TO THE RESOLUTION OF PROBLEMS. 
123 
that is, 6ax 2 — За\т + 
^ C2 n J) 
4 гг c ; therefore x 2, —. 
4 . 
ax 
~Y~ 
c 
6 a 
Jl 
8 
; and consequently x — ~ . 
v 6a 
known. 
——f , whence, z, r>,and ?/, are likewise 
8 48 
The same otherwise. 
Let the sum of the two means rr s, and their rect 
angle — r\ so shall the sum of the two extremes — a 
.— s, and their rectangle also r, {by the question) : 
from whence, and Proh. 68, it is evident, that the sum 
of the squares of the means will be =. s' 1 — 2r, and 
the sum of the squares of the extremes rr a — s\ z 
— 2r ; also, that the sum of the cubes of the means 
will be = s' — Злу, and that of the extremes =r a — sf 
— 3r x a~—r s: by means whereof, and the condi 
tions of the problem, we have given the two following- 
equations, 
viz. s 2 + a — s] z — 4r = b,or2s 2 —2as—4r=b— aa; 
and s 3 4- —3га — с, огзas*—3a z s—3ar~c—rt 3 : 
divide the former by 2, and the latter by 3a, and then 
subtract the one from the other, so shall r — —■ 
о 2 
+ - c L whence the value of s ( — — 
3a 2 
v”- 
— c lfL _}_ 2r 4- by the first equation) is also 
2 4 
CL 
given, being (when substitution is made) — 
■v/— - 
Y 12 
+ 
2C 
3a ’ 
PROBLEM LXXIII. 
Having given the sum (a), and the sum ofi the squares 
fbJ, of any number of quantities in geometrical pro• 
gression; to determine the progression.