
(5) Let y=the number of seconds required for the second
hand to gain a position bisecting the other two hands.
Whilst the second hand is travelling over y spaces the hour hand will
travel over ^ spaces and the minute hand wiU travel over ^ spaces,
therefore at the required time the minute hand will be 30 f ^ spaces
from the top of the dial and the hour hand will be 7i + =^ spaces from
the top of the dial ; hence we have —
1,427 _ 75
^ 720 ~ 2
Answer. — 18ff seconds.
SOLrTIONS. 141
(c) Let z be the required time in seconds.
.=30+1
/^=30
60
^=30f.
Answer. — 30f seconds.
72. Since one tap will fill the cistern in 8 minutes and the other in
24 minutes, the two together should fill it in 6 minutes, but it is stated
that the two together take only 5 1 minutes, which can only be accounted
for by there being a leak, which, of course, will show less effect when the
water is flowing in more quickly.
Let 1st tap (when there is no leakage) fill cistern in oo minutes.
,, zna ,, „ ,, „ ,, ,, y >»
Let leakage when no tap is running empty cistern in z „
Therefore, if 1st tap and leakage are running together f J of the
\^ . z/
cistern will be filled in 1 minute, whence by problem —
JM ■ .a.
Similarly, when 2nd tap and leakage are running,
l_l = i (2)
Similarly, when both taps and leakage are running,
Adding equations (1) and (2),
112 1
^ + y^ = 6 • • W
Subtracting equation (4) from (3) —
z~66
• 2! =66.
That is to say, the leakage would empty the cistern in 66 minutes.
142 SOLUTIONS.
73. It must here be assumed that the clock proceeds at a correct
rate during the 4 hours.
The last time the clock strikes there will be (5 f m) strokes ;
thus the total number of strokes will be (2+w) + . . . + (5 + w)
= 144w, which by problem equals (^m + n) ; therefore —
14i\4im=m\ny
or 14+3w=w (1)
Again, by problem (5+m)2— 41=^ (2)
(5 + w)2— 3w— 55=0,
or ^2+ 7^—30=0,
(7w + 10)(?w3)=0,
whence m=S,
and n=2S.
That is, the clock is 3 hrs. 23 mins. fast.
Answer. — 3 hrs. 23 mins.
74. When the goods are placed in the side requiring extra
15
weighting he sells T^ths of a pound, and when placing goods in the
15
other side it would evidently require a weight of zr^ths of a lb. to
result in an exact pound of goods being weighed, therefore a weight of
14
one pound will result in ^^ths of a lb. of goods being sold. Therefore,
15
total goods sold = — f
14 15
15"^ 14*
2^ lbs.
Answer, — 22^ lbs.
75. In the first innings he must make his previous average, together
with an additional 2 runs for each innings, including the one now
being played.
Let a denote the total runs obtained previous to the match.
„ w „ „ number of previous innings.
„ iT „ „ „ „ runs obtained in first innings j
SOLUTIONS. 143
then a?= +2(72 + 1).
n
Similarly, 80=^+4(^ + 2),
= — +4(;2 + l)+4,
n
=2^+4,
whence iP=38.
That is, he made 38 runs in the first innings, and 42 in the second innings.
Answer. — 38 and 42.
76. Let cc, y, and z denote the number of pounds, shillings and
pence; where z<(V<12, then employing the usual monetary notation —
£ . s. d.
X y z
reversing z y x
subtracting x—z z—x^
but z—x is negative, we must therefore adopt the usual artifice of
borrowing and carrying, whence the difference becomes —
reversmg
adding
or
77. Let X denote half the distance from P to Q, expressed in miles.
Then by problem —
X \^ X 1x
10 "^ 60 "^ 15 "^ IT
Multiplying by (60 X 11) Q^x\ 165 + 44;r = 120^
x=V6\
2^=33.
Answer. — 33 miles.
£
s.
d.
x—z—\
19
12x+z
12x+z
19
x—z — 1
11
38
11
12
18
11.
144 SOLUTIONS*
78. Let X denote the number of miles ' from A to B. By problem
we are told that the first train travels at the rate of 2 minutes per mile,
and the second train at the rate of  minutes per mile.
The first train is six miles out when the second train starts, and since
the second goes ( « + « ) niiles before meeting the first train, it is evident
that the first train has travelled (9—6— J miles, whilst the second
train has been travelling (9 + 0) miles. The time taken by first train
to travel/ —6— j miles is ( —6— « ) >< 2 minutes.
The time taken by the second train to travel ( « + « ) miles is
X  minutes.
2i
Hence the equation —
S«D'<^=(i+i)'<
3^7+3
xl^=
4
Answer. — 55 miles.
79. Since the second train takes longer to pass the two men, and
yet at the same time it is required to prove that it will overtake and pass
the first train in 36 seconds, it is therefore evident that the length of the
second train must be greater, and also that it must travel faster than
the first train.
It must also be assumed that the respective rates of the men and
trains do not vary.
Let length of the first train =Z. Let length of the second train = Z'.
„ rate per sec. of ,, „ ^t. „ rate per sec. of „ „ =t'.
V >) ;> ?> 7> jj man=ii?. ,, ,, ,, ,, ,, ,, man =y»
, .=36 (a)
SOLUTIONS. 145
The time taken by the trains in passing each other will be the sum
of their lengths divided by the difference in the rates at which they are
travelling, viz. : —
l+V 
t't
Since we are told that the time taken in passing is 36 seconds,
the problem, therefore, resolves into simply proving the equation —
t't
Dividing the time taken by the first train in passing the first man it is
evident that the length of ground covered is Z+35^, and dividing this
by the time taken, we have the rate at which the train is travelling.
Therefore, ____=^=____ (^)
similarly, ^^_^_^^_ (y)
therefore, from equations ()8) and (y), we have —
V 1
t't=— (8)
VI ..
= 45 36 ' ^'^
from equations (8) and (e) we have reducing to common denominator —
315Z'308Z'=396Z385Z
substituting —1 for V in equation (8),
we have t'—t=  — —, — ,
7 X 44 35
_^
~14*
18
Also, l'\l=l;
., „ Z'+Z 18 1 
therefore, _ = _Z^_Z,
=36.
Q.E.D.
146 SOLUTIONS.
80. Let x^ denote the number of men who formed in a solid square.
By problem, the number of men in the front row of a hollow square,
four deep, is 2^ + 4; there will therefore be 8 rows of 2a^f 4 each, and
8 rows of 2cc—4i each, that is, a total of —
8(2^+4) + 8(2^4) = 32a:.
Hence the equation —
x^=S2x,
07=32,
a?2= 1,024.
Answer. — 1,024.
81. Denote the four numbers by o:— 3y, or— y, o:+y, 07 + 3y.
Then by problem —
(or3y)(o7y)(o:+y)(o7 + 33/) = 384,
that is, (072%2)(a72_y2)=384 . . . . (1)
Also (ar3y)2+(o7y)2+(ar+y)2+(o? + 3^)2=120 .... (2)
From equation (2) —
072+5^2=30,
072=305/ ... (3)
Substituting (30— 5y2) for 072 in equation (1) we have —
(305y2_9y2) (S05ff)=S84<,
900600/484^4=384
7/_50y2+43=0 (4)
From equation (4) we have by formula —
50±'v/2,500l,204
= 1 or y.
Substitute the root y2=i jn (3), then
^2=305,
=25
07= ±5.
14
SOLUTIONS. 147
Hence, taking 07= d=5 and y= +1, the four numbers are —
(53), (51), (5 + 1), (5 + 3),
that is, 2, 4, 6, 8 ;
or (53), (51), (5 + 1), (5 + 3),
8, 6, 4, 2.
2, 4, 6, 8;
Ansioer. —
_8, 6, 4, 2.
82. Let the digits be a: and y.
By problem —
(10a;+y)(10y + ^) =8,722 (1)
Also, it is stated that if {10a: iy) be divided by (ar + lOy), then the
remainder, viz. (9^—%) or 9(^— y), will be an integer less than 10,
it is evident, therefore, that (.r— y) = l; hence, substituting (y + 1) for a?
in equation (1), we have —
(lly+10)(lly + l) = 8,722,
121^2+ I21y8,712=0,
y2+y_72=0,
(^8)(y + 9)=0.
Therefore, y=8 and x=9.
(98
Answer. — <
(89
83. (a) Let cc denote the lefthand digit, and y denote the
number represented by the remaining digits, then by problem —
10y + a:=3(100,000.r+y),
or 7y=299,999ar,
le., ^ y=42,857^.
Now X and y are positive integers, x being less than 10 and y less
than 100,000, and by trial it is seen that £c may equal 1 or 2.
(h) The number resulting from the smaller solution is 142,857.
If the operation is performed the resulting No. is 428,571 = 3 x 142,857.
repeated „ „ 285,714=2x142,857.
» „ again repeated „ „ 857,142=6x142,857.
and so on.
Answer.— 142,857 and 285,714.
L 2
148 SOLTTTIONS.
84. Let the ages of the three children be x, y and z.
Then by problem —
^+y + ^ + 9=(y + 9) + (2! + 9) (i;
^+y + 2; + 12=(ar + 12) + (2^ + 12) .... (2)
.r+y + 2; + 15= (07 + 15) + (y4 15) .... (3)
From equation (1) ir= 9
„ (2) y=12
„ (3) ^=15
Therefore the ages are 9, 12 and 15 ; also the father's age being the
sum of the three ages is 36.
An8wer.—% 12, 15, 36.
85. The whole course is 440 yards, therefore,
A runs 440 yards,
B „ 425 „
C „ 400 „
I
It is evident that C's rate is — — that of B, therefore if B runs
425
25
440 yards he should give C — X 440 yards start, that is 25if
yards start.
Ansiver. — 25^ yards.
86. Let the original expenses be represented by £ir, therefore, the
/*»
cost per head is £ — .
oi)
By problem —
X 07 + 300
30  35 •
35orl,050=30a; + 9,000,
5a?= 10,050,
07=2,010.
Answer.— £2,010.
87. Let the three sums be represented by £o7, £y, and £z,
then a:+y^z=Sn , (1)
r
SOLUTIONS. 14D
Also after the first game, the three shares are '
{xyz), 2y, 2z.
After the second game,
2{xyz), 2y{xy^z), 4^.
After the third game,
4^{xyz), 2{2y{xy^z)}, 4^z2{ooyz)2y+{xy\z),
or ^{(jc—y—z), Gy—2x—2Zf *lz—x—y.
And by problem 4(a:— y— a)=w (2)
Gy—2x—2z=n • (3)
^z—x—y=n (4)
13
From (1) and (2) 2x=—n,
13
or '*^="8^'
substituting in (1) y + 2;=— n, (5)
8
17
» (3) Zyz=—n,
adding 4iy= —n^
o
7
from (5) ^=o^
. ISn 7n 4tn
Amwer. ^, g , ^.
88. Let gross income be represented by £x.
By problem —
9 226
.X ^=10664375,
2,034a?=10664375 x 240,
339ar=10664375x40,
a?=12583.
Answer.— £1,258. Qs. 8d,
150 SOLUTIONS.
89. Let A's age=a?, and B's age=y.
It is {x—y) years since A was B's present age, and at that time
B's age was y—{x—y)=2y—x years.
We therefore have from problem —
x=^{2yx),
4iX=.Qy,
3
In 24 years' time B's age will be y + 24. The present age of A is a?,
and half this age is ; it is [y—j:) years since B was that age, and
therefore at that time A's age was x— (y— ] = (— —y \
We therefore have by problem —
y + 24=2(yy).
Substituting the value of x found above,
y=16.
Therefore from above x=24i.
Answer. — A is 24 ; B is 16.
90. (a) See Arts, (348) and (350).
(b) 5 Halfcrowns =25 Sixj)ences.
6 Florins =24 „
29 Sixi^ences =29
40 Coins. Total value =78 Sixpences.
Each person is to have 39 Sixpences made up of 20 coins.
Let first person have x Halfcrowns, y Florins, and (20—x—y)
Sixpences. Then the second person will have (5— a?) Halfcrowns,
(G—y) Florins and 29— (20— a?—y), that is, (a7+y + 9) Sixpences.
SOLUTIONS. 161
Therefore by problem,
5a; + 4y+ {20xy) =5(5^) +4(63^) + (9 + ^+^),
4a? + 33^=39 . . . . ' (1)
x—1
Dividing by 3, ^+yH ^~—^
o
x—1
X and y must be integers, and consequently — — — is also an integer.
o
x—\
Let ——= any integer =^ (say)
o
a?=3^ + l.
Substituting 3^H1 for a? inequation (1)
 12^ + 3y=15,
y=5— 4p,
giving ^ successive values, it will be found that the required values are
or 1, and that consequently
a:=l, or 4;
and y=^j 01* 1
'Each person may have either —
Answer. —  1 Halfcrown, 5 Florins, and 14 Sixpences ; or
, 4 Halfcrowns, 1 Florin, and 15 Sixpences.
91. («) /Sfee Arts. (348) and (350).
(J) Assuming that both halfcrowns, and sovereigns are to be
used on each occasion.
Let X denote the number of halfcrowns, and y the number of
sovereigns.
By problem —
+y=100.
X
y is necessarily an integer, therefore,  is also an integer, say p.
X
x=8p.
152
SOLUTIONS.
Taking successive values of ^, it will be seen that x and y have
both positive values for all values of p from 1 to 99 inclusive. There
are therefore 99 ways in which the distribution can be made.
Answer. — 99.
[Note. — The solution to this problem can, of course, be seen by
inspection without the aid of any algebraical symbols].
CHAPTER VII.
After reading the solutions of one or two elementary examples of
problems in Permutations and Combinations it is easy to solve others of
an exactly similar nature, but later on, when a larger assortment of
problems is presented, students often find it difiicult to determine the
exact nature of the problem under discussion. We therefore set out, as
concisely as possible, the principal cases which may arise, together with
their respective formulae.
(1) Permutations of n dissimilar
things r at a time denoted by '^Pr
=<wl)(^2) . . . Otr + 1)
(1) * Combinations Qin dissimilar
things r at a time, denoted by —
^(?i— l)(w2) . . . {n—r\\)
^C.=
1.2.3
\n^r
\n
m—r \r
(2) Permutations of n things all
at a time, when p are alike of one
kind, q^ of another, r of another, and
so on. Formula —
\p \^ r
{2a) * The number of ways of
dividing n things into on groups
containing p things in one group,
q things in another group, r things
in another, and so on (where, of
course, n=p\q\r\&,c.').
\n
Formula
l£ \g t
[* Note. — It will be seen that {2a) is an extension of (1), that is to
say, in (1) we have the case where a number of things are divided into
two groups containing r and {n—r) things respectively, whereas in (2)
we have the general case where n things are divided into several groups
containing ^ . 2' . r . . . things in each group.]
SOLUTIONS,
153
(3) Permutations of n things
r at a time when each may be
repeated 1, 2, 3 ... r times in any
arrangement. Formula w**.
(2 J) The number of ways in
which mil things can be divided
into n groups, each containing
m things. Formula —
{\m\'^\n
(3) Combinations of n things
^ at a time when each may be
repeated 1, 2, 3 ... r times in any
selection. Formula —
w(w + l)(^ + 2) . . . {n^r—1)
\n+rl
\r\n
—n+rin
(4) Permutations of n things
T' at a time when p are alike
of one kind, q of another, s of
another, and so on. Formula, see
Todhunter's Algebra, Art. (812),
viz. : — The coefficient of x'^' in the
product of —
(1 + Pa7 + P2a72+ . . .
+ BP^^) (1 + Pa? + P2^2 _j_ _ . p(Z^g)
. . . (1 + Pa7 + P2a?2+ . . . P«a7«)
when P, P2, P3 . . . have been changed
into 1, [2, 3 . . . respectively after
the product has been formed.
(4) Combinations of n things
r at a time when p are alike of
one kind, q of another, s of another,
and so on. For formula see Art.
(190), also Todhunter^s Algehra^
Art. (811), viz. : — The coefficient of
x'*' in the product of —
(1 + 074^24. . . . +07^) (1+^
+ 0^2+ . . . +07?)
. . . (l + a7 + ;r2+ . . . +^)
(5 ) Total number of permutations
by taking some or all of n dissimilar
things. Formula 2**Pf, which will
(5a) Total number of selections
by taking some or all of n dissimilar
things. Formula 2»Cr=2'»— 1.
154 SOLITTIONS.
be found to be the integer lying
, ( , n\l) (55) Total selections by taking
between e\n and \ e\n \ ii . j. , , . „
^— \ ^— n ) some or all out oi p + £4^^ + &c.
Since things when ^ are alike of one sort,
^wp =^j.u_i_i. , 2411 . ^^^ another, r of another, and s are
~ ~ ~ all of different kinds. Formula sum
I 1 1 1
^l+jj+2.. + u 2 =(p + l)(^ + l)(r + l)2«l.
I I 1 1
•=\n\ e— ■ ,
=:e\n\l\ ——
+ s
^■1
— . . , ad
(^ + l)(w + 2)
+
inf.^
■!
=e^
1+4.+ '
n\l ' (m + 1)2
+ ...!• approx.
=dw approx.
' n
There are also examples in which certain groups have first to be
selected, each group being permuted afterwards ; the grand total being
the sum of the number of permutations obtained from all the groups.
92. The definition of Permutations and Combinations given in
Golenso's Algebra appears to be the most complete, viz. : — The
Permutations of any number of quantities are the different arrangements
which can be made of them, taking a certain number at a time together.
