Q.:A geometric series has first term a and common ratio r. Its sum to infinity is 12. The sum to infinity of the squares of the terms of this geometric series is 48. Find the values of a and r.

Ans.:From text book: a = 6, r = 1/ 2

S¥ = a/ 1 - r

Eq.:

S¥ = a/ (1 - r) = 12

Attempt:

a = 12 ( 1 - r)

a = 12 - 12r

S¥ = a/ 1 - r

a^2/ (1 - (ar^2/ a^2)) = 48

(12 - 12r)/ (1 - ((r(12 - 12r))^2/ (12 - 12r)^2)) = 48

(12 - 12r)/ (1 - (12r - 12r^2)^2/ (144 - 288r + 144r^2)) = 48

(12 - 12r)/ (1 - (144r^2 - 288r^3 + 144r^4)/ (r^2 - 2r +1)) = 48

(12 - 12r)/ (1 - (r^4 - 2r^3 + r^2)/ (r^2 - 2r + 1)) = 48

(12 - 12r)/ (1 - (r^2 - r^2 + r^2) = 48

(12 - 12r)/ (1 - r^2) = 48

(12 - 12r) = 48 (1 - r^2)

(12 - 12r) = 48 - 48r^2

1 - r = 4 - 4r^2

4r^2 - r - 3

(4r + 3)(r - 1)

r = -3/ 4orr = 1 … from answer above, though, r should = 1/ 2.

Can anyone help me spot where I have gone wrong? Thank you.