(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

Eq.: S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

3. The attempt at a solution

S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex] = 12

a = 12 ( 1 - r)

a = 12 - 12r

a = 1 - r

S[itex]\infty[/itex] = [itex]\frac{a^2}{1 - r^2}[/itex]

[itex]\frac{a^2}{1- (a^2r^2/ a^2)}[/itex] = 48

[itex]\frac{r^2 - 2r + 1}{1 - (r^2((1 -r)^2))/ ((1 - r)^2)}[/itex] = 48

[itex]\frac{r^2 - 2r + 1}{1 - ((r^2 - r^3)^2)/ (r^2 - 2r + 1)}[/itex] = 48

[itex]\frac{r^2 - 2r + 1}{1 - (r^6 - 2r^5 + r^4)/ (r^2 -2r + 1}[/itex] = 48

[itex]\frac{r^2 - 2r + 1}{1 - r^4 - 2r^4 + r^4}[/itex] = 48

[itex]\frac{r^2 - 2r + 1}{1}[/itex] = 48

r^2 - 2r + 1 -48

r^2 - 2r + 1 -47

(r - 47)(r - 1) ...?

from answer above, though, r should = 1/ 2.

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

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# Homework Help: Sum to Infinity of a Geometric Series problem

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