The Combinations of any number of quantities are the different sets
that can be made of them, taking a certain number together, without
regard to the order in which they are placed.
The demonstrations given in Arts. (141) and (144), or Arts. (143)
and (144) are evidently the best for this question, as too much time
would be taken up if the demonstration given in Art. (155) were
employed.
r
SOLUTIONS. 155
93. In every group of r things the same 'p things are to occur.
Therefore the result is evidently the same if we take the 'p things away,
and then choose {r—'p) things out of the remaining {n—p) things.
This must be done by the method given in Art. (155) and not that
of Art. (144). The reason lies in the presence of the word Find.
When told to find a certain result, the Student must go back to first
principles and demonstrate each step.
It is, of course, fatal to miss the point of the question, and merely
state: — "this can be done in ^^^C^p ways," while the use of a
demonstration involving other undemonstrated. formulae should also be
avoided. A satisfactory answer may also be given as follows : —
We have (r—p) things to choose out of (n—p) things, the first
thing can therefore be chosen in (n—p) ways, leaving {n—p — 1) from
which to select the second thing. With each selection of one thing we
can associate a second in {ii—p){np—l) ways, but we see that any
two given things will appear in two separate selections, i.e.^ if a be
chosen first and h be chosen second, we shall also have h chosen
first and a chosen second. Therefore the number of distinguishable
selections of two things will be
Similarly with every possible way of selecting two things we may
XI • J • (n—p)(n—p—l){n—p—2) ^ , , „
associate a third m ^^ ^^— —■ — — ways. But, as betore,
2i
we see that any three given things will appear in three separate
selections, for, considering the three things «, h and c, the selections of
two things will include as separate selections ah, ac and he. When,
however, we add c, h and a to these respectively, we do not obtain three
new selections but one, repeated thrice.
Therefore three things can be selected in —
(^np){npV){np 2)
2.3 ^ '
Proceeding in this way, we see that {r—p) things can be selected in —
{np){npl)(n'p2) . . . {nprp^l) ^
2.3 . . . (rp) ^ '
(n—p)(n—p—l) . . . (« — r + 1) 1^— i?
I.e., m ^^ — — ^^ — ^^ ~ = ■ — ^=r^ ways.
\r^p \r—p \n—r
\np
Answer. — , . .
\r—p \n—r
156 SOLUTIONS.
Seven people can be arranged in a line in [7 ways, but in such a case
it will be considered a different arrangement if the first person is shifted
to the end of the row ; in the case of a round table, however, where it is
to be understood that positions at the table are not distinguishable, it is
evident that there can be no possible rearrangement, such as above
described. Thus, let the seven persons be represented by the first seven
natural numbers, when arranging them in a line, the following will
be considered as seven different arrangements —
12 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 12
4 5 6 7 12 3
5 6 7 12 3 4
6 7 12 3 4 5
7 12 3 4 5 6
but when arranged in a circle, all of these seven different arrangements
will be considered as identical. We see, therefore, that we are not
concerned with the relative positions of the people to the table, but
with the relative positions of the people to each other.
It will be recognised, therefore, that the number of arrangements in
a line are seven times as many as when in a circle, and since the number
of arrangements in a line is 7, the number of arrangements at a round
~7
table or in a circle will be ^ = 16=720.
7 "
Answer. — 720.
94. («) See Art. (151).
(b) The Committee must consist of either
6 British and 2 Canadians,
5 ,, }j 3 „
^ 55 ?J 4 ,,
and can thus be chosen in the following number of ways : —
6 0*2 6
6C6X8C2=,^h7^.77^= 28
16 8
16 8
1414
Answer. — 1414.
SOLUTIONS. 157
95. I£ we consider one link to be stationary, we can then arrange the
remaining 11 in 11 ways (as explained in Solution 93).
But when the links of a chain are arranged in any one particular
order, it is evident that, by turning the chain over, a different
arrangement will be presented, in which the links formerly appearing on
the lefthand side of the fixed link will appear on the righthand side,
and vice versa ; hence the distinct number of arrangements of 12 links
will only be onehalf of 11, or 19,958,400.
Answer,— 19,958,400.
96. («) See kxiiclQ (155).
(h) Ahracadabraca is composed of 13 letters, of which 6 are
a's, 2 b's, 2 r's, 2 c's, and 1 d.
113
.*. No. of words is ~ by Art. (151)
13.12.11.10.9.8.7 ,^ ,^ ^^^^
— =13x12x6,930,
2.2.2
=13x83,160,
=1,081,080.
Answer.— 1,081,080.
97. {a) See KtHcXq (151).
(J)) Here there are three distinct cases to be considered.
1st. When the first and last consonants are both t.
2nd. „ „ „ „ neither t.
3rd. „ first or last consonant is t.
(1st) The two fs can only be selected in one way, and the
18
remaining 8 letters may be permuted in r^ ways =10,080. Total —
I? If
10,080 ways.
(2nd) The first and last letters can be selected in ^02 or 3 ways, and
each can then be permuted in 2 ways, and the 8 letters can be permuted
in ,, ^ ._ or 5,040 ways. Total— 30,240 ways.
[2 2^
158 SOLUTIONS.
(3rd) The t can occupy first or last place in two ways, and the
other end letter can be selected in three ways, while the 8 letters can be
permuted in t^ or 10,080 ways. Total— 60,480.
Grand total— 100,800.
Answer.— 100,800.
98. Five counters can be chosen in ^Cs ways, i.e., 126 ways.
In each of these 126 selectioDs there will be nine digits, of which
four are zero.
The lefthand digit must not be zero, and that space can therefore be
filled in only five ways, and the remaining eight places can be filled in
18 18
^ ways. Thus there will be 5 x ^ different arrangements of each
selection ; therefore the total number of different ari'angements will be
126x5x^.
= 126x5x1,680,
=1,058,400.
Answer.— 1,058,400.
99. {a) See Arts. (143) and (144).
(Jb) The result here will consist of the product of three separate
items, thus : —
(1) The thre.e people who wish to travel with their back to the
engine have five seats to choose from, and can, therefore, be arranged in
^Pa ways, that is —
5.4.3=60 ways.
(2) The two persons who wish to travel facing the engine have five
seats to choose from, and can, therefore, be arranged in ^P2 ways, that is
5.4=20 ways.
(3) The remaining five people have five seats to choose from, and
can, therefore, be arranged in ^Ps ways, that is 15 = 120 ways.
Since any arrangement in any one of these three groups can occur
with any two of the other two groups, we see that the total number of
arrangements is —
60x20x120=144,000.
Answer. — 144,000.
SOLUTIONS. 159
Note. — A not unusual mistake is to argue thus : — Considering first
the seats nearest the engine. The first seat can be filled in 8 ways, that
is, with any one of the three who wish to sit that side, or with any one
of the five who do not mind which side they sit, similarly —
2nd seat can be filled in 7 ways.
3rd „ „ „ 6 „
4th „ „ „ 5 „
5th „ „ „ 4 ,,
Now coming to the other side.
The five seats can be filled in 5 ways, that is, with either of the two
people who wish to sit that side, or with any one of the remaining.
Giving an incorrect answer of 806,400.
The fallacy, of course, lies in saying that we can fill the first seat in
8 ways, second in 7, third in 6, thereby assuming that we may or
may not use the three people who wish only to sit that side, and then
in filling the remaining 2 seats we assume that the three said people have
already been disposed of.
This latter line of reasoning is not possible, and the Student should
carefully notice the fallacy.
100. Two sopranos can be chosen in ^€2 ways, with each of which
two contraltos can be associated in *C2.
.. Two sopranos and two contraltos can be chosen in «C2X*C2 ways.
Similarly, we may show that a choir of eight voices, two of each kind,
can be chosen in «C2 X *C2 X ^€2 X ^€2, or —
alcd{aV) Q)V) { cX) (dl)
abcd(al){il)(cl){dl)
16 ^"y^
Answer. —
16
101. The first prize cr.n be given in 5 ways, and the second can
also be given in 5 ways. /^
.'. The first two prizes can be given in 5 x 5, or 5^ ways.
Also the third prize can be given in 5 ways ; .. the first three prizes
can be given in 5^ x 5, or 5^ ways.
160 SOLUTIONS.
Similarly, it may be shown that all the 7 prizes may be given in
57, or 78,125 ways. Art. (152).
Answer.— 78,125.
102. (a) See Art (155).
(b) The first side can be chosen in 7 x 4 ways, and
„ second „ „ „ „ 6x3 „
giving a total number of sets 7x4x6x3=504, but each couple is
included once in the first side, and again in the second side, and the
correct number of games is, therefore, —=252.
Answer. — 252.
103. (a) See Solution No. 92 and Arts. (143) and (144).
(b) See Art (147).
112
The first parcel can be formed in ^^r ways, and each time we form
the first parcel we have 9 balls left, from which we can form the second
19
and third parcels in ijj ways.
112 19
Therefore the number of parcels = ;=• x m^
1 3 1 9 [4 [5
I? 1^1?
=27,720.
Answer. — 27,720.
104. The number of ways in which mn things can be divided into
\mn
m srroups is 7 , ~~i  See Note to Art. (147). Now each of these
^ ^ {«.}'^w
groups in any one division is distinguishable from all the rest in the
same division. Each division can therefore be distributed among the
m persons in [^ ways.
\mn
Therefore the total number of ways= ^rr" .
SOLUTIONS. 161
105. (a) /S'<se Article (151).
(5) To find how many different numbers can be formed is
the same as finding the number of ways in" which seven places can be
divided into two groups containing three and four places respectively.
17
This can be done in ^==35 ways.
If we imagine these 35 numbers set out for addition in the ordinary
way, it will be evident that each column will contain 6*s and 2's in the
ratio of 3 : 4, that is, there will be fifteen 6's and twenty 2's in each
column. The sum of the different numbers is therefore —
6,666,666 x 15 + 2,222,222 X 20
= 1,111,111x130
= 144,444,430.
Answer. — 144,444,430.
106. (a) /S'ee Article (141).
(b) The number of ways in which 10 similar presents are to
be distributed amongst 6 children, so that each child receives at least one
present, may be determined as follows : —
Suppose the 10 presents to be placed in a row, then by placing
5 marks of partition amongst them, we shall create 6 parts, each of
which will contain at least one present.
There are 9 spaces in which the 5 marks can be placed ; they can
19
therefore be distributed in ^€5 ways, that is, in .Fr: = 126 ways.
54
See Proposition XXV., WhitwortJi's Choice and Chance.
Answer. — 126.
M
162 SOLUTIONS.
CHAPTER VIII.
107. (a) See Arts. (178) to (181).
a: must be less than unity, h and k must be positive integers.
m
= (l + 2^)K16 + 3.r)Hl^)'
= (l + .r)ri6^+^xl6tx3^Vl + 2.r)appro^
= (l + ^)(2+~)(H2a.)
« 105
108. («) ^S'e^ Arts. (178) to (181).
Q.KD.
Kii)a2)(i 3)^,
1.3.5 /xY
, , 1 /■x\^ 1.3 /.r\3 1.3.5 /a?\
Before applying this formula to the evaluation of '^101 it must
first be put into the form of ^1 + a; where x<l.
^101=:^^100xyH:j^
= 10{ 1 + 005  000,012,5 + 000,000,062,5 }
= 10050,000,6 000,125
= 10049,875,0.
Q.i:.D.
I
SOLUTIONS. 163
109. (a) See Avis. (178), (179), (181).
(b) An important method of obtaining successive approxima
tions is here introduced, which will be of frequent use in practical work.
Let N=(«— ^)4 so that a7<l (1)
then N=«44«3a? + 6a2a?2 (2)
Now iV<l, therefore, as a first approximation, let us discard all terms
involving the square and all higher powers of ^, so that —
Now let 0:^=0;— —~ • (4)
, 16«4_6a4f6N
4«
^~ 10«^ + 6N '
Substitute in (1) N?=« ~ ^
_ 3a^+5N
~^5^3N*
Q.KD.
110. The first portion of this question is dealt with fully ,iu
Art. (172). We would, however, direct especial attention to the last
line of that article, viz. : — " It will be found best to work each example
independently of the general formula." This advice will be appreciated
by those who notice that the positions of a and cc are reversed. The
resulting formula will therefore be —
a:
>r.
BP The second portion of the question would appear to present no
difficulty, in view of the fact that it is a direct application of the above
164 SOLUTIONS.
formula, where ^=7, rt=3«, ^=7i, i.e., the (r + l)th term will be greater
than the rth term so long as —
8
56J
It must be particularly noted that it is impossible to say which term
is the greatest until the values of a and h are known.
111. (a) See Arts. (178) to (181).
(5) These two infinite series are obviously expansions by the
Binomial Theorem, but this point is not always suggested by the form
of the question ; the following characteristics of expansions of binomial
expressions, having fractional or negative indices, should therefore
be remembered.
1st. — The numerator of the coefficients of the successive powers of cc
consist of factorial terms, the common difference of the factorials being
the denominator of the index.
2nd. — The denominator of the coefficient of x'^ is the product of \r
into the rth power of the denominator of the index.
3rd. — The terms of the series are all positive when the index and the
second term of the expression are negative. They are alternately
positive and negative when the index is negative, and the second term of
expression is positive.
[Note. — The above will be readily understood if the form of the
coefficient of the general term, as set out in Example 2, Art. (182) is
borne in mind.]
The first series may be written in the form of —
^ 3 1.3/3V 1.3.5/3V , p
1+8 + 1:2(8) +1723(8) +^"'
, 1 3 1.3/1 3V 1.3.5 /I 3V . .
i+2s + r:2(2i)+f:2:3(2i)+^^"
and is then easily recognised as the expansion of —
(D"'=(r'
SOLUTIONS. 165
The second series may be put in the form of —
= (l^) ' Art. (186).
=2.
That is, the sums of the two infinite series are equal.
Q.E.D,
112. {a) Since
Therefore
(10.)^=!+ (2) (..•)+ ^^^^^ C^)^
^ (2)(2l)( 2^^_^^3^^^_
These expansions are valid only when a;<l and x<\ respectively.
[Note. — It is advisable to set out the working in full as above,
otherwise errors in signs, &c., are very likely to occur.]
{h) In both series the denominator of the index is 3, but no
power of 3 is contained in the first series, and must therefore be intro
duced ; thus : —
1 1^ 1.4.7 1.4.7.10
■^ 4 "^ 4.8 "^ 4.8.12 "^ 478::T2J^ ■^'^''•
14 \ tl
106 SOLUTIONS.
. . , 2 2.5 2.5.8 2.5.8.11
Again, 1+ 6 + 6712 + 6712X8 + 6:i2.T8724 +^^
2 5 2 5 8
, 2 1 3*3/lV 3*8*3m3 „
^+32+T2"y +^(2J+^^•
=(^D"'=(2)"'^^^)*=^^•
That is, the two infinite series are each equal to \^4!.
113. (a) See Art. (174).
(h) Let—
+ .70 + r73 . Art. (316) ;
Q.E.D.
(a;l)(;r2)(^3) xl x2 ' ^3
^^2_.3^_.l^A(^2)(^3) + B(^'l)(^^3) + C(a'l)(^2)
By i^utting in succession a;— 1=0, a: — 2=0, 07—3 = 0, we find that —
A=?,B=ll,C=^;
therefore
a;2 + 3.r+l 5 11_ 19
{xl){x2)(x3) 2(^1) 072 ^ 2(073) '
11 5 19
or, rearrangmg
<•!) '<'" <'3
=¥('ri<>"Mr'
The sum of the coefficients of 07*^ in these expansions is
11 AV 5 19AN^
2^
11 X 3*^+' 5 X 2» X 3«+i19 X 2''
/1\^ 5 19 /1\*^
•(2) 26(3) ^^'■^'''^'
6^+1
11 X 3«+i— 5 X 2« X 3»^+'— 19 X 2"
^^i^u'^r. g^^,
SOLUTIONS. 167
114. (a) See Art (172).
(h) The expression must be adjusted before it can be
expanded, thus —
\a—xj V a^—x^ J
= (« + a7)(«2^2)i
^ X \fx\^ l/^\3 1.3 (xy 1.3 /^V .
a • 2'
115. («) The general term is '•C,.a'«''a;'*,
^^ „(^_l)(„2). ■■(«>■+ 1) ^„_^^
r
/^^e Art. (166).
(J) Substituting the values of «, or, ;j, and r in the above
expression, we have —
/ 3 V^ /3 \» 1'7 1
The 10th term in f 2 +  o^ J is i' Cg x 2« x (^ ^ j or =^ •  (3;r) ».
17 1
That is, the coefficient of x^ is rrj^ ' o ' ^^'
17 1
/ 1N2M + 1 / In
The expression f ^^ j = x'^^^ + ^ ( 1 \
]_n2» + 1.
We therefore require to determine the coefficient of x^^'~'^'^ in the
\ N2M + 1
ex
/ iNiiM + l
pansionof [} ^ , Art. (167).
Suppose that x'^^^^^ occurs in the (^ + l)th term.
The (i? + l)th iQvm=''^+'Cp( ^^
2tl + lC^(^l)p^2i5.
168 SOLUTIONS.
But by hypothesis —2p — 2r—2n,
or p=n—r.
And the coefficient of iv^^^^ is 2n+ic,,_,,( — l)wr^
or ^= (^lyir
^^ \nr \ n + r\l ^ ^
Unless (w— r) is a positive integer there will be no term containing
^2r+i in the expansion.
