1. ## Additional Sum to Infinity of a Geometric Series

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

Eq.:
S¥ = a/ 1 - r

Attempt:
S¥ = a/ (1 - r) = 12
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/ 4 or r = 1 … from answer above, though, r should = 1/ 2.

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

2. ## Re: Additional Sum to Infinity of a Geometric Series

The problem is, you can only use the formula $\displaystyle \frac{a}{1-r}$ for the sum of the geometric serie if $\displaystyle |r|<1$ but is this given?

3. ## Re: Additional Sum to Infinity of a Geometric Series

Originally Posted by GrigOrig99
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
Eq.:
S¥ = a/ 1 - r[B]

$\displaystyle \frac{a}{1-r}=12~\&~\frac{a^2}{1-r^2}=48$

4. ## Re: Additional Sum to Infinity of a Geometric Series

img.top {vertical-align:15%;}$\frac{a}{1-r}=12~\&~\frac{a^2}{1-r^2}=48$
Actually, I have already factored that in the attempt; see the bold reference following:

S¥ = a/ (1 - r) = 12
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

Is this still incorrect?
Thank you for responding.

5. ## Re: Additional Sum to Infinity of a Geometric Series

Originally Posted by GrigOrig99
Actually, I have already factored that in the attempt; see the bold reference following:
S¥ = a/ (1 - r) = 12
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
First of all you really need learn to post is symbols. You can use LaTeX tags

$$\sum\limits_{n = 0}^\infty {ar^n }$$ gives $\displaystyle \sum\limits_{n = 0}^\infty {ar^n }$ .

In the sum of the squares $\displaystyle \sum\limits_{n = 0}^\infty {a^2 r^{2n} }$ the first term is $\displaystyle a^2$ the ratio is $\displaystyle r^2$.

Therefore if $\displaystyle a+ar+ar^2+ar^3+\cdots$ is the geometric series then
the series of sum of its squares is $\displaystyle a^2+a^2r^2+a^2r^4+a^2r^6+\cdots$.

In the second sum, the first term is $\displaystyle a^2$
and the ratio is $\displaystyle r^2$.

Therefore, the sum of the second series is
$\displaystyle \frac{a^2}{1-r^2}$.

I really think that your lack of ability to use symbols is a hindrance to your understanding.

6. ## Re: Additional Sum to Infinity of a Geometric Series

You have this equation:
$\displaystyle \frac{a}{1-r}=12 \Leftrightarrow a=12(1-r) \Leftrightarrow a^2=144(1-r)^2$ (1)

You have also given:
$\displaystyle \frac{a^2}{1-r^2}=48$ (2)

Substitutie (1) in (2):
$\displaystyle \frac{144(1-r)^2}{1-r^2}=48$
$\displaystyle \Leftrightarrow \frac{144(1-r)^2}{(1-r)(1+r)}=48$
$\displaystyle \Leftrightarrow ...$

Work this out and you'll find $\displaystyle r$.