120 
TUE APPLICATION OP ALGEBRA 
s 1 — 2r (by the last problem), and we shall have r — i 
— b, and s z — 2r — a, whence, by adding the double 
of the former equation to the latter, s 2 — 2.9 — a + 2b; 
and consequently i — \/a + 2b f I f 1. From which 
r (— b f .9) is likewise known; and from thence the 
numbers themselves. 
problem lxx 
The sum (aj, ami the sum of the squares (b ) of four 
numbers, in geometrical progression, being given; to find 
the numbers. 
If .r and y be taken to denote the two middle numbers, 
the two extreme ones, by the nature of progressionais, 
will be truly represented by and 
Put the sum of the two means zr .9, and their rect 
angle r r; so shall the sum of the two extremes 
^ be — a — s, and their rectangle also r: r 
(by the nature of the question). But (by Problem 68) 
the sum of the squares of any two numbers whose 
sum is s, and rectangle r, will be — ss — 2r; and (for 
the very same reason) the sum of the squares of our other 
two numbers (whose sum is a — s, and rectangle r) will 
be r a—i) 1 — 2r. Therefore, by adding these aggre 
gates of the squares of the means and extremes together, 
we get this equation, viz. s z + a — a]* — 4r — b. 
Moreover, from the equation, — 4- — — a — s, 
y X 
we get x 3 4- y 3 — xy / a — s = r x af—s: bnt (by the 
same Prob.just now quoted) x i Py 1 —s 3 —3sr; therefore 
S' 3 
3sr — ar —sr, or r — ——-—; which value be- 
2i 4- a 
.3 
ing substituted for r, in the preceding equation, we 
have s* 4- a—¿j 1 — b. This, solved, gives 
1 2.9 f a 0