124 
THE APPLICATION OF ALGEBRA 
Let the first term be denoted by x, the common ratio 
by 2, and the given number of terms by n: then, by 
the conditions of the problem, we shall have 
x 4- xz 4- xz 1 + xz 3 4- xz* ... 4- xz n — 1 — a. 
x r + a?V + xV 4- + x 2 z* ... 4- x'z 2 "- 2 = b. 
Multiply the first equation by l — z, and the second 
by 1 — z z \ so shall 
x — xz n — a x 1 — z, and 
x z —x z z 2n — b x i — z\ 
Divide the latter of these by the former ; whence will 
be had x 4- xz n = x l + z: let this equation and 
the first be now multiplied cross-wise, into each other, 
in order to exterminate x ; so shall a x 1 4- z* = 
— x I f ; x 1 4- s 4- z* 4- z 3 ... z"—K 
a 
If n be an even number, put 2m =r n; then our last 
equation, when multiplication by l 4 z is actually 
made, will stand thus,-T- x 1 4- z 2m z l 4 2; 4 2; 1 
.... 4- 2z 2m 2 4- 2z 2 ”'- 1 4- 22«; which, divided 
by z m , becomes-,-- x —— f z m — — -4 ? j_ 
b — m ' jn—l X- 
jn—2 
+ -v + 
b 2 + 2z 4- 
O r X 
....42 2 
m—2 
4- 22«— 1 4- z m . Let s be now put ( — —— 4- ~) — 
the sum of the halves of the two terms of the series 
adjacent to (2) the middle one; then, the rectangle of 
these quantities being l, the sum of their squares (or 
halt the sum of the two terms of the series next to 
those) will be = s 1 — 2 {by Problem 68); and the sum 
( ~r + ~ 3 ) °f half the two next terms to these last — 
s l — 3s, &c. &c.