\2n + l
Answer.— ■ , —  (—1)''^.
\n—r\n+r + l^ ^
116. Expanding the expressions, we have —
I
{l + (i)(3..)+ LIK J^)(_3^).
r, 4^ 1.2/4^\2 1.2.5/4a7\3)
(, 3^ 1.3/3.r\2 1.3.5/3a:V)
■^L3"2"Uy ■^~~j3 U; ~2'72~W
1.2.5/4Y1 „
3(3JJ^
, 17 144 + 243128
= 1 + — 0? +
+ 13
6 ■ 72
5,832 + 10,9353,456 + 5,120
+
17 _^259 , . 18,431 ^
1,296
SOLUTIONS. 169
This example has been given in full in order to show how a correct
solution is dependent on, and is rendered easy by, a careful attention to
small details ; errors are always likely to arise where an attempt is made
to condense two or more steps into one.
The expression — — — must first be resolved into partial
fractions, as follows : —
12^ + 3^2 A B C , ^ ,^,^,
so that l2a: + 3a72=A + B(l2^)+C(l2^0'
= A + B + C(2B + 4C)^ + 4Ca:2,
whence equating coefficients of like powers, Art. (311), we have —
A+B+C=l
2B + 4C=2
4C=3
and C=7, B = — , A=
4' ^ 2' ^4
1207 + 3^ ^ _ If 3^ __2 3 j
(12^)3 ~ 41 (12^)3 ~ (120^)2'^ l2^r
The coefficient of cc^ in the expansion of the expression  —
I JL JLOS )
is, therefore, one quarter of the algebraical sum of the coefficients of x"^
in the three expressions on the righthand side of the identity, that is —
_ 2^/ 3(^2+3yz + 2)4(^ + l ) +6^
~ 4 1 2 /
2» ( )
= g 3^24.5^,^_8
=:2«3{3;t24.5^.8}.
170 SOLUTIONS.
Or the expression may be written —
1 2x Zx^
{\2xY {\2xy^ {\2xy'
and the required coefficient is the sum of the coefficients of the (7i + l)th,
«th, and (;? — l)th teinns respectively in the expansions of these
expressions, that is —
1.2 1.2 ^ 1.2
Answer.— 2^^3(3^2 + 571 + 8}.
117. (a) See Art. (158), Example 2.
(b) See Art (176).
We may, however, consider the question as one in Finite Differences.
Since A**CV=**C,._i when An=l (see Finite Differences, paragraph
43), and from formula (10) we have —
™+«C,=>^C,. + m«C,._i+ '^^^l ^ ^C^_2 + &c.
I_
='^^Co^C,.+ "^C,^C^_i + "*C/Cr24 &c.
Put m=n=r, then 2/iCn="Co'^Cn + ^*Ci^^C,,_i + »^C2^C,,_2+&c.
= 0'Coy+Q''Ciy\C'C2y+&c. Art. (i7o).
_ \2n
Q.E.D.
118. («) /S'^e Art. (158), Example 2.
(h) Adopting the method employed in the preceding solution —
2>iCr = ^^CVCr + ^Ci^Cr 1 + ^C2^C,,_2 + &c.
or, writing n—2 for r,
2^C«_2 = ^Co^Cn2 + ^Ci»^Cn3 + ''O^'^Cn, + &c.
=^Co»*C2+*^Ci^C3 + ^C2^C4 + &c. Art. (170).
\2n
\n—2 \n + 2
Q.E.D.
SOLUTIONS. 171
Again,
1 2 "^ 8 • • ^^ ^^ n{l
\n 1 1^ 1 w 1 I'*
~/^0 2 ;il 1^3 ^z2 2 '••^^ >* « + l0^
=^{ f^  5^rg + 5^2^ ^^ }
_ 1 f l » + ^ 1^+1 MJ I
~ ^7+1 1 ^ ~ \n^l \2 "^ ^I^2]3 ~ ^"^^ j
nf i
+1
Q.MI).
119. (a) /^ee Solution No. 117.
2;i
Answer, — 7T=r„ •
^i=4i = !l"Ci^ + '*C2^' ■ ■ ■ +(l)"''C„a:»}{l + ^}3
The coefficient of x^ in the expansion of (1 — .r)** is ( — l)i'^Cp, and
the coefficient of a:^'~P in the expansion of (1 + ^^)"^ is —
^ ^ \2
n cc)^^
The coefficient of x^^ in the expansion of — — is therefore the
(1 + a^y
sum of the expression —
172 SOLUTIONS.
when ^ is given all integral values from to w inclusive, and r may
have all values from {n\V) to infinity ; i.e.^
(1)^{
2 + '' —2— + ^ 2 + • • •
+ ^C^^^ + 2C2}.
120. (l+^)^ = Co + Cia; + C2^2_.c3^3+ . . . +c,,^^ ... (1)
Squaring, (1f a?)2^={<?o + Cia7 + C2^'^+ . . . +CnCC^^}^
writing 1 for a:, 2^^= {co + Ci + C2+ . . . Cn}^
= Co^+Ci^+C2^+ . . . +c,,2.2(CoC, + CoC2+ . . .
+ CoCn + CiC2+ . . . + Cw_iC,j).
Transposing and denoting the sum of products by S,
2S = 22^(C0^+Ci2+C22+. . .+C^2) ... (2)
To sum the series ^0^ + ^12+ . . . +c^2^
we have (l\x)^=Co^CiX + C2X^+ . . . +Cna:'^ (3)
also (a7 + l)»^=Co^^+Ci^^> + 6?2^^2+ . . . +c^ ... (4)
Multiplying together the series on the righthand sides of equations
(3) and (4), we see that in the product the coefficient of x^ is the
series Co^+Ci^\ . . . +Cn^.
But the product is equal to (1 + ^)^"', and the coefficient of ^** is
\2n
^^= , therefore —
1^ l^t \2n
\n \n
Substituting in equation (2)
\2n
2S=22^ ^=^
\2n
whence S = 22»~ 1 — ^tF^, 
2(^2
Answer. — 2^^~
\2n
2{\ny
SOLUTIONS. 178
CHAPTER IX.
121. See Arts. (449) to (454).
Also TodTiimter's Algehra, Art. (726).
The probability of an event happening is the measure of our belief
that the event will happen, where it must, of course, be noticed that the
sum of the probabilities of the happening and not happening is certainty.
122. (1) See kvis. (449), (450) and (452).
(2) A may win two successive matches in many different
ways, which for convenience may be divided into two groups : —
(«) A may win 1st and 2nd.
„ „ 1st, lose 2nd, then win 3rd and 4th.
„ „ 1st, lose second, win 3rd, lose 4th, then win 5th and
6th, and so on.
Probability of these exclusive events : —
/ a \2r ah ( ah \^ ^ ^
\«+J
ah
{a+hy
a^ + h^ + ab
(h) A may lose 1st, then win 2nd and 3rd.
„ „ 1st, win 2nd, lose 3rd, then win 4th and 5th.
„ „ 1st, win 2nd, lose 3rd, win 4th, lose 5th, then win
6th and 7th, and so on.
Probability of these exclusive events : —
«2 h r^ ah ( ah
(« + 6)2 (« + 5)
h r a^ \
r ab f ab Y n
L (^+^^ "^ \("^+5)"2/ + • • J
174 SOLUTIONS.
The total probability is, therefore,
~ \a + bAa^ + b'' + abJ *
Q.E.D.
(3) If A wins the first game, the probability of his winning the
next game is z .
^ a + b
If he loses the second game, his probability is then I = — — )
^ ^ ^ a + b\a^ + b^ + abj
as shown in the second part of the preceding argument. Thus his total
chance is —
a b / a^ \
a + b "^ a + bW+b^ + abJ
_ d^ + ab^^a^ + a^b
~ {a + b){a^\b^+ab)
_ a(a+b)
a^+b^ + ab'
Q.E.n.
123. (a) Probability that only one ringleader will be drawn is —
12 X 138 _ 12 X 138 X 2 _ ^2^
i5«C2 ~ 150 X 149" ~ 3^725 •
552
Answer. — tt^ttz •
3,72d
(5) Probability that first and second men are both ringleaders —
_ 12 11 __22_
~ l50 ' l49 ~ §7725 •
22
Answer. ~^.
(c) Probability that neither of the men are ringleaders —
138 137 3,151
150 149 ~ 3,725'
Answer,^ pS
SOLUTIONS. 175
It is evident that the three results as above comprise every possible
result, and must be equal to unity. To prove our results we have,
therefore, merely to sum them, and see that the total is unity.
552 + 22 + 3,151
3,725
= 1.
124. The total number of ways in which the twenty persons can
be arranged in a straight line is j20.
If four given persons are placed together, they can rearrange
themselves in [1 ways.
Considering the group of four persons as one, we can arrange them
with the remaining sixteen persons in 17 ways.
There are thus ^x[17 ways in which the twenty persons can be
arranged so that the four given persons shall always be together.
14.117
The required probability is evidently ~~
14
20.19.18
1
5.19.3
1
285'
Q.E.D,
125. The three events are mutually exclusive and the sum of their
probabilities must equal unity, that is, if the three probabilities are
represented by p, q, and r, we shall have —
p + q + r=l (1)
Also from the problem —
^+i=ro + l ^2)
Subtracting (2) from (1) wo have —
11
^'=30
Therefore the odds on C are 11 to 19.
Answer, — 11 to 19,
176 SOLUTIONS.
126. If the three marked balls are drawn, then only two unmarked
balls are drawn, and it is merely necessary to find the number of ways
in which the two unmarked balls can be drawn and compare this with
the number of ways in which five balls out of twelve can be drawn.
Two unmarked balls can be drawn in ^€2 ways.
Five balls (marked or unmarked) can be drawn in ^^C^ ways.
Required probability = —
_9.8 1.2.3.4.5
1.2 12.11.10.9.8
1
22*
Ansiver. — —
127. A's chance is
B's ,
3 3
7 + 3 ~I0'
5 5
>5
11 + 5~16'
7 7 ■ i
13 + 7 20'
Since only one competitor can win, the above probabilities are
mutually exclusive, and therefore their sum must represent the
probability that one of the three competitors A, B, or C, will win —
10 "^ 16 "^ 20 ~ 80
Answer. — —
oO
128. {a) The number of ways in which the 5 men can speak is 5.
The number of ways favourable to the event is [4, for we must consider
A and B as taking the place of one speaker, A delivering the first
portion, and B concluding the speech.
Answer. — .
SOLUTIONS. 177
(b) If B agrees not to speak until after A, it becomes
necessary to obtain the probabilities separately. Let ^i, ^o, ^3, &c.,
denote the probability that A will speak 1st, 2nd, 3rd, &c., and p' denote
the probability that B will speak immediately following :
thus, if A speaks 1st, Pi= ^ and p\ =  p^p\ ~ 5 * 4 ~ 20 '
2nd, i?2=g andy2=g
, 1 1
1
15'
3rd, ^3=^ and/3=2
, 1 1
1
10'
4th, ^4=andy4=l
, 1
1
5'
V ^ 1 1 1
^^^20+15+10+
1
5'
_3 + 4+6+12
" 60
_ 5
""12*
Answer. — —
129. As is well known, the peculiarity of the recurring decimals
representing any proper fractions with a denominator 7 is that they all
consist in periods of the 6 digits set out in the question, thus : —
i= 142857 1= 285714 f= 428571
1= 571428 f= 714285 f= 857142.
The number of possible arrangements of the 6 digits is 6, and since
the arrangements representing the required fractions are 6 in number it
follows that the required probability is t^
_1
_ 1
~120'
Therefore the odds against are 119 : 1.
Ansiver. — 119 : 1.
178 SOLUTIONS.
130. 80 is the sum of 40 different pairs of positive integers, viz.,
(1 + 79), (2 + 78) .. . (38 + 42), (39 + 41), (40 + 40). The product of
the first eight of these pairs is in each case less than 600, and the
product of each of the remaining 32 pairs is greater than 600, and
therefore favourable to the event.
32 4
The required probability is .*. — or 
Answer. \
5
131. Let p be the probability of throwing an ace, and let q be the
probability of not throwing an ace, so that p\q=.\. There must be
either 0, 1, 2 ... or 35 aces thrown, and the respective probabilities are
^^Gop^q^^^ ^^Cip^g^, ^^(j2p'^q_^i &c., or the successive terms in the expansion
of (^+i?)35.
Now 2'= and P—^y
hence the most probable number is that of which the probability of the
happening has the largest numerical value, that is to say, it is the term
/5 1\35
in the expansion of ( ^ + ^ ) which has the largest numerical value.
Now '^CrP'^q'^'^>^Cr\p''~^q_'^~'^^^,
SO long as {n—r + V)p> rq^
or {n\l)p>r{p\q).
But p\q=l, therefore the terms will increase numerically so long
as {n\V)p>r^ that is to say, the (r + l)th term is the greatest when r
is the greatest integer in {n + l)p^ and when (n\l)p:=r the probability
that {n\X)p aces will be thrown is equal to the probability that
{{n\V)p^l} aces will be thrown, and these two numbers are the most
probable. In the present case (n + l)p = 6, therefore the most probable
number of aces is 5 or 6, either of which is equally likely to be
thrown.
Answer. — 5 or 6.
Note. — An answer frequently met with is as follows : —
The most probable number of aces is 6. But this answer is never
supported by any algebraical proof, and is only mentioned here as it is
obviously based on a misapplication of Bernoulli's Theorem (Art. 469),
which is only true when the number of trials is indefinitely increased.
SOLUTIONS. 179
132. (1) See Arts. (449), (450), and (452).
(2) Having found that the first card is the ace of hearts,
there are 12 cards left in that set which may be chosen from the 51
remaining cards of the pack in 5^Ci2 ways. If 3 of the 12 cards are
specified, there are only 9 of the set remaining to be chosen from 48,
which can be done in '^^Cg ways, therefore the required probability is —
10.11.12
51.50.49,
44
17.49.5
44
4,165
94 +
s
The second convergent to this continued fraction is — Art. (334),
95
and the true value lies between —  and — . Art. (335).
94 95
Q.E.D.
133. The number of possible ways of choosing one card from a
pack is 52.
The number of ways of choosing a spade is 13.
13 1
Therefore required probability = — =  .
Similarly, the probability that the next card chosen will be a
heart = — .
51
12 6
Similarly, that the third card will be a spade = — = — .
1 13 6
180 SOLUTIONS.
134. The possible number of ways of drawing two cards is
5202=1,326.
(a) 20 can be made up by 10 + 10 in ^€2 = 6 ways,
or by 94llin4CiXi6Ci = 64 „
Total of favourable ways =70
70 35
Required probability = ^^^ = — .
Answer. g.
(5)
20 can be made up in 70 ways (see above).
21 „ „ 64 „ i.e., by (11 + 10) in ^eCi X'^Ci ways,
22 „ „ 120 „ i.e., by (11 + 11) in 16C2 ways.
254
254 127
Required probability = j^^g = qq^
127
Answer. —
135. (1) The number of ways in which one player can have all the
trumps depends on the number of ways in which the remaining 39 cards
can be divided amongst the three other players ; that is, the number of
ways in which 39 different things can be divided into three equal groups,
139
which is .,—. : also the 52 cards can be divided between four players
(13)3
152
in 7~ ways. (See Note to Art. 147.)
Therefore the probability that one particular player will have all the
39 (113)4
trumps is r~: • tz~ , and the probability that this particular player
(13)3 1 52
will be chosen when one is picked at random is  .
1 39 (\lSy 139^
.. the total probability is  . ^3 . ^ = ^^ .
(2) The number of ways in which a player can receive
13 cards none of which are trumps is ^^Cis, and the number of ways
SOLUTIONS. 181
in which he can receive 13 cards without any restriction is ^2(3^3 .
Therefore the probability that one particular player will have no trumps is
390,, 139 113 139
—  =  — X ,'^^ , and the probability that this particular player
^^Cia 26 113 52
will be chosen is  .
4
1 39 39
••• Total probability =.^.j^.
(3) The probability here required is that the player shall
hold cards of 3 suits only : there being no restriction as to which suit
shall not be held, the value of this probability is evidently 4 times that
found in (2), that is —
1 39 39 _ (39)2
^ ^ 4 ^ 1 26 ' 52 "^ 2652 *
139 113 (39)2 (139)2
Answer. — —  —
452 ' 42652' 26 [52
136. (1) See Art. (462).
(2) If the event fail r times it happens (n^r) times, and the
required answer will be found by substituting (n—r) for (r) in (1).
137. (1) ^^e Art. (463).
(2) The number of possible ways of drawing 4 cards from a
pack is ^^Ci .
The number of ways of drawing a card from each suit is 13'*.
Therefore the probability of drawing one card from each suit is —
134 ^_^ 1.2.3.4
= 134 X
52C4 52.51.50.49
133
17.25.49
2,197
20,825
1
1052
''+2 +
182 SOLUTIONS.
The first two convergents of this continued fraction are  and ^.
Art. (334). And therefore the true value of the fraction is less than ^
and greater than ^^. Art. (335). Much more, then, is it greater
than j^.
Therefore the probability lies between \ and y^^, and the odds
against lie between 8 to 1 and 9 to 1.
Q.E.I).
[If we reduce ^, ^, aVsVs^ ^ ^ common denominator, it is at once
evident that gV^V l^^s between ^ and ^ ; but had we been required to
express as simply as possible the limits between which the required
probability would lie, it would have been necessary to have made use of
the theory of continued fractious : the above demonstration has therefore
been inserted.]
