Additional Sum to Infinity of a Geometric Series

**GrigOrig99** · August 10th 2011, 01:37 PM

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.

**Siron** · August 10th 2011, 02:06 PM

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

**Plato** · August 10th 2011, 02:07 PM

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]

Here is your mistake.

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

**GrigOrig99** · August 10th 2011, 03:38 PM

Here is your mistake.

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.

**Plato** · August 10th 2011, 04:07 PM

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

[tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives $\sum\limits_{n = 0}^\infty {ar^n }$ .

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

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

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

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

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

**Siron** · August 11th 2011, 02:44 PM

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

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

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

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

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