Question

The su three numbers in g.p. is 21 and the sum of their squares is 189 find the numbers.

Answer 1

Sum=21=A + Ar + Ar^2

sumsquares: 189=A^2+(Ar)^2 + (Ar)^4

21=A(1+r+r^2)
189=A^2(1+r^2+r^4)

squaring the first equation:
21^2=A^2 (1+r+r^2)^2
441=A^2 ( (1 +2r+3r^2+2r^3+r^4)
441=A^2 (1+r^2+r^4)+A^2(2r+2r^2+2r^3)
441=189+A^2*2r(1+r+r^2)
441=189+A^2*2r(21/A)
441=189+42A
solve for A. Now go back to the first equation and solve for r.

Answer 2

A(1+r+r^2) = 21
A(r^3-1)/(r-1) = 21
A^2 (r^3-1)^2/(r-1)^2 = 441

A^2 (1+r^2+r^4) = 189
A^2 (r^6-1)/(r^2-1) = 441

Now divide and you get to cancel a lot of factors, winding up with

(r^2+r+1)/(r^2-r+1) = 441/189

cross-multiply and clean things up, and you end with

2r^2 - 5r + 2 = 0
(2r-1)(r-2) = 0
r = 2 or 1/2

A(r^3-1)/(r-1) = 21
A(7) = 21
A = 3

or

A(-7/8)/(-1/2) = 21
A(7/4) = 21
A = 12

check:

3+6+12=21
9+36+144=189

12+6+3=21
144+36+9=189