,«« XX 1 n—\
138. Here»=; .. q= .
n ^ n
Let xn be the number of trials, then the probability that it happens
at least once is to equal the chance that it will not happen at all, and to
equal .
\\xn
^ =(•?)
Taking logarithms and multiplying by —1,
loge2 = ir;i{logeW— loge(w — 1)}
=Hi + 2^^ + 343+*''} ^^M224).
X X
but all terms involving , — , &c., vanish when n is very large.
loge2=^=693 . . .
e.e., the number of trials =693 . . . y.n.
Q.E.D.
SOLUTIONS. 183
139. (1) See kvt. (462).
(2) The number of ways in which the sum of the readings
in one throw with two dice may be 9 is the coefficient of x^ in
{x^x'^+x^ + x^ + x^ + x^y. See Art. (466), Example 2.
{x\x'^^x^^x^ + x^ + x^y=xH^\
=x\l2x^ + x^^){lx)^.
The coefficient of x^ is easily seen to be (8 — 4) =4.
[Note. — This result could have been more easily obtained in this
particular example, but in more complicated cases the above method will
be found necessary, and it has therefore been deemed advisable to set it
out it full.]
The number of possible readings from two dice is 6^=36.
4 1
Therefore the probability of the sum being 9 is — =  . The
probability of obtaining this result in exactly 3 out of 7 trials is found
by taking the fifth term in the expansion of ( f ) , see Art. (462),
8^
which is 7C4 X — .
84
Answer. — 7C4X — .
140. (1) See Art. (462).
(2) The number of ways in which 12 may be thrown with
3 dice is the coefficient of a?^^ in —
(a?! f ^2 _^ a;3 + a:4 ^ ^,5 __ ^6)3
that is in —
x^l j =a;3(l3^6 + 3^i2_^i8)(i_^)3.
It is easily seen that the coefficient is (55— 30) =15,
Therefore, since 12 can be thrown with three 4's in only one way,
the required probability is — .
Answer. — — .
15
184 SOLUTIONS.
141. (a) Let A, B, C denote the three men.
Q
A's chance of drawing a black ball the first time is .
8
A's chance of drawing a black ball at his second attempt depends on
A, B, and C having previously failed, and therefore —
/5\3 3
= (8)^8
o , /5\3 /5\^ 'i
Thus A's total chance = o ] ^ + ( o ) + ( o ) + •  adinfA
Similarly, B's total chance = (g)(g) { ^ + (s) + (s) "^ * ' * "^"^ *^^' j '
 "■■ ' =(D'(i){'Kfi'HD" •■"■/(■
Therefore the ratio of their respective chances is —
Q.E.r).
(J)
A's chance of drawing a black ball first = « "I" o * ;^ * « ' ^
_53, 54323
" " " " ~8*7"^8'7'6*5*4
543.543213
^ O >> »> 5> 5J
8 7 6^8 7 6*5'4*3
. . n. o „ ,, }, ,,
27
56
t* s „ „ „ ,,
18
"56
^ s „ „ „ „
11
"56
Therefore the chances of A, B, or C winning are as 27 : 18 : 11.
Q.E.D.
SOLUTIONS. 185
142. See Art. (456).
fir The sum of the numbers thrown by A may range from 2 to 12.
The sum of the numbers thrown by B may be 24, 19, 14, 9, 4.
We are, therefore, only concerned with the probabilities that B will
throw 24, 14, or 4, and A 12, 7, or 2.
Now the probability that B will throw 24 is ( j ,
and „ „ „ A „ „ 12 is Tj ;
therefore the probability that both events happen ~ Tfi ' oft *
14 /1\4
Similarly, the probability that B will throw 14 is = ( „ 1 >
and „ „ A „ 7 is 32(^iJ;
therefore the probability that both events happen is
16.36 16.36
Again, the probability that B will throw 4 is () ,
and „ „ A „ 2 is Tj ;
therefore the probability that both events happen
Total probability
16.36
1+36 + 1 _ 38 _ 19
16.36 ~ 16.36 ~ 288*
Answer.— ^
143. The number of possible ways of throwing two dice is 6 x 6=36.
7 may be thrown with 1 and 6 in 2 ways.
7 » ?5 2 ,, 5 ,, 2 „
7 may therefore be thrown in 6 ways.
Therefore the chance of throwing 7 with two dice is — =  .
^ 36 6
186 SOLUTIONS.
If A does not win the first time it will be necessary for all four —
A, B, C, D — to fail once, in order that A shall have a second throw, the
probability of which is ( ^ ) • Similarly the probability of A having a
third throw is (  ) , and so on. Therefore A's total chance of winning
is represented by the series —
1
6 216
^ .5V 671
©
Similarly, it may be shown that the respective chances are : —
1
lf/5\ /5\5 /5\9 , . .^ /5\ / 6 \ 180
='—=5(©+a) +©+'•■''}©
/5V/ 671'
@
1
1(/5V /^5V /^5n^o , . ^) /5V/ 6 \ 150
<" ■• =5(U) +{8)+(5) + *./}=(s)
(!»
671'
If/^^V /^^V . /5^" ^ • ^) /'^Vf 6 1 125
^ ^ » =6{(6) + uJ + u) ^ • • • ^'' '^f'He) \r7^j=e7v
216+180 + 1 50+125 ^ ,, ...
[Note. — ^;^ =1, thus provmg the correctness
of our results.]
216 180 150 125
Answer. en ' 671 ' 671 ' 671 *
144. Let a? denote the number of times 5 must be thrown ;
then (100— a?) denotes „ » 3 „
and 5iP + 3 (100  a;) = 400.
Therefore 2a;=100.
ar=50.
SOLUTIONS. 187
It is, therefore, evident that in order to make up a total of 400 there
must be 50 throws of 5, and 50 throws of 3.
The probability that 50 fives will be thrown is the term containing
(  j in the expansion of [  + o ) » *^^ ^^^ ('^^2),
2
where  is the probability of 5 being thrown in any one trial.
o
" ^ 3
1100 /2y^/'iy^
Therefore the required probability = . r^" [ q ) ( 3 )
(2)^1100
(2)so100
Answer. —
(3)i««{50}2'
145. (a) A pair of dice can be thrown in 36 ways.
 =5 ways.
8 can be thrown with a 6 and 2 in 2 ways
>J )? 1» *^ 5) " >> ■^ »
» »» » 4 „ 4 „ 1 way ^
Therefore the probability of throwing 8 is ,
9 can be thrown with a 6 and 3 in 2 ways
4 1
)» » » ^ ?} 4 ,, J „
Therefore the probability of throwing 9 is
V =4 ways.
36 9
Therefore the probability of throwing 8 the first time and 9 the second
5 15
time = —  X  = — — 
36 9 324
Answer. A.
188 SOLUTIONS.
(h) The probability of throwing 9 the first time = r.
y
The probability of not throwing 9 the first time and then throwing
8 5 10
8 the second time = ;; ^ — = — .
9 oo 81
Total probability required =+— = —
9 81 81
19
Answer. — — 
81
3
146. The probability of A winning any game is ,
o
2
therefore „ „ B „ „ .
Also A and B want 3 and 2 points respectively.
Therefore, as shown in Art. (4iQQ),
27 X 11 297
54 625
3\2^ 4 X 82 328
A's chance is (^)'{l + 3Q}
That is, the odds against A are 328 : 297,
or 33 : 30, nearly,
or 11 : 10, „
Q.E.D.
147. The number of possible ways of placing 3 letters in 3 envelopes
is 3P3=3=:6^ viz., as follows: —
All right in 1 way.
All wrong „ 2 ways.
1 right and 2 wrong „ 3 ways.
2 1
Hence the required probability =  = 
6 3
Answer. .
3
[Note. — This question is fully discussed in Art. (483).]
SOLUTIONS.
189
148. Let the large circle with radius four inches, viz. : A to F,
represent the bottom of the saucepan, and the small circle with
radius of halfaninch, viz : A to B, represent the hole in the saucepan.
The question supposes that the coin always falls quite flat, and
therefore the centre of the
coin must always fall within
the large dotted circle, having
a radius of three inches, viz. :
A to E. If the centre of the
coin falls within the dotted
circle whose radius, A to D, is
one and ahalf inches, the edge
of the coin will overlap the hole.
The chance of the coin drop
ping clear of the hole must
evidently be the same as the
ratio of the area between the
two dotted lines as compared
with that of the whole circle having the radius A to E. Now the area of a
circle varies as the square of its radius, therefore the ratio required
IS
(3)>
That is, the odds in favour of the coin falling clear of the hole
are as 3:1.
Q.ED.
149. Each aperture of the netting contains 9 square inches.
If the centre of the ball passes within half an inch of the wire, the
ball will touch the wire ; there is, therefore, a space of 4 square inches
only within which the centre of the ball must pass in order not to touch
the wire, and since there are 9 square inches in each aperture, the
4
probability of the ball not touching the wire is  .
Answer.
150. See Art. (464).
190
SOLTITTONS.
151. 33 can be formed by 10 . 10 . 10 . 3 in 4 ways.
10 . 10 . 9 . 4 in 12 „
10 . 10 . 8 . 5 in 12 „
10 . 10 . 7.6 in 12 „
10 . 9 . 9 . 5 in 12 „
10 . 9 . 8 . 6 in 24 „
10 . 9 . 7 . 7 in 12 „
10 . 8 . 8 . 7 in 12 „
9 . 9 . 9 . 6 in 4 „
9 . 9 . 8 . 7 in 12 „
9 . 8 . 8 . 8 in 4 „
Number of favourable ways = 120
The number of possible ways =10'*. Art. (152).
120 3
Therefore the probability = — ^ = — ,
and tbe expectation =£ 1,000 x
120
10,000
=£12.
Answer.
250
, £12.
The problem might also be solved as follows. See also Art. (466),
Example 2 : —
The number of ways in which 33 can be drawn is the coefficient of
a?33 in the expansion oi (x^ + x^ { a;^ + . . . +a:^^y^
i.e., of ic*(l + a^ + iv^{ . . . +a:^y
or of the coefficient of cc^^ in the expansion of —
(l4a?^o+6a720_&c.)(la:)4
SOLTTTTONS. 191
The terms required in the second factor are those involving
^729, a?^9, and o:^, which. will be multiplied into 1, (— 4:r^<^), and (601:^)
respectively. ^
The coefficients are —
32.31.30 22.21.20 12.11.10
(3 ' I ' I '
or 4,960, 1,540, 220,
which, after multiplication by +1, —4, and \6 respectively, give —
4,9606,160+1,320=6,2806,160=120.
120 3
Therefore the probability = — — = ^— r.
162. 7 can be thrown with 6 and 1 in 2 ways.
5 „ 2 „ 2 „
4 „ 3 „ 2 „
Total, 6 ways.
® ii ti »i ^ ?) ■^ j» ^ >i
5 ,, 3 ,, 2 ,,
4 „ 4 „ 1 way
Total, 5 ways.
Also total number of throws =6^=36.
/»
.. chance of throwing 7=.
36
8i
36'
6 25
Therefore A*s expectation is ^ x 1,000 i — x 500,
oh oh
5 2*1
and B's „ „  X 1,000 fx 500.
.'. A should pay B  of tha difference between the two shares,
=i(£27. 15s. 7d.)
= £13. 17s. 9^d.
Answer.— £13. 17s. d^d.
192 SOLUTIONS,
153. See Art;. (466).
The probability that A will win is the sum of the probability that he
will win the first three games, or that he will win the fourth, and
at least two of the preceding three games.
.. A's probability=Q3 + 3C,Q4=(^l)3(^l + 3 ^ ^),
_1 5_^
~8^2~16'
5 11
and B's probability =1— :r^ = ^
[This can of course also be shown in the same way as has been done
for A.]
5
.*. A's expectation is — of £16 =£5.
B's „ ^ of £16 =£11.
lb
Ansiver. — £5, £11.
154. 10 can be thrown with 4 dice, as follows : —
6
. 2
. 1
in
12
ways.
5
. 3
. 1
12
4 ,
. 4 .
, 1
6
4
. 3 .
, 2
24
5 .
, 2 .
. 2
12
3
. 3 ,
. 3
4
2
. 3
. 3
. 2
6
2
. 2
. 2
. 4 „
Total
4
80
ways.
Or we can obtain this result as explained in Art. (466) by taking
the coefficient of x^^ in the expansion of —
la;6\4
/ 1—x^V
(x^x^^x^\x^ + x^[x^Y={x——\
=x^(lx^y{lx)\
where the coefficient of x^^ is —
_4_j_ 84=80,
SOLUTIONS.
19S
The number of ways in which 4 dice can be thrown is 6*, that is
80 _ 5
L296~8i'
1,29G ; therefore the probability of throwing 10 with 4 dice=
Similarly it may be shown that 10 can be thrown with 3 dice in
27 ways, and since the total number of ways in which 3 dice can
be thrown is 216, the probability of throwing 10 with 3 dice is —
27 _1
216~8'
Therefore the total chance of A throwing 10 before B
76 7
6 7\2
= 8il^
■+8l^
8
5
= 8i^
1
1 76
"7
^81
'8
5
81.2
= 81^
29
_10
""29*
V81 Sj
\ ... ad. inf.
Therefore A's expectation is £^= £345= (Ss. lO^d.
andB's „ £=£655=13s. li^.
•A
Answer. — Qs. lO'id. and 13s. 1^^. respectively.
155. See Arts. (467), (468), &c.
The observed event, viz., 2 white balls and 1 black having been
drawn may have arisen from 6 different a priori causes, viz. : —
1 black ball and 7 white balls. Therefore Pi= 8(^)8
P2=28(i)8
P3=56(i)8
P4=70(i)«
P,=56(i)8
P6=28(i)8
The respective probabilities that the event would happen as a result
of these causes are —
2 ,
, balls „ 6 „
3 ,
„ 5
4 ,
» 4
5 ,
,, 3
6 ,
„ 2
3
15
^^=8' ^^=2^'^^^
15
28'^^^
3 15
7'^^=56'^«^
28
194
SOLUTIONS.
Hence we have the a 'posteriori probabilities of the 6 different
causes, as follows : —
8x
Qi=
8
«i+2«2^+^«i^^°'+^<'56+2«28
= 8x
3
8
^96=
1
'32'
Q2=
=28 X
15
28'
^96=
5
'32'
Q3 =
=56 X
15
28
^96=
10
'32*
Q4 =
=70 X
3
7
r96=:
10
'32'
Q,=56x^f96= ^
32'
Q«=28x 1^96=^^.
If Qi represent the true cause of event, the probability of
drawing 1 white ball
Q2
Q3
Q4
Q5
11
Therefore the required probab
4
5'
3
5'
2
5'
1
5'
5'
lity—
4 3 2 1
=Qi + Q2X+Q3X g+Q4XfQ5X 
_16
~32
_1
~2'
Hence the chance is even.
Q.^.D.
SOLUTIONS. 195
[Note. — It may be pointed out that since nothing is known as to
the colour of the 8 balls, except that each is equally likely to be black
or white, the withdrawal of 3 will in no way affect the question, and the
next draw is equally likely to be black or white.]
156. (a) See Arts. (470) and (471).
(b) The a priori causes of the event are as follows : —
2 sovereigns and 3 shillings, therefore Pi = 10(i)5
3 „ „ 2 „ „ P2=10(i)s
4 „ „ 1 „ „ P3= 5ay
5 P.— ^iV
The resjiective probabilities that the event will follow from the above
causes are
The a postei'iori causes of the event are therefore as follows : —
lOxA 1
Qi
1 o
10x^V + 10x^+5x^+i§ 8*
o o 1
Similarly, Q2=  , Q3=  and Q4= g .
The value of the purse according to the above reasoning is, therefore,
i(£2. 35.)hf(£3. 25.)+f(£4. l5.) + K£5)
= i(£2. 3s. + £9. 6.?. 4 £12. 3s. + £5)
=i(£28. 12s.)
= £3. lis. 6d.
157. (a) See Art. (456).
(h) A priori the probability that the white ball would be
drawn was j^ , and therefore the probability that it would not be drawn
was^.
That is, P,= ^V, P2=,^.
The probability of A and B both telling the truth = x Y^=p\
„ „ „ the same untruth —
=ix^x(i)2=^2
196 SOLUTIONS.
It must here be noticed that, if A and B both tell an untruth, it is
very unlikely that they will both mention the same ball ; they each have
9 lies to choose from, and the probability that they will both choose the
same one is (^y. Hence the probability that A and B both spoke the
truth, and consequently that a white ball was withdrawn,
^ Pi;?i ^ ToXf XtV
f+ix(i)2
243
244'
Answer.
243
244'
[Note. — It should be carefully noticed that f X j^ is the probability
of A and B telling the truth in any given case, but in the above example
the fact of their both having made the same statement entirely alters
the probability as to the correctness of their statement.]
158. The a priori probability that the event did happen is
, and the probability that the 13 witnesses have all told the
1012+1
truth is (^ j .
By similar argument to that in the preceding solution, the proba
bility that the event happened is —
^2+1 uo.
9\i3
1012+1 uoy 913
101
1 /^yS 1012 /ly^ 913+1012
Hi VioJ "^1012+1* vloy
Dividing numerator and denominator by lOi^, we have —
(•9)i3 + l'
By contracted multiplication, we find that "913= 254.
