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Math Help - Geometric Progression

  1. #1
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    Geometric Progression

    Hi guys,
    I am a little stuck on where to begin here and any help would be great.

    This is the question I am faced with:

    The sum of the first three terms of a geometric series is . The sum of the first six terms is . Find the first term and common ratio.
    Now, the formula I have been using recently for working out the sum of a geometric series to n places is:

    S_{n} = \frac{a(1-r^n)}{1-r}

    But I am unsure how to use it in this context.

    Thanks for any help!
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  2. #2
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    Hi

    You know

    S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}

    S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}

    Dividing gives

    \frac{1-r^6}{1-r^3} = \frac{91}{64}

    Using R = r^3 you will find a quadratic whose unknown is R

    You will get R then r and finally a
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    You know

    S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}

    S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}

    Dividing gives

    \frac{1-r^6}{1-r^3} = \frac{91}{64}

    Using R = r^3 you will find a quadratic whose unknown is R

    You will get R then r and finally a
    There is no need to solve a quadratic since

    \frac{1-r^6}{1-r^3} =1+r^3= \frac{91}{64} \Rightarrow r^3=\frac{27}{64} \Rightarrow r=\frac{3}{4}
    Last edited by mr fantastic; September 13th 2009 at 04:37 AM. Reason: Restored original reply (please don't edit/delete a reply after it has been quoted).
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Hi

    You know

    S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}

    S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}

    Dividing gives

    \frac{1-r^6}{1-r^3} = \frac{91}{64}

    Using R = r^3 you will find a quadratic whose unknown is R

    You will get R then r and finally a
    Could you please explain the steps you took to divide those equations?

    Thanks for you help
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  5. #5
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    Quote Originally Posted by luobo View Post
    There is no need to solve a quadratic since

    \frac{1-r^6}{1-r^3} =1+r^3= \frac{91}{64} \Rightarrow  r^3=\frac{27}{64} \Rightarrow  r=\frac{3}{4}
    Yes

    I realized this after switching off my computer !
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  6. #6
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    \frac{a(1-r^6)}{1-r} = \frac{3367}{512}

    \frac{a(1-r^3)}{1-r} = \frac{37}{8}

    Dividing is the same than multiplying the first equation by the inverse of the second one

    \frac{a(1-r^6)}{1-r} \: \frac{1-r}{a(1-r^3)} = \frac{3367}{512}\:\frac{8}{37}

    a and (1-r) are both at the numerator and the denominator, therefore they disappear

    \frac{1-r^6}{1-r^3} = \frac{91}{64}
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  7. #7
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    Thanks for the help guys, much appreciated
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