TO THE RESOLUTION OF PROBLEMS, 
127 
I 
•U » 
mn of : à 
№ wn ones, 
i » be an odd 
'-LrM 
so s + s 3 — 3s — c x 1 + s — 2, or s 1 — 2s — c X 
s z — i, by case 2. Lastly, if n — 9, then m — 4, 
and therefore l p s 1 —2 -W 4 —4s*-p 2 — cxs + s 3 — 3s 9 
or s 4 — 3s 7, -p l zz c x s 3 —- 2s, case 1. 
PROBLEM LXXIV. 
Having given the sum f aj, and the sum of the cubes 
(b J, of any number of terms in geometrical progression ; 
to determine the progression. 
By retaining the notation in the last problem, and 
proceeding in the same manner, we here have 
a — x + xz P xz 1 
. + x 
>n—1 — 
-, and 
ic. 
we sn 
m oAdnum- 
the fcCOftl 
, have 
c into i + 
<o tar conti* 
best oi them 
ber of terms. 
equation Je* 
if. 
. j r or 
-jj» 3; 
b — x 3 -p x 3 z 3 P x 3 z 6 . . . -\-X 3 Z 5n — 3 — 
n3 „371 
i&f 
Theorem 9. Sect. 10). 
Divide the last of these equations by the former, so shall 
b z z — l x z 3n — 1 % z 2n ■ + z n -p 1 
—= x X --= -t= - = x 7 x - -— be- 
a ~3 j y rpi j 2 -f- 2 -f- 1 
cause 
— 1 
— __ s 2 —J— z -p* 1, * and 
3n 
1 
2 1 
-p l). Let this equation, and the square of the first 
~2 n 
a 2 — x z x % ——Η 1 , be now multiplied, cross- 
z — 2z "t 1 
wise, in order to exterminate x ; whence will be had 
s 2ft -P z 11 + 1 , 
2 . - " WIIICII3 
2 -+- 2 + 1 J 
b Z 2n — 22” + 1 a ^ Z* 
— X r = a X 
a z — 22+I 
the numerators being divided by z n , and the denomi 
nators by z, will stand thus, 
2 + 
z n + 1 + 
b x 
a* X 
Put (as 
2 ■+■ 
z -p 1 -p