^, p (9)^' 254 5
Therefore, ^^y^^^ = ^354 = 7 "PP""^'
Q.E.B,
SOLUTIONS. 197
159. (a) Since the coin is known to have an obverse and reverse,
and also to be dynamically perfect, the fact of its having always come
up heads in 10 successive throws must not in any way alter our
judgment, and we must still believe that the probability of the next
throw giving a head is  .
Answer. — 
(b) In this case there is a possible cause for the succession of
heads which have occurred, viz., that the coin may have been made
with two sides alike.
The coin may have been made with two heads or two tails, therefore
the probability that the coin was made with two heads
Adopting the usual notation, we have —
2,000,000 '*
Probability that the coin was made correctly
_ 999,999 _
~ 1,000,000 ~ ^*
Probability that the event would happen if first cause existed
=pi = l.
Probability that the event would happen if second cause existed
'l\i«
=Gy'
Hence we have —
1
^iPi _ 2,000,000
i?iPi + P2P2 1 /^1^'V. 999,999
©
2,000,000 ' \2J 1,000,000
29
29 + 999,999
512
1,000,511
512
Hence, since Qi + Q2=l, and Qi=
Q2 =
1,000,511'
999,999
1,000,511'
198 SOLUTIONS.
Therefore the probability that the next throw will be a head is
512 , 999,999 1
1,000,511 ' 1,000,511 2
1,001,023
2,001,022
•5002559.
Answer.— 5002559.
160. Before the event there are 5 possible hypotheses, viz., there
may be 4, 3, 2, 1, or white balls.
Adopting the usual notation, we have —
p.=a)s P2=4(i)s P3=6a)^ P4=4(i)s p.=iiy.
If there are 4 white balls, then the probability of drawing 3 red
balls is — — — . After the balls have been replaced, the probability of
4
drawing 1 white ball is — , therefore
^'"loCa^ 10~15
By similar reasoning, we have
7 7 7
Hence
^,= ,^3=^,i^4: ^^^
1 80 ^ 4 105 _, _ 6 112 _, 4 84
Pii^i=T«T^i7^' ^2i?2T^r^7^, t^3i?3:[7;7^7T7:, i#
and %Pp
whence
16 1,200' '^'~16 1,200' '^' 16 1,200' '^' 16 1,200'
1,508
16x1,200'
Qi Q2 Q3 Q4
80 ~ 420 672 336 1,508
The required probability is the sum of the probabilities that there
were either 4 or 3 white balls in the bag, that is —
^ _ 500 _ 125
^^ "^^'1,508 377
125
Answer. — 7—.
377
SOLUTIONS. 199
161. The required probability is the sum of the probabilities that
there will be n, ^ + 1, ^ + 2, . . . n + m sixpences, each multiplied by
the probability that there will then be a sixpence drawn.
By question the probability that there will be ii, ^ + 1, &c.,
sixpences, is in each case , therefore the required probability
m\L
_lj'7i n+l n + my
~ m\\\n\7n n + m' ' ' n\m)
((n + 7ii) (n + m\ 1) n(n — 1)
H1
1
_ 1 ^ n{n\m) + {m\l){n\m)—n{n — l) ]
"m + ii 2{n + m) I
2{n + m)
_ 1 ( n(m + l)h(nhm)(m\l) )^
""wtl( 2(n + m) )
2n\m
Q.E.n.
162. There are four cases to be considered, viz. —
That there are only 2 spades, the probability of which == P^
„ „ „ 2 hearts, „ „ =i=P2
„ „ „ 2 clubs, „ „ =i=P3
„ „ „ 2 diamonds, „ „ =i=P4
Probability of the event stated if ] ^ ^ ^^^ _ 3 x 3 x 2 _
\ 2 spades
there were only j "C5
^, . 1x3x3
2 hearts =— ^— = ^2
^ , , 3x1x3
2 clubs =—^—=^3
3x3x3
2 diamonds = — — = p^,
Thatis ^,= ^, ^2=j5^, i^3=j^, P.= ^^
200 SOLUTIONS.
2 1 1 ^
Therefore Qi=]^, Q2=^, Q3=^, 04= =•
^~ ^ . . _ [ 1 spade, 1 heart, 1 club and 3 diamonds,
remaining cards are )
Q2= „ ,, „ ,, 2 spades, 1 club and 3 diamonds.
Q3= „ „ „ „ 2 spades, 1 heart and 3 diamonds.
Q4= ,, „ ,, „ 2 spades, 1 heart, 1 club and 2 diamonds.
The probability that of the two cards drawn, one at least will be a
4 4 4 3
diamond =Qi x  +Q2X  +Q3X  +Q4X 
0000
_2 4141433
_S_ 4^ ^ ^
"" 35 "*■ 35 "^ 35 "^ 35
CHAPTER X.
163. (a) Finite Differences, (1), (2), (10), (54).
(^) » » (6).
{0) „ „ (19).
164. (a) Finite Differences^ (4).
{h) „ „ (27)(29).
165.—
_ 2a? + 7
^''~ 5^72+ 26a? + 5
2^7+7
(507 + 1) (a: + 5)
8(507+1 07 + 5)
therefore,
. _if_il__^l_ + _i L_l
"^ 8(507 + 6 5o7 + l 07 + 6 07f5)
Q.KD.
SOLUTIONS. 201
and
_lj 11 _ 11 11 1 1 1 I
_ 1 ( 650 2 \
~8 ((5a7+ll)(5^ + 6)(5a7 + l) (a;+7)(a: + 6)(a; + 5)j
_ 1 / 275 1 j
~4l(5a; + ll)(5a7 + 6)(5a; + l) (a; + 7)(a; + 6) (;r + 5)/"
_ 1 r 275 , 1 \
166. (a) Finite Differences, {^^).
2^^=^(0711) (^ + 2) . . . {x\Qn — l)
then —
= {cG^l){x + 2) . . . {x^m)x{x^l){x^2) . . . (x + m1)
= (0711) (07 + 2) . . . (07 + m — l){07 + W07}
=w(o7+l)(o7H2) . . . (x + m—l).
Again —
A%^.=7?i{(o7 + 2)(o7 + 3) . . . (o7 + w)(^+l)(^+2) . . . (o7 + ml)}
= 7?i(ml)(07f2)(07+3) . . . (07 + Wl).
Proceeding in this way, we have —
/^^Ux=m(m — l){m—2) . . . {m—n + l){x + n){xhn\l) . . . (o7 + m— 1)
until n=m, when ^^Ux=\fn,
when n>m ^^Ux = .
(b). Finite Differences (19).
167. (1) Finite Differences {l^), {20).
(2) Finite Differences (40).
r(l) 1^
Answer.— < {iym{m^l) . . . (m + w1)
1^ C^)
07 (07 + l) (0712) . . . (071^81^ — 1)
202 SOLUTIONS.
168. (a) Finite Differences {21) {ZO).
In formula (11) write n for w, for a, and 1 for h,
(b) Differencing, we have —
Wo= 89,685 691 9
2^1=88,994 700 9
«^2= 88,294 709
«^3= 87,585
then, by formula (11),
U9=Uo\ QAuo + 36 A%o
=89,6856,219324
= 83,142.
Answer.— 83,142.
169. Finite Differences (28) or (29).
Having demonstrated formula (10), write a for h and n for ma,
so that w=  .
a
170. Finite Differences (32), (33) and (34).
171. {a) Finite Differences (32), (33) and (34).
After demonstrating the formula, let A=l.
(5) Applying formula (a) Finite Differences (58), we have —
A^z^i = «^9 — 82^ 4 28^7 — 56effi f 70% — 56e^4 + 28«^3 — 82^2 I «^i
=0,
transposing and dividing by 70,
56 (t^4 4 %) — 28 {u^ + e^;) + 8 { u^ + e^) — {ui + Uq)
Q.F.D.
70
SOLUTIONS. 203
172. («) See Solution (171).
(6) The symbols ^x+2i ^x+\ and Aj; are not defined, and must
be understood to represent —
^Ux+2i ^^a;+i and ^Ux
Second differences being constant, third differences will equal zero.
Also the third difference of a function is the second difference of the
first difference of the function, Finite Differences (17) ; therefore,
applying formula (12),
A%^. = A*'( A^^a;) = ^«^a;+2 — 2 AWa?+ 1 + Az^a; = 0.
That is, Aa.+22Aa;+i + Aa;=0.
Q.E.D.
173. {ci) When m consecutive equidistant values of a function are
given, we can obtain (/« — 1) orders of differences. If any order of
differences not greater than {m—1) is constant, we can accurately
represent the function as a rational integral function of x, by means of
formula (10).
If, however, we obtain (m— 1) orders of differences, we may
assume the last order to be constant, provided it is relatively small
com^Dared with the function. Finite Differences (21). In this case
formula (10) will only furnish an approximation to the general form of
the function.
(6) Herew2=2, «^3=14, «^4=:40, %=86,
therefore Aw2=12, A%=il!, A%2=(>;
also
. ,,.^ ra;2)(a;3)^^ (a;2)(a;3)(a;4) ^^
= i^2+(^2)A«^2+ ^ ^ ^A22^2h ^ ^^ 1^ ^ ^A%2
\l 3
=2H12(a;2) + 7(a?2)(ar3) + (a?2)(a?3)(a:4)
= 12a? 22 + (a;2 5a? + 6) (a? + 3)
=a;32a72 + 3a:4.
Answer. — x^—2x'^\Zx—^,
204
SOLUTIONS.
174. (a) After differencing, we have —
Awo=— 139, A2e^o= — 65, A%o=4.
Applying formula (11),
^
= 2,844
139 2 ^^ , 6 ^
5 ^25 ''+125^
=2,844278 + 52 + 192
=2,849392 278
=2,821592.
3
Answer.— 2,821592.
(5) The method explained in Finite Differences (21) is best
adapted to the present case.
We may therefore proceed as follows —
X
Wx
A%
A%a;
A%j
1
1
2
6
2
3
2
6
6
3
5
13
33
8
20
12
4
18
24
5
38
6
71
62
30
7
133
92
36
8
225
128
42
9
10
353
523
170
The accuracy of these values will be tested if we obtain u\q by
means of formula (10),
thus
9 8 9 8 7
Wio=l + 9x 2+^x0+ ^^ x6
=523. ^^^
Answer.— 1, 3, . . . 71, 133, 225, 353, 523.
SOLUTIONS.
205
175. By problem,
+ 8x + K
+ &C.
+ Bx+^l + K+l+ ■ '  +^Un2
=nBx+{nl)Sl+{n2)^,+,+ . . . +8^+„_2
But the coefficient of 8^ is a recurring series, whose scale of relation
is (1^)2, Art. (398).
The sum of the series is therefore —
therefore
Ar— w8j
K
(^n_l)_^(^_l) (^_i)s
Q.JE.D.
or
176. (a) Finite DiJ'erences (53), (54), (56).
(i) Finite Differences (58).
(c) By Finite Differences (58), formula (a), we hav(
U2
6
4 xl94s877 194,772
6
:97,456.
Answer. — 97,456,
206 SOLUTIONS.
177. Let log350=e^o, log351=?^i, log352=?^2, log353=:?^3,
log354=W4, then, proceeding as in solution (176), we have —
Us= ^
_ 254900 + 152792 4 10181 24 + 254407
_ __ .
_ 2037231 1018124
~ 4
1019107
4
=254777.
Answer. — 2*54777.
178. Let log422=wo, log423=«^i, log424=?^2, log425=i^3
log421684=2^;^,, therefore ^= — 316. Then—
Auo= 0010279, A2mo=  0000024, ^^Uo=  0000001,
then by formula (10) —
on... 316x1316^,
= •6253125 316 x 0010279 316 x 658 x 0000024
= 6253125  0003248  0000005
=•6249872.
Answer,— 6249872.
179. Let Uq, U2, U4, Uq denote the given quantities, so that the
required value is Ur, ; then —
Awo= 9467971; A2wo= '2542232 ; A%o= '0665146
By formula (11) —
5^ 5.3_ 5.3.1_
=34371090+23669928 + 4766685 + 0207858
=63015561.
Answer.— 63015561.
[NoTE.^The correct value is 630094. The comparatively large
discrepancy is due to the neglect of differences above the third order.]
SOLUTIONS.
207
180. These values may be ascertained by the method explained
in Finite Differences (62) to (65), or by subdivision of intervals (66).
The amounts must, of course, be first expressed as decimals of £1, and
we have ^0=247916, Awo='4416, ^'^Uq='1^, A%o=0583, whence
Si^o='07913, 82z^o= '00413, 8%o= '00046, and the contributions for ages
20 to 25 inclusive are found to be 247916 ; 255830 ; 2*64156 ;
272942 ; 282235 : 292081.
Ansioer.
£2.
9a.
Id.
£2.
lis.
2d.
£2.
12s.
lOd.
£2.
145.
Id.
£2.
IQs.
6d.
£2.
18s.
&c.
5d.
181. Denote the given values by Uo:Q, 2lo:4, «^0:8; ^2:0j «^2:4,«^2:8;
Ui:o, U4:4, U4.Q, and proceed as explained in Finite Differences (74).
Then the required value will be denoted by ^3:1, and by formula (11).
1 3
32
=17176+ (1073)
3^
32
(105)
Similarly
and
= 17176 268 010
= 16898.
^2:1 = 16307,
?^4: 1 = 15685.
Now interpolating for 1^3 ..1 employing the values just found —
3 3
^3:1 = ^0:1+ o^^"^^^" Q^^'0:l
=16898887012
= 15999.
Answer, — 15999.
208 SOLUTIONS .
182. Awo=4473, A^Uo=01G9, A%o=0017.
By formula (10),
e..=9545 + a:(4473) ^^^^ (0169) + ^(^l)(^2) ^.^^^^^
= 15000,
that is, we have a cubic equation from which ic may be determined ;
thus —
•5455=^{4473+00850006}a;2{0085+*0009} + a73{0003}
= (•4552)07 (0094)072+ (0003)073.
Now 07 lies between 1 and 2, and is evidently very near to unity, it
may therefore be denoted by (1 + 8) where 8 is small, and therefore its
second and all higher powers may be neglected, whence we have —
•5455 = (^4552) (1 + 8)  (^0094) (1 + 28) + (0003) (1 + 38) ,
or
•0994= (4373)8
8= •2273. \
If we wish a closer approximation we may include 8^, and may
assume that 8^= (2273)8.
Ansiver.— 12273.
183. (a) Finite Differences (Qiyill).
(5) Denote the given values by Wo, z^i, W3, z^e? and let u^
denote log 282.
(20)(2l)(23)(26)= (2) (1) (1)(_4)= 8
(0l)(03)(06)(20) = (l)(3)(6) (2) = 36
(l0)(l3)(l6)(2l)= (1) (2)(5) (1) = 10
(30) (31) (36) (23)= (3) (2) (3)(l)= 18
(60)(6l)(63)(26)= (6) (5) (3) (4) = 360,
u^ Uq Ux Us Ue
therefore 8 ="36 + 10 + 18 " 360'
Multiplying up 45z^= — 10% + 367^i + 202^3 — Uq
=  244720 + 881532 + 49036024564
= 1102608,
whence 2^=24502.
Answer. — 24502,
SOLTJTIONS. 209
184. {a) Finite Differences (67)(69).
(6) Denote the given quantities by z^o, «^i5 ^4} «^5) and let u^
denote the required probability, then, as shown in Finite Differences (70),
(3_0)(3l)(34)(35)= (3) (2) (l)(2)= 12
(0_l)(04)(05)(30) = (l)(4)(5) (3) =60
(l_0)(l_4)(l5)(3l)= (1) (3)(4) (2) = 24
(4,_0)(4l)(45)(34)= (4) (3) (!)(!)= 12
(5_0)(5l)(54)(35)= (5) (4) (1) (2) = 40,
therefore 12 =  60 + 24 + r2io'
and 10«^3= —2uo+5ui + 10^4—82^5
=  196856 + 491675 + 980080293631,
whence «^3= 98127.
Answer.— 98127.
[Note the values will be found in Table VII. in A Short Collection
of Actuarial Tables].
185. («) Finite Differences (67)(6d).
{b) Denote the given quantities by Uq, Ui, u^, ^3, and let Ux
denote the term whose position is to be determined, then by problem
Ux= 233333.
By formula (22)—
_ (^l)(^2)(^3) (ic0){pc2){x2.)
^•^ ^\0l)(02)(03)'^^'(l0)(l2)(l^
(^0)(^_l)(^3) (.;0)(:.l)(;r2)
■^ ''(20)(2l)(23)'^ '(30)(3l)(32)
cc^—Qx'^+llcc—Q 0^35^2 + 6^
= Uo ^u, 2
"^ 2 +^^ "6
233333=230103+ (02171)^ (00055)a:2+ (00001)^:3.
210 SOLrTIONS.
Neglecting
x'^ and x^^ we have, as a first approximation —
•03230= 021710;,
or
0?= 1*4878 approx.
Assume
0:2= (14878)0?,
then
•03230= (02171)0: (00055) (l4878)o;
= 020890;
and
o;= 1^5462 approx.
That is,
233333 =log 215^462 approx.
True value,
=log 215444
Answer. — 215*462 approx.
186. If the series formed a geometrical progression with a common
ratio r, then by Finite Differences (7),
U.T L.U:,
= (rl)
In such a case it would be advisable to form a new series consisting
of the logarithms of the given series. The logarithm of the required
term could then be interpolated, and the antilogarithm would be the
required term.
187. Mnite Differences (80)(82).
188. (a) Finite Differences (78), (79).
W ,, ,, (47)(51).
189. {a) Finite Differences (87).
