Sum to Infinity of a Geometric Series problem

[odolwa99]

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$\infty$ = $\frac{a}{1 - r}$

The Attempt at a Solution

S$\infty$ = $\frac{a}{1 - r}$ = 12
a = 12 ( 1 - r)
a = 12 - 12r
a = 1 - r

S$\infty$ = $\frac{a^2}{1 - r^2}$

$\frac{a^2}{1- (a^2r^2/ a^2)}$ = 48
$\frac{r^2 - 2r + 1}{1 - (r^2((1 -r)^2))/ ((1 - r)^2)}$ = 48

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

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

$\frac{r^2 - 2r + 1}{1 - r^4 - 2r^4 + r^4}$ = 48

$\frac{r^2 - 2r + 1}{1}$ = 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.

 

[HallsofIvy]

Okay, if "S" is a geometric series:
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$
with a and r, then the sum of squares,
$$\sum_{n=0}^\infty a^2 (r^n)^2= \sum_{n=0}^\infty a^2(r^2)^n$$
is also a geometric series, with $a^2$ and $r^2$
So, as you say,
$$\frac{a}{1- r}= 12$$
and
$$\frac{a^2}{1- r^2}= 48$$

To solve that, I recommend dividing the second equation by the square of the first. (Recall that $1- r^2= (1- r)(1+ r)$.)

 

[jaumzaum]

Hi odolwa99

I think there is a much easier way to solve the problem

Like you know the s infinity sum of a geometric serie of ratio <1 is a0(1-q) where a0 is the first term and q is the ratio

In our case the infinity sum is a/(1-r) = 12 -> a = 12(1-r)

Now see: The sum of the squares is also a geometric serie, first term a² and ratio r², so the sum is

a²/(1-r²) = 48 -> a² = 48(1-r²)

So 144(1-r)² = 48(1-r²) -> 3(1-r)=(1+r) -> r = 1/2

 

[odolwa99]

Ah, fantastic. Thank you very much. Btw, I'm getting used to latex so that's why the question has changed apppearance, and thanks to everyone for your help.