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Thread: Additional Sum to Infinity of a Geometric Series

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    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.
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    MHF Contributor Siron's Avatar
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    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?
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    Re: Additional Sum to Infinity of a Geometric Series

    Quote Originally Posted by GrigOrig99 View Post
    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.

    $\displaystyle \frac{a}{1-r}=12~\&~\frac{a^2}{1-r^2}=48$
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    Re: Additional Sum to Infinity of a Geometric Series

    Here is your mistake.

    img.top {vertical-align:15%;}
    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.
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    Re: Additional Sum to Infinity of a Geometric Series

    Quote Originally Posted by GrigOrig99 View Post
    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 $\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.
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  6. #6
    MHF Contributor Siron's Avatar
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    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 $\displaystyle r$.
    Last edited by Siron; Aug 11th 2011 at 02:58 PM.
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