In formula (28) substitute o? for a, so that
. . • . ^(w— 1) .
Ux + Ux+i^ . . . +Ux+n^\=nux\ 12 — i^Ux
. n{n—l)(n—2) ^^
SOLUTIONS, 211
(b) If 2^a; = ^\
then «o=0, Awo=l, ^^Uo=l4, A%o=36, A%o=24.
Mnite Differences (48),
and
0. + i. + 2^+ . . . +«^=(. + l)0+ (^^1>1+ (^ + 1)^(^^) 14
li
(„+l>(«l)(«2)(«3)
+ ]5 ^*
o
__ 6^51 15^^^^ + 10^3—^
" 30
~ 30
Answer. ^+1)(2. + 1)(3.^ + 3.^1)
30
190. The square of the ni\i term is (3w — 2)2=97i2_l2^f 4, and
the series will be found by giving n all values from 1 to w in this
expression.
The sum of fche series is therefore—
:.{.. *fl'}
\4<n
Q
=zn{n^l){2n\l)—Qn(n^l){4<n
_ ^(6n23^ — 1)
~ 2 *
Answer. —
212 SOLUTIONS.
191. The mileage run in the aggregate is evidently the sum of the
series 2.3.4 + 3.4.5 + 4.5.6+ . . . +28.29.30, but—
that is —
o. . .N, ..X 28x29x30x31 1x2x3x4
2M^+l)(^+2)= ^
=203x9306
=188,784.
Answer.— 188,784 miles.
192. (a) Differencing, we have —
15 4
16 9 4
2 15 13
' ^\ ,1
By formula (28), ^^
n(n\) ^ n(nV)(n2) ^^
U0 + U1 + U2+ . . . +Uni=nuo\ ^—^ — Az^oH rs ^%
Z o
= {0n\15n^15n + 4in^12n^\Sn}
= i {4^3+37124
n(n + l)(4^—l)
6
n(n\l)(4!n—l)
Answer. — — ^^ h:
(h) See Solution (190).
:S«ar2=i7j(w + l)(27i+l).
Answer. ±±3p±}}
b
SOLUTIONS. 213
193. (a) Finite Differences (41) when Ux={ax\h).
1 —an
Ux'UxJtl'Ux+2 ' • ' 1^x+ni '^x'^x+i'i^x+2 • • • '^x+n
therefore, by integration
1 =s ^ +c,
^x'^x+l ' • • ^x+ni UxUx\\ . . . Ux[n
or, making the alterations suggested in Finite Differences (100)
or
—a{n—l)UxUx\]. . . . Ux\n2 ^x^x+i ' • • Ux+ni
=c ^
UxUx+i . . . Ux+n1 «(^ — l)^.i;«^a?+i • • • Ux+n^2
(b) The general term of the series
_1_ 1 1
1. 5. 9 ■*"5. 9. 13*^9. 13. 17 ^*
1
IS
then
(4^ + l)(4ar+5)(4a? + 9)'
(4a; + l)(4^ + 5)(4a?f9) 8(4a;+l)(4a?+5) '
therefore
Snl 1 _A 1
" (4a? +1) (407 +5) (4a: +9) 8.1.5 8. (4w + l)(47i + 5)
It is important to note that the relation
UxUx+i • ' ' Ux+n1 {Ux+nl — Ux}UxUx+i • . • Ux+n2
only holds good when Ux is in the form («a7 + &).
194. (a)—
^UxUx+iUx+2 • • • Ux+mi= (.^x+m—1^x)Ux+iUx+2 • • • Ux+mi
=amUx+iUx+2 • • . ^x+mli
therefore, by integration
UxUx+iUx+2 . . . l^x+mi = (^^^Ux+iUx+2 • . . Ux+mi + ^1
^xi^x'^x+i ' ' • 1^x+mi yr^ , rt
or — =2,UxUx+i . . . Ux+mi + G.
a{m\l)
UxlUxUx+i . . . Ux+m1 ^
Answer. 7 — r C.
a{m\l)
214 SOLUTIONS
(b) The general term of the series 1. 4. 7 + 4. 7. 10 f etc.
is (3a:+l)(3a?+4)(3a? + 7),
therefore —
S(3.+ l)(3. + 4)(3.+ 7)= (^m + m^+m^ + 7) ^^^
12
and
IJ!
2.1.4.7
12
12
^ws^t^er.— — {(3w2)(3;^ + l)(3/^+4)(3w+7) + 56}.
195. (a) See Solution (193).
(b) The series — + —— +  +&c., may be written
J. 4 d.5 4.D
3 4 5 „ n+2
2.3.4 3.4.5 ' 4.5.6 ' (w+l)(^ + 2)(^ + 3) *
The general term may also be expressed as —
1 1
+
(w + 2)(w + 3) (n + l)(n + 2){n + S)'
Also —
(n + 2){n + 3) ' (w+i)(^i + 2)(^ + 3)
=Ci ^ .
n + 2 2(^ + l)(w + 2)'
therefore —
(^i + 2)(w + 3) '(w + l)(?i + 2)(w + 3)
11 1
= o +
3 2.2.3 ^ + 3 2(w + 2)(^ + 3)
_ 5 2n\4i + l
■" 12 ■" 2(^ + 2)(;i + 3)
_ 50^+2)(yt+3) 12/^30
~ 12(^f2)(w+3)
— ^(5^ + 13)
~12(w + 2)(^ + 3)"
w(5w + 13)
Answer. —
12{n+2)(n+S)'
SOLTITIONS. 215
196. (1) See Solution (192).
Answer. — .
b
(2) The general term of the series is w(w + 2)(27j + 5),
or 2^(w + l)(w + 2) + 3^(^ + 2),
or 2w(7^ + l)(7^+2) + 3^0^ + 1) +3w,
whence :S;'^(^ + 2)(2w45)
= 5?27i(7i + l)(7i + 2)+S"3w(w + l)+S;'3w
^<^ + l)(^ + 2)(^h3) , ^n(n + l ){n + 2) . ^ n{n^l)
= ''^'^{(^ + 2)(^ + 3)+2(^+2)+3}
=»_(^{,.+7»+13}.
Answer. — ^r
2
..* 197. (1) The nth term oithe series is (3^2 + 5^—2) ; therefore the
'sum of n terms of thefteries is —
•T 2«3^2 + S«5/j:S?2
. • " ^^ n{n + l){2n+l) ^ ^njn^l) _^^^
• 6 2
2
=7i(w2_^4^J + l).
Answer. — w(n2 + 4^ + l).
(2) The series may be written —
32 42 52 (w + 2)2
1.2.3.4 2.3.4.5^3.4.5.6^ " * ' n{n^l){n\2){n + Zy
the nih. term is
w(^ + l)(7i + 2)(w + 3)
216 SOLUTIONS.
+
3 /
(n + 2)(n + S) (^ + i)(^ + 2)(^+3)
4
H
n(n{l){n + 2)(n + 3)
The sum of n terms of the series is, therefore,
(^ + 2)(^ + 3) ' ^'{n+l)(n + 2){n + S)
7i(w+l)(?i+2)(7J+3)
/i 1 1 3 f 1 _ 1 ^
""l3""/i + 3j "^212.3 (^+2)(w + 3)j
4 r_l 1 \
"^311.2.3 (n^l)(n + 2)(n+S)j
112 1 3 4
3 ' 4 ' 9 n + S 2(^H2)(m + 3) S{n + l)(n + 2){n + S)
29 6(^ + 1) 0^H2) +90^41) +8
36~ 6(w + l)(7^ + 2)(w + 3)
29 Qn^ + 27n + 29
36 ~ 6(^ + 1) (7i + 2)(w+3)'
29 6^2^27^+29
Answer. — — i —
36 6(w + l)(?ij2)(7i + 3)*
198. (1) The general term of the series is n{n\l)(n + S), or
n(n\l)(n\2)+n(n+l).
The sum of n terms of the series is, therefore,
%^n(n + 1) (n + 2) + 2X^ + 1)
_ ^(^ + l)(^ + 2)(^ + 3) 7i(M^l)(7i + 2)
= ^n{n+l)(n+2)(Sn+lS).
12
Answer.— —n(n + l){n + 2)(Sn+lS).
SOLUTIONS. 217
(2) The general term of the series is —. ~^ — — 
n^ + 2n + l 1 . 1
n{n + l)(n + 2)(n + S) (n + 2)(n\S) (n + l)(n + 2)(n\S)
1
The sum of n terms of the series is, therefore,
= /I _ J_U ^1^ ^ \
l3 w + 3j 212.3 (^+2)(7i+3)/
3\l.2.3 (7i4l)(w + 2)(w + 3)/
17 6(;t+l)(^ + 2) 43(^ + 1) + 2
36 6(7iHl)(w+2)(w + 3)
17 6^^2^2171+17
36 6(7j+l)(7i+2)(w+3)'
17 67i2+21w+17
Answer. —  —
36 6(n + l)(n\2)(n + S)'
(3) See Art. (386).
199. (1) The series may be written as
and is a recurring series whose Scale of Relation is (l—a^y. Art. (398).
Therefore the sum to infinity = ^^^£t^; . Art. (325), (326).
J 07(1 + ar)
Answer. — i e •
(107)3
(2)
+ TT^T^ + TTT^ + Tl^ + . . . «i inf.
8.18 10.21 12.24 14.27
= ^{4^ + 5^ + 0^7^+ •••"^^W
_1/ 5 6 ._J^. ^ + 4 ^
^ 6\4X6"^5.6.7^6.7.8 ^ '"^ (^+3)(;2+4)(7t+5) "^ * ' """^ ''''^7 *
218 SOLUTIONS.
The general term may be written —
I + I .
Therefore the sum of the series ad inf. is —
^ +^r
615 2 4.5/
3^
80
200. (1) Differencing the series, we have —
X Ux AUx A%x A^Ux
108
Answer. — — ,
80
1
1
12
36
2
13
4S
144
3
61
192
4
253
Here we notice that the first order of differences forms a Greometrical
Progression whose common ratio is 4, then we may assume that the
general term is «x4^~^ + c, where c is independent of a:. Art. (401).
To determine a and c we make a: equal to 1, and 2 successively,
then Ui= 1= a\c
iC2=lS=4ia + c,
whence a=4, c= —3.
The sum of n terms of the series is, therefore,
=:s;*4x4«'i:S?3
= l(4i^+'4i9n).
o
Answer. —  (4w + 1 _ 4 _ 9^) .
o
SOLUTIONS. 219
(2) —  + — + —  4 . . . +
1.2.3 ' 2.3.4 ' 3.4.5 " ^ n(n\l)(n\2)'
^'n(n + l)in + 2)~2\T:2~' (n + Ij(7i + 2)]
(n + l)(n + 2)2
4i(n + l)(n2)
4(7^ + 1) (^ + 2)
Answer. —
4(w+l)(^ + 2)
201. (1) Extracting the leading differences of 0^ from the table
of differences of 0, we can then employ formula (28), providing we
remember that n^ will become the (n + l)iih. term of the series, which
commences with 0^,
then 5?a;3=(7i+l)0+ ^^^<l+ ^^—^ Q
If. l£
^ ______ 6
_ n%n+iy
" 4
Answer. — \ — I .
( 2 I
C9\ n(n—l)—2nx\ic^ x
{n—x){n—\—x)~ (n—x)(n—\—o^'
The sum of n terms is, therefore,
2?l^5?7 77^—1 ^
{n—x){n—\—x)
220 SOLUTIONS.
1 1
{—£c n—x)
=«+!+ "^
n—x
n{n—x + l)
Answer, —
n—x
202. Expressing ^2, 2^3, «^4, &c., in terms of Ui and its leading
differences, we have —
UiX + {ui + ^Ui)(x^ + {ui + 2 A?^i + A%i)ip3 4 . . . a^ i;?/.
=Wi{a?+a:2 + ir3+ . . . adinf.} + j\ui{x^+2x^+Sx*+ . . . ad inf.}
+ ^^Ui{x^ + Sx^+ . . . adinf.}+&c.
CHAPTEE XI.
203. (1) See Art. (69). Also Finite DiJ'erences (87), and
Solution (201).
The nth. term of the series is n^=Un.
Then, by formula (28),
n(n — 1)
r
Differencing, we obtain —
2^1 = 1, Awi = 7, A%i = 12, A%i=6,
so that— ^
SOLUTIONS. 221
' ^ 2 ^ 3 ^ 4
= 7i— ^ +w(wl)(;i2)^
= {14?^10^?^3 + 2w213;^ + 10}
= 7{w3^2?i2+7^}
4 '^ ^
^f7i(n^\^Y
Answer. ' '
_ ( n{n + l) y
(2) Let 1—pcc—qa:^ denote the Scale of Kelation, Art. (322),
then 290 + 30^+20^=0 ^ ^'^^ ^ ^f^'~ ^
and — 30 + 20p3^=0,
whence S'=10, ^=3.
The sum to infinity is ^"*"^~^^~^^/ Art. (326).
32907
(15^) (1 + 207)'
329o7
Answer. —
(l507) (1+207)
[Note. — The form of this expression shows that the series is
convergent only when o? lies between  and — r .
5 2 J
204. By the Method of Differences, we find that the second order
of differences forms a geometrical series, whose common ratio is 2.
X Ux AUx A^Ux
1
4
1
2
2
5
3
4
3
8
7
8
4
15
15 .
16
5
30
31
6
Gl
222
SOLrTIONS.
"We may therefore proceed as in Art. (401), thus :
Let Uaf=a.2^^\ba:^c
then ui—a\i + c=4i
U2=2a + 2b\c=5
%=4« + 35 + c=8,
whence a=2, 6=— 1, c=3.
Therefore Ua;=2.2^^—a: + S
=2^—07+3,
and the nth term is 2**— w+3.
The sum of n terms, or 2>aj= (2»*+^2)  (^^±1)!^ _,. 3^
3:^2**+'
(^l)(7^4)
Answer. —
(^2»^_(^_3)
[Note. — The series is also a recurring series, whose Scale of Relation
is l—4<x + 5ic^—2iv^. It may therefore be dealt with as in Arts. (324)
to (328)].
205. Let l—pcc—qo!^ denote the Scale of Relation ; then —
14531;?6^=0
and 643145^31^=0,
whence ^=7, g'=— 12.
The generating function is —
611^ _ 6lla?
l7ar+ 12^72 ~ (l3ar)(l4a7)
_ 13 7_
""14^7 1—3^'
The nth term is, therefore, (13 x 4*^^—7 X 3'»0^^"^
SOLUTIONS. 223
Hence, by Art. (328) , the sum of n terms is —
13 X — 7x — ^.
1—4^7 1— 3a:
Answer.
' (13x4'*i7x3«0a?»i
l_4w^w 1 — 3»*a:'^
1— 4a: 1— 3a:
206. Let 1— ^a:— g'a:^ denote the Scale of Relation,
then 95_17^_ 2^=0
and 461 —95p—l7q=: 0,
whence ^=7, 2'=— 12.
The sum to infinity is  — —  —
^ 1— 7a: + 12a:2
_ 2 + 3a:
""(l3a:)(l4a:)
__ 11 9
""l4a: 1—307*
The nih. term of the series is the sum of the wth terms in the
expansion of each of these expressions, and is
(11 X 4»i— 9 X 3^1)^''"*
= (llx4»»— S^^+O^'^^
!(llx4'*^— 3**+')^***
2 + 307
(l3o:)(l4o7)*
207. (1) Let 1— jpo7— ^'072 denote the Scale of Relation,
so that —23+ 5p— q=0
and —77 + 23^ + 5^=0
whence i?=4, q=—S,
Therefore the Scale of Relation is 1— 4a7+3a:2.
224 SOLUTIONS.
Then S=l5a?23^2_77<y3_ _ .
— 4^S= — 4a: + 20a?2+92a:3+ . . .
3ar2S= Sx^15a^ . . .
.•: S(l4a7 + 3a:2) = l9a;
1— 9a?
and S= — — ^ , .^ r =The Generatins: Function.
(1— 3a:)(l— a?) °
Resolving this expression into its partial fractions —
l9a? _4_ _ 3
(l3ar)(la;)~"r^ l3a?'
and the general term, e.^., the (w + l)th term, of the series is, therefore,
(4—3x3^)0?'*
or (4— 3^+0^''
]^ 9^
Answer. — — — tt^tz r; (4i—S^+^)x^.
(1— 3a7)(l— a?)
(2) The ^th term of the series is evidently (n+l)(n+4i), or
^2+5^ + 4, and we may employ the formulas for the sums of the 1st
and 2nd powers of the natural numbers.
It will, however, be easier to employ Finite Differences, thus : —
X
«»
AUx
A%a;
10
8
2
1
18
10
2
28
Then, by formula (28),
^ , nn—1^ n.n—l.n—2^„
or, substituting numerical values,
= lQn\^n.n—l + n.n—l.n—2
o
o
= {^3+9^2+20^}
■»(n4
•4)(« +
3
5)
•
Answer, —
^(^+4)(;i45)
3
SOLUTIONS. 225
208. (1) The series is a recurring series with a Scale of Relation
(l^)3=l3:r + 3a;2^3^ Art. (398). Therefore, if S denotes the sum
to infinity, we have —
S=2 + 12^ + 30ar2f 56ar3+ . . .
— 3a:S=  6^—36^:2—90073— . . .
+ 3^28= 6ar2 + 36^3+ . . .
a?3S= —2x^— . . .
whence (1 — ^)3S=2 + 6a;,
r. 2 + 6^
and
(1^)3*
2 + 6a:
Answer. — — . ^ .
