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odolwa99
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Homework Statement
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 textbook: a = 6, r = 1/ 2
Homework Equations
Eq.: S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]
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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