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Math Help - Geometric progression

  1. #1
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    Geometric progression

    If the ratio of the sum of the first 6 terms of a G.P to the sum of the first 3 terms of the G.P is 9 , what is the common ratio of the G.P ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    If the ratio of the sum of the first 6 terms of a G.P to the sum of the first 3 terms of the G.P is 9 , what is the common ratio of the G.P ?
    S_6