(lar)3
(2) This is a series of factorial terms, and can be solved
as shown in Solution (198). The sum to n terms being —
(3^1) (3y?H2)(37? + 5) (3^ + 8) + 1.2.5.8
12
_ (Snl)(Sn + 2)(Sn + 6)(Sn\H)\1.2.6.8
12
CHAPTER XII.
209. («) See Arts. (38), (51), (61), and {6Q, sect. 4).
It must be noted that a definition in order to be satisfactory must
be complete. It is therefore necessary to define the phrases common
difference and common ratio.
r
{b) SeeAris. (40), (55), and (57).
/ N 4 5 4 5 „ , ^
W 7  72 + 73 74 +&«•' «^ '^/
226 SOLUTIONS
11. , . A 5/1 1
23/1 1 , , . ,\
= y(^^ + ^3 + &c., ad.mf.j
1
23
7
7
1
1
72
23
7
23
7
48
l~'48*
Answer. — — .
40
210. (a) See Art (40).
(b) Let d^ denote the common difference of the completed
series, which will consist in 12 terms —
then —5= the 12th term
=8 + 11^,
or II '
and the required means are —
75 62 49 36 23 10 3 16 29 42
n' H' li' 11' li* n* 11' 11 ' 11 ' HT*
75 62 42
Answer. —
11' 11' 11
211. (a) See Arts. (47), (48) and (49).
(&) Let a denote the first term, and d the common difference
of the series —
then the ("2")*^ *^^"^ ~^ "*" (~2 ^j^'
also the wth „ =a{(m—l)d.
But if the yY'r^ ^®^'"^ ~^'
SOLUTIONS. 227
then, by giving n odd values, we see that if
^ = 1 the 1st term =1
^i=3 „ 2nd term =3
^=5 „ 3rd term =5
Therefore, if ^=(2??? — 1) the wth term =2w— 1,
that is «=1 and J=2.
Therefore, the sum of m terms = — {2+(w — 1)2}
Q.E.D.
212. By problem S = l+ (H^) + (l + 2^)H . . . ^{1 + nlh)
= '^(2+^^31^).
If n be odd, let it be denoted by 2wjf 1, where m is an integer,
then it is required to find the value of k such that
S + 3 (2m + l^^ o ,x «) 1
xv = I — 2 — (2 + 2w^) +3 I^T = an mteger = p suppose.
(2w + l,^ ^ ,, ^1 1 (2wil)(l + w^)+3
_ 2m + 4 + ^^(27w + l)
~ 2k ^
m + 2 m .^ . ,
+ (2^ + 1)=;?,
so that
k ' 2
w + 2
but
p'^(2m\l)
2(m + 2)
2p—m(2m+l)
n1
2
n1
~~2~
2(71 + 3)
2p^n
4ip—n{n—l)
228 SOLUTIONS.
Therefore, in order that S + 3 may be divisible by 2k for all odd
rt /„ _i_ Q\
values of w, Jc must be of the form ^^ —— , where v mav have
4ip—n{n—l) ^ ^
any integral value.
213. Let a denote the first term, and d the common difference, so
that the three quantities are —
«, a\d, a\2d,
,, a + d a\2d
then >
a a^d
so
 a + d a\2d
long as IS positive,
a a\d ^
i.e.^ so long as {a\dY—a{a\2d)
a{a\d)
d^
or so long as
a(a + d)
The numerator is always positive ; we are therefore only concerned
with the sign of the denominator, which must be positive.
'J'he necessary conditions are therefore as follows —
(1) If a and d be of the same sign they may have any value.
(2) If a and d have opposite signs, d must be numerically less
than a.
214. (a) This identity follows from the definition of an Arithmetic
Mean. Art. (43).
It may also be proved as follows : —
Let a denote the first term and d the common difference.
Then the (m + n) th term = « + (m + w — 1) ^
and the (m—n)th „ ^a^ {m—n—l)d.
Then the sum of the {m\n)i\i and (m—n) ih terms
=2a\(2m2)d
=2[a+(ml)d]
= twice the mth term.
SOLUTIONS. ^29
(b) Let S denote the sum of the series, then
S=l + ll + lllh ... to w terms
= 1 + (1 + 10) + (1 + 10 + 102)+ . . . +(1 + 10+ . . . 410^^0
_ 1021 1031 10^1
~ "^ 101 101 "^ • • • "^ 101 '
or 9S= (101) + (1021) + (1031)+ . . . +(10'^1)
= 10+102+103+ . . . +10^*;^
,^10»*1
= i{10''+'109«},
y
or S=^{W+'109n}.
81
10w+i_i09;i
Answer. — .
81
215. (a) See Articles (38), (51), (61), and (62).
2 12^
(b) If ah J bacj ci be in Geometric Progression, we have the
relation —
2_ 2
&ac=(«!c)6, Art. (53),
or b(>i^=ac
or b^ = {acY^,
and, since «, J, and c are positive integers, b must equal ac ; therefore —
and 5*"=(«c)«"c"
or 5a»c»=i(«c)&".
2^2
That is, if a, b, c be positive integers, and «&, bac^ cl be in Geo
2 12
metric Progression, then «6», Sa'c, c&« are also in Geometric Progression.
230 SOLUTIONS.
216. (a) If a, b, c, d are in Geometric Progression, then
6 _ <?_ d ___
a b c '
Then *4I = '^. Art (12).
That is, (b + cy=(c\d)(a + b),
or 2(b + cy=2{c + d){a + b),
and
(^a+by+2(b\cy + (c + dy=(a + byi2(c + d){a\b) + {c + dy
= {a\b + c + dy\
Q.E.I).
(b) This is a recurring series in which the Scale of Relation is
(107)2. Art. (398).
Let S denote the sum to n terms, so that —
(— 2ar)S= —2a;^'4<x^6af^— . . . —2n—liv^—2na;^+^
therefore —
(l2a7 + ar2)S= £c—n + l£c''+^ + nx^+^
' (1=^0^ —
_ x(l—x'^)—nx»+'{l—a:)
x(l—a;'')—nx''+'(l—a;)
Answer. ^^—y^
SOLUTIONS. 231
217. Tzr + —p= + p= + &c.
111
+ ~ /m + T. 7^. + . . . +
Ha/2 (1+^^2)2 (1+a/2):^ *" (l+^^2)'^
1
1
(1+ v'l)**
l+^2l ^ 1
H^^2
 ^ f (l+A/2)^l ^ I+a/2]
~l+^/2( (Ha/2)^ ' ^/2 )
_ ^2{ (1+^2)^^1}
~ 2(1+ V'2)^*
_ ^/2(a/21)^{(14 ^2)^1}
_ _
V2{l(v^iy*}
Answer.—  'v/2{l — (^/2— 1)»*} .
2
218. By problem a — = h ,
or 1 + r =  ,
That is, 5 is equal to the sum of the first two terms of the first series.
219. (a) See Art. (55).
(&) ,, ,, (57).
^3^ SOtiUTIOKS.
(c) Let a denote the first term, and d the common difference ;
then by problem a+(p — l)d= . . (1)
and a+(ql)d= (2)
(l)(2) (^_^)^=:^,
^ 1
or d= —.
M
From (1) a^^ = ,
1
or « = — .
M
The »^th term —a\{pq'X)d= — ^—
V9. M
=1.
The sum of pq terms =:^i — + 1 1
220. (a) Since a and 6 are the first two terms of a Harmonical
Progression,  and  are the first two terms in an Arithmetical
Progression, whose common difference is = — r .
b a ah
The wth term in the arithmetical progression is —
 + n—l —  =
a—b n—la—n—2b
a ah ah
Therefore the nth. term of the Harmonical Progression is
ah
w— la— 'W— 25
SOLUTIONS. 233
(b) If a, b, c, d are in harmonical progression, the 4th term
ah
3«25'
while
b= ^'''' . Art. (64).
Therefore
, 2a^c a^c
a\c '6a{a\c)—4^ac
2a^c
~Sa^—ac'
By problem,
d b
a c'
f.hnf i«
2a^c
2ac
a{Za^—ac) ~c(« + c)'
fW
c
1
Za?—ac a\c
or «cf c2=3«2— «c
(c«)(c + 3«)=0.
But, by hypothesis, {c — d) is not equal to zero. Therefore —
(c + 3«)=0.
221. («) ASfee Art. (65).
b\c
(b) By problem, a= 2 (1)
and h=^c (2)
squaring (2) b^=:ac
or c=— (3)
from (1) c=2ab ......*... (4)
(4) (3) 0=2«5
=2a^abb^
234 SOLUTIONS.
b— —
Substituting in (4) G=2a 1
a + b
_ 2ab
222. {a) See Art. (253).
(V) The wth term of the series is
Q.E.I).
a + n—lb
The sum o£ n terms of the series is —
111 1
the
a a + b a + 2b ' ' a + n—lb'
By Art. (253) i j + ^ + — ^ + . . . + 1 is
arithmetic mean of the quantities, and this is greater than their
harmonic meau. Art. (65).
Thatis, 1^4 J+ ... + L^]> L=,
n{a a\b a + n — lbl 2a + n—lb
11 \ 2n
+ y+ ... +
a a + b ' a + n—lb 2a\n—lb'
Q.E.D.
223. (1) Let a denote the first term and d the common difference,
then P=a+(^l)rf (1)
and q=:a\(q—l)d (2)
therefore F—Q,= (p—q)d,
, PQ
t.e.f d=i .
pq
Substituting in (1) ^=a+{pl) ^^ ,
.=p?:=^(pQ)=Q^:::^p^.
And the series is q ^^ P ^j, q^^ P ^^1, &c.
I p'q p—q) I p—q p—q)
SOLUTIONS. 235
(2) Let a denote the first term and r the common ratio,
then P=«/i'i (3)
and Q=ar3i (4)
P
therefore —r^rv^
and r
:pi
substituting in (3) P=«[ — j ,
pj
therefore
Pp3
1 1
1^ /73
iJg
And the series IS p^ » Vp^2 ' ^^•
(3) Let a denote the first term and d the common difference
of the Arithmetic Progression formed by the reciprocals of the
required Harmonic Progression —
then by problem P= ; — , (5)
and Q=_^^ (6)
therefore ( ^ ~ S') <^= ^ ~ K
P C2
~ PQ '
Q— P
so that dr=i
PQ(p?)'
substituting in (5) a=\ ^^^^
_ T(pl)q(q l)
 Fq(pq) ■
236
And the series is
SOLUTIONS.
&c.
Answer.
V p — q p — qJ \ p—q p—qj
rQ^h va rQ^2^ pq ^
PQ(p5)
PQ(i^?)
P(^_l)_Q(^_l)' p(^_2)_Q(^_2)
, &c.
224. {a) This is equivalent to inserting 5 arithmetic means between
 and — , thus: —
5 10
Let d denote the common difference;
1 1
Then
10 5
+ Qd,
or
D
__ j^
~~60'
Therefore the arithmetic series is
and the harmonic
1 11 10 9 8 7 1
5' 60' 60' 60' 60' 60' 10'
5, 5A, 6, 6, 7i, 8t, 10.
^wsw?er.— 5, 5^, 6, 6f, 7^, 8f, 10.
(5) If a, 5, and c be three positive quantities in harmonical
progression, then —
h is the harmonic mean of a and c,
~2~
and
, arithmetic
'v/ttc „ geometric
SOLUTIONS,
But
or
(a + cy>4iac,
or
a^^c^>2ac,
also
aob^,
or
2ac>2b^.
237
Much more then is a^\c^ > 2i^
Q.E.D.
225. {a) See Article (68), or the sum may be obtained by the
Method of Finite Differences, as in Solution (192).
(h) The arithmetical mean of the quantities is
{a\cc^h\x\ . . . jw + ^l, Art. (253),
n
or ^%aAx, Art. (521).
n
Again, the arithmetical mean of the squares of the quantities is —
1
11
{(« + .r)2+ (5 + ^)2+ . . . J^{n + xy]
or {^a'^^2tax + nxn.
n
Therefore the difference between the square of the arithmetical mean
of the quantities and the arithmetical mean of their squares is —
^« + a: i%a'^\2^ax\nx^\
In n J In n J
or
and is thus shown to be independent of x,
q.E.D,
238 soLrTioNs.
226. (a) See Art. (253).
then ar2— 2^7— 3=2y^2_j_2y^fy,
or —
(l_2y)ar22(l+y)a:(3+y)=0,
2(l + .y)d.^/4(l+y)^+4(3+ y)(l2^)
whence a; 2(l2y)
_ 2(l+y)zfc2A/48yy2
2(T=2^ •
Since a^ is real the expression 4i—Si/—i/'^f or (4+y)(l— y) is positive,
that is, the factors are either both positive of both negative, whence
y must lie between 1 and —4, and these are the greatest and least values
of the expression respectively.
Answer. — 1, and— 4.
227. Let P9 denote the population nine years ago ; then by
problem^
11,000=10 X P9 11 X 10,000,
^ 121,000
= 12,100.
Similarly, the population 8 years ago was —
^ 12,100 + 11x11,000
P3_— ^ .
=13,310.
The population in each successive year is therefore shown to form a
series, as follows :
10,000; 11,000; 12,100; 18,310.
11,000 12,100 13,310 11
And
10,000 11,000 12,100 10
That is the population increases in geometrical progression, the common
ratio being w) • '
Q.JEJ.D.
SOLUTIONS. 239
CHAPTER XIII.
228. (a) See Art. (220).
(b)
e1 _ {elf _ e22g + 1
e + l" e^—1 ~ e^—1
e— 2 + e»
?— «i
K^^2+r^H
K'M}
1 1
fl 1 ) (1 1 1 )
{2+i+}^{ri+r3+g+}
Q.RD.
229. (a) See Arts. (223) and (226).
,,. ,101 ^"^100
~100
therefore, giving a: the value — — ,
log.^=2{j^ + 3— 3+&e.}
99 1100^3(100)3
=2{01 + 000000333+ . . .}
= •020000667.
240 SOLUTIONS.
If we multiply this quantity by the modulus, Aits. (216), (225), we
shall obtain —
log 10^ =43429 X 020000667
= 008686179.
(•020000667
Answer. — \
•008686179.
230. {a) See Arts. (220) and (223).
(5) In the expansion of a^ , x is not limited.
„ ,, loge(l)ir), X must not be greater
than unity.
(c) See Arts. (224) and (225).
In Todhunter's Algebra we have the formula —
This formula is obtained by direct substitution of m for in the
formula given in Art. (226). Although not rapidly convergent, it is of
considerable importance when the logarithms of (^n\V) or (w — 1) cannot
be readily ascertained.
W log.3=2{? + ^QV&e.}
=2{5 + 04166 + 00625 + 00112 + 00022}
=2 X 54925
= 10985.
Answer. — 1099.
231. (r/) Seekx^. (223).
(ft) logo: log (^fl)log(a7l) = log
2 »v • / 2 ''^ ' 2 "iF^.x
SOLUTIONS. 241
1 + ^ x^
*If in the expansion of log——, Art. (226), we write — —  for
i — X x^ — 1
l\x 1
, and therefore ^r^ —  for or, we have —
1—x 2^2_i
=loga?log(a?+l)log(a:l).
* [Note. — In effecting this transformation, it must be remembered
that x^ is being employed as a new symbol, and xx^ must not be
written x^. This point will be rendered clearer if we first write
for , and afterwards replace n by xJ]
n^—1 1—x
If in the expression logo?— log (.r + I)— log (a;— 1) we write
(?2 + l) for X and multiply throughout by ( — 2), we have —
log (^ + 2) — 21og {n\\)\ log n,
but, by Art. (224),
log (. + 2)log (. + !)= ^^  — + —^3 &c.
and log(. + l)log(.)=^ + — ^— + — ^3+&c.
therefore —
log(. + 2)21og(. + l)+log.= 2{2^^— 4^
232. loge(l + ar)=a7 +1 ~I "^'^^•'
writing  for x.
subtracting —
H loge(l + ^)— logi^^j=loge^
242 soLrTioNS.
It must be noted that this expansion is only true when x is unity ;
for if X is less than unity  must be greater than unity, and the
X
expansion of loge( 1+ j does not hold.
A valid expansion for loggo? is that given in Solution (230).
233. {a) See Art. (224).
/y»2 ^3 /y»4
log.(l + l) = (lg + gg+..
= 0^.3^+ «
Also we may write loge(l + ^) = ^ ~ (2 "" 3 ) ~ U ~ 5) * ' •
12^4^ ...... (2)
Hence, adding the equations (1) and (2), we have —
=loge4.
234. {a) loge(la7 4^2)=iog^ ^
Q.KD.
l\x
=log,(lfa:3)log,(l + ^)
/ , ip6 \ f X^ X^ \ '
= r2+)(^2+3 ■•■) ;
a:2 2x^ x' x^ 2.r'i
Answer. x+  +  + j  .
SOLUTIONS. 243
= 10(logaOlog,8) + 3(log,— log.^j, Art. (224).
= 10(log,10log,8)f3(loga28loga25)
;=101oge5 + 10 log^230 loge2 + 21 loge29 log^S
=loge5 + loge2
=iogao.
q.B.D.
235. {a) See Art. (223), noting that the positions of x and y have
been reversed.
(&)
l+ax 1— cm;
logJ(l + «.r) ^« (la^) 2« H^^I«g^(^+^'^)+^l«g^(l^'^)
l + ax (
~2^
'( a?x^ a^x^  ^ 1 — axf a^x^ a^x^ . )
= l22:2 + 2:3*'= + It 2:2 + 2:3 *"l
f X ax^ r?2^3 r «^2 ^2_^3 ^3^ ^
■'"i~2~272~2:3j \ T~2:2~2:3~ ""•/
= <^^^J) + «^<2^2T2^ + 2^)+*'=
.2 /y3/»«4
ax^ a'^x*
= T + 3^+*<=
^ Q.^.JD.
244 SOLUTIONS.
236. {a) See Articles (223) and (224).
{l)
=log,(lar)+log,(lf2^)
/ x^ x^ x^ ^^ „ \
5 7 17
2 3 4
(c) The rth term will thus be
Q.E.D.
r r
Anstoer.— — f( — iy+^2^ — l\
237.—
whence
2«^h2'^i = 10.
2^(1 + 20 = 10
2^ = 10 x
o
or log 2 = log 10 + log 2 — log 3
= llog3 + log2
log 2
5228787
~ 30103 "^
= 273697 approx.
Answer, —
273697 . .
SOLUTIONS. 245
238. (a) iS'ee Arts. (199), (206), (216) and (217).
(b)
Since log 49= 6901960,
log 49=l6901960=log 72=2 log 7
log 7= 8450980 (1)
Again, log28=log7+log4
14471580= 8450980 + 2 log 2
2 log 2= 6020600
log 2= 3010300 (2)
Given 2^=35
a; log2=log35
=log5+log7
log5 + log7 logio— log2+log7
x=
log 2 log 2
1 + '8450980 ^
From (1) and (2) = ,^^^— 1
18450980
•30103
a?=51292.
Answer.— 51292.
246
SOLUTIONS.
CHAPTER XIV.
In the following examples, the tables of Logarithms and Anti
Logarithms published in the Short Collection of Actuarial Tables have
been employed.
239. (a) /S'ee Arts. (209), (210), (211).
(8345)3 X ^7^89
(*)
log
(233)^fv^4696
=3 log (8345) +i log(789)^ log233i log4696
=3 X 92145 f i X 8971 fi x
23674  X 267174
]
=276435 4 •4485533820 + 89058
=4^10348 33820
=376528
= log 5,8246.
Answer. — 5,824*6.
240. (a) See Art. (206).
(b) After the first sale he has S9^ gallons of wine, he then
adds halfagallon of water and obtains a mixture, each gallon of which
contains — wine.
80
After the second sale he has 39^ gallons of mixture, he again adds
halfagallon of water and obtains a new mixture, each gallon of which
79
'79\2
contains ^ of the first mixture, or ( — j of
oO \.80/
wine.
Proceeding in this way, it is seen that after the nth. sale and addition
of water, he will have a mixture, each gallon of which will contain
''X of
.8oy
wine.
SOLUTIONS. 247
Let ^ denote the exact number of times this process must be
repeated, so that the mixture consists of wine and water in equal parts —
79
or ^ log— = log 2,
•3010
whence ^=
190311'8976
•3010
•0055
= 54^.
That is, he may repeat this process 54 times before the amount of
wine in the mixture is less than half of the whole.
Answer. — o4.
241. The theory of proportional parts depends upon the assumption
that the first differences of logarithms to the base 10 are constant. We
have shown in Finite Differences (25) that if the number, the logarithm
of which is to be obtained, be equal to or greater than 10,000, the
numerical value of the second difference will not exceed 000,000,004,343,
therefore the use of first differences will give accurate results in the case
of sevenfigure logarithms for numbers greater than 10,000.
242. The assumed value of the quantity is 1201, and the error in
the common logarithm of this assumed value is —
Iogiol201logiol2,
or logiol,201logiol,200, Art. (210).
But log,o(^+l)log,o^=2/x{^^ + ^^^3+&c.} Ait. (226).
248: SoLUTtoKs.
If, therefore, w= 1,200, we see that the error will be denoted by —
•86859
= 0003618.
That is, the common logarithm of the assumed value will be
•0003618 in excess.
Answer. h '0003618.
243. (a) See Alt. (199).
(b) See Arts. (216) and (217).
^ /(7839)^r^i43^
C^; °^Vl235f (00037)^
= ilog{ (7839)4 X (1435)^ X (l235)ix (00037)*}
= ^ {4 X (189425) ~ ^ (2^15680)  (09165) + ~ (?5682) }
= ^{75770107840917l8333 + 1894}
= i {7766425034}
= i X 52630
5
= 10526
=log 11288.
A^ifswer.— 1129.
15
244. At the end of the first year, — of the trees will be standing ;
15 15
at the end of the second year, f^ ^ r^ of the original number of trees
— 1 of the
original number of trees will be standing.
SOLUTIONS.
249
Suppose that onethird of the original number will be standing at
the expiration of x years, then
/15Y_ 1
viey ""3
or a?{logl5— logl6} = — log3
or a:{log3 + loglOlog2— 41og2} = — logS
or
X
log3
51og2log3— loglO
477121 3
15051500 47712131
4771213
0280287
= 1702.
That is, the number of trees will be reduced to onethird of the
original number early in the 18th year.
Answer. — 18 years.
245.
log 3,479 log 3,478 = 0001248= A
By Proportional Parts, log 34,782=log 3,478 +  A
o
=45413546
and
log (34,782)^= 9082709.
Also
289= (17)2
therefore
(289)S=(17)t
and
log (289)1=1 X 12304489
o
= 16405985
But log(34,782)i= 9082709,
therefore log (34,782)i + log (289)§=25488694
=log 35389.
Answer.^ 353 89,
250 SOLUTIONS.
CHAPTER XV.
246. (1) See Art. (224).
(2) Let i denote the rate of interest, per unit, per annum,
then 1 will amount to (l + e)** in n years. Art. (232).
Let X denote the number of years in which a sum of money will
double itself, at compound interest, so that —
or a?loge(He)=loge 2
= •69314718. . .
( P ^3 ^
or x\ *— 2 + o &c. [ = 69314718 . . .
•69314718 ...
.2+3&C.
By actual division we see that
:2+3&c.
1 1 »■ „
therefore
.= C693...){,+2B+M
•gqo
= r^ + 346 . . . i(057 . . . ) +&C.
Now i is very small, therefore the terms involving the first and all
higher powers of i may be disregarded, and
•693 ... ^^ . , ,
ir= . V '35 approximately,
69
or, replacing 100 x t by r, x= h "35 approximately.
Q.E.D.
I
SOLUTIONS. 251
247. (a) Let V denote the present value of £P due at the end of
n years, that is, V invested at r per unit, per annum, must amount to
P at the end of n years, or
P
therefore V =
(1 + ^)^
If interest be convertible halfyearly, we may consider V to be
invested for 2n periods at  per period, whence
(J) v=
Answer.— F{l\r)\ vfl\ ^
10,000
■2»
(l04))2o
logV=420 1og(104)
=420 X 0170333
=4 340666
=3659334
=log 45638 + (^gxl)
=log 4563878
therefore V=4563878 ;
= £4,563. 17s. 7d.
Answer.— £4,563. 17s. 7d.
248. (a) See Solution (247).
(b) In 12 years' time the value of the estate will be —
^=10,000, Art. (240).
04
Let V denote the present value of this amount, then —
10,000
"~(1 •04)12'
252
But
therefore
Also
SOLUTIONS.
104
104 23x13
therefore
100 100 '
log 104=3 log 2 + log 13— log 100
= 90309 + 111394342
= •0170334.
log V=412 log 104
=4 2044008
=37956
= log 6,246;
V= 6,246.
Answer.— £6,246.
249. See note to Art. (231).
Three days' grace must be added to the 57 days. Therefore banker's
discount —
fin
=34532916 X —  x 04
ODD
1381316 X
165758
12
73
73
=22706
= £2. 5s. 5d.
Answer. — £2. 5s. 5d.
250. Let S denote the sum ; then Sni will denote the simple
interest on the sum.
One half of the harmonic mean between S and Sni is —
S'^ni Sni
S(l + ni) l + ni
=True simple discount. Art. (231).
263
ANSWERS,
CHAPTER I.
1. y4 + 2y2a;2_y2^2_.^4 + ^2^2, 2. 1,024.
4. (1)^^; (2)^3. 6. 2a:^4>x^\al,
occ + 1
7. (a) (x^a:^ + l){x* + x''+l)(a:^{l), iv^x^ + l, (b) 1.
• ^"^ (a:a){a—h){ac) "^ {a^b){ba)(bcy (a:c)(ca){cb)
15. 1, , , , &c. 16. ^^.
CHAPTER 11.
17. (1) ^=4; (2) 07=3 or 2, y=2 or 3.
18. ar=^(l±V'5) or±«, y=^(l=F^5) or =F«5, 2?=0 or —a.
19. (1) ^=; (2) a?=4or2.
20. (1) x=.abc, ^=ab\ac + bc, z=za\b\c\ (2) ar=4.
11 5
21. (1) x=za + b,y=b', (2) ^=y, ^^ or 4.
22. (1) ^=\ or 7; (2) ^=6, 8, or (7± a/39) ;
(3) a:=±3 or±2, y==F2 or =f3.
264 ANSWERS.
75 15
23. (1) ..= , y=^; (2) .r=±(5±«).
24. (1) a?=4; (2) ^=7 or — 5, y=5 or — 7.
25. (1) ^^=^±3y' ^^^ ^=^ ""' ^' ^=^ ^^' ^•
^^ ,_, 7 _7±v'l3±12^/l .^. 1±a/5
26. (1) ^^="2 o^* 2 ' ^^ "^"^ 2 •
27. (1)^=1; (2) a7=±2 or d=l, y=±l or ±2.
28. (a) .r=(2+ ^3) or 2(2+ ^^3) ; (5) a^=2, i, ^, or ^.
29. (i) ^=2, y=4 or 5, z=o or 4 ; (ii) .r=3, 22, 41, &c., ^=4, 27,
50, &c. 30. (h) 07=5, 9, &c., y = l, 4, &c.
CHAPTER III.
32. (h) The value of the expression must lie between 4 and  .
4
CHAPTER IV.
39. (b) £2,915. 19s. lOd. 40. 1703355.
41. 33580246913. 42. £487894, £33. 13s. 10^^.
43. £1. 6s. Od. in excess. 44. £21.
45. 915064 inches. 46. 28691948 lbs. troy approx.
47. £104. 10s. 0^.
CHAPTER Y.
48. ^/=6.
49. The value of gold and copper varies as n^.q: 240 m^. p.
50. 4 gallons from 1st vessel, 3 gallons from 2nd vessel.
52. (b) 10 or 2. 53. 29 % and 71 %. 54. 30 miles per hour.
55. £43. 17s. 2id. 56. 325 inches.
CHAPTER VI.
57. 720 days hence ; 3,600 days. 58. £40. 59. 79 days.
a +a ( V aa! J ) a Ha ( V aa! J )
61. He gains £5. 15s. 4\d. per cent. 62. 10^. per lb.
ANSWERS. 255
63. £40. 64. Is. Sd. 65. £1,249. 12s. 0^.
66. £476. 85. Sd. 67. 133i per cent.
68. Gross income in first year is £2,320.
69. The amounts of 5 per cent, and 6 per cent, stock were in the
proportion of 7 to 5. 70. £1,875.
71. (a) 7f I seconds ; (h) 18iif seconds ; (<?) 30f seconds.
73. 3 hours 23 minutes. 74. 2^ lbs. 75. 38 and 42.
77. 33 miles. 78. 55 miles. 80. 1,024.
2, 4, 6, 8.
81
■ {.; J: _:; J;
83. 142,857 and 285,714. 84. 9, 12, 15, 36.
85. 25^ yards. 86. £2,010. 87. ^, ^, ^,
88. £1,258. 6s. Sd. 89. A is 24, B is 16.
90. (b) Each person may have either 1 Halfcrown, 5 Florins, and
14 Sixpences; or 4 HalfCrowns, 1 Florin, and 15 Sixpences.
91. (b) 99.
CHAPTER VII.
93. , '^, ^ — ; 720. 94. (h) 1,414. 95. 19,958,400.
\r—p w— r ' '
96. (h) 1,081,080. 97. 100,800. 98. 1,058,400.
99. (*)mooo. 100. "MfrlMrlKEzrW^).
16
101. 78,125. 102. (b) 252. 103. (b) 27,720.
105. (b) 144,444,480. 106. (b) 126.
CHAPTER VIII.
112. (a) l + 2.'r + 3.r2 + 4.T3, .r<l; l + .v'^\~, w<l.
11 X 3^+^5 X 2'' X 3^+^ — 19 X 2^
n. m,*l4Q\l(S)\^.
256 ANSWEES.
117 1 2w + l
116. W nC,a«r.r, (J) L^ . _ . 3., _L.^___^(_1)„..
lie. 2«3{3;i2_^5w+8}.
119. W^i^a' W (ir{^V*^C2 + ^Ci^'+'C2 + &c.}
120.
2w
2(1^)2
CHAPTEK IX.
123.
, ^ 552 ,,, 22 , ^ 3,151
(") 3,725' ^^) 3,725' ^'^ 3,725*
125. 11 to 19.
126.
1. 1^7. g. 128. (.)^^;
(*) i4
129. 119:1. 130. ^. 131. 5 or 6.
5
1 13 6 .o. . X 35 ,^^ 127
4' 5i' 25 ^^^ (^) 663' ^'^ 663
39 13 ((39)2 (39)2
^^^ (^) "4i52"' ^^) ij26j52' ^^^ J26l52
Q4 1 1 Q
139. (2)7C.x. 140. (2). 142. ^^,
216 180 150 125 ,^^ (2)«>[100
143. — T, 77^, ;T;rT, t;^ 144.
671' 671' 671' 671' ' {^y^{\^^'
1« «34' (*) 81 "'• 3 ^*« 9
151. ^^ £12. 152. £13. 17s. 9H 153. £5, £11.
243
154. 6s. lOi^. and 13s. li^. respectively. 157. {h) — .
159. {a) i; (5) '5002559. 160. g^.
ANSWERS. 257
CHAPTER X.
1 / 275 1_ I
• 4l(5a7 + ll)(5a7 + 6)(5^ + l)"^(a?47)(a; + 6)(^ + 5) j*
167. (1) \n , (2) ^^^_^^^^_^2),..~(^ + m + nl)
168. (h) 83,142. 173. (b) ^3_2^2^3^_4.
174. («) 2,821592 ; (5) 1, 3 . . . 71, 133, 225, 353, 523.
176. (c) 97,456. 177. 254777.
178. 6249872. 179. 63015561.
180. £2. 9s. 7d., £2. lis. 2d., £2. 12s. lOd., £2. 14s. 7d., £2. 16s. Qd.,
£2. 18s. 5i., &c. 181. 15999. 182. 12273.
183. (h) 24502. 184. (5) 98127.
185. (5) 215462 approx. 189. (^) <^il)(2.+^l)(3.^+3.l)
190. ^ ^ ^ . 191. 188,784 miles.
192. («) <^ + l)(4^ l) . (J) <^ + l)(2^ + l).
193. ^ ^
40 8(4;e + l)(4;2 + 5)*
194 r^") ''^x\UxUx+i . . . e^a;+w i _p
«(w + l) '
W T^{(3^2)(3;i4l)(3^+4)(3^ + 7) + 56}.
195. (5)
12
^(5;^ + 13 )
12(7^ + 2)(w^3)*
196. (1) K^ + l)(2^^ + l) . ^2) ^(^ + 1)(^H 77^ + 13)
197. (l)»(«^ + 4„ + l); (2) i 6«H27»+29
36 6(« + l)(« + 2)(«+3)"
"»■ « i(K.«)(..+«), m S3^ti^iL_^
258 ANSWERS.
200. (1) l(4^+l_49.^); (2) <^ + ^)
201. (1) {^f; (2) ^^.
202 .,_+A.,^3^,+A%.^^— ^3+&c.
CHAPTER XI.
203
(1) {"^2— 1 = (2)
204. 2»— (w3),
(l5a;)(l + 2a7)
2^+2 (w1) (714)
205. (13 X 4^17 x3^0^^^\ 13 X  — , 7 x VA—
1 — 4a; 1—3^7
206. (11x4*^13^+0^*^' ^^
(1 3a;) (14^)
on« n^ 2 + 6:>; (.S«l)(3« + 2)(3« + 5)(3«+8) + 1.2.5.8
CHAPTER XII.
. X 23 „,„ „^ 75 62 42
209. (.) ^. 210. W rVU^
217. J^/2{l(•^/2l)»}.
223. (1) {qP^p^^\ i(iP=?pi=?\,&,., {2)\p^y\
^ P1 Pi> ^ Pi Pl> ^ '(Pffi) '
f^P^yh PQ(yg) PQ(i>g) „
224. («) 5, SyV, 6, 6f, 74, 84, 10. 226. (J) 1, and 4.
ANSWERS. 259
CHAPTER XIII.
229. (5) 020000667, 008686179. 230. {d) 1099.
^^^ , . . ^2 2^3 ^ ^5 2^6
^ ^ 2^3^4 5 6
236. (c) ^' r(l)^+i2^'l\ 237. 273697 . . .
238. (h) a:=51292.
CHAPTER XIV.
239. (5) 5,8246. 240. {h) 54. 242. '0003618 in excess.
243. (c) 1129. 244. 18 years. 245. 35389.
CHAPTER XV.
I+2J ' ^^^ £4,563. m. 7t^.
248. (h) £6,246. 249. £2. 5s. 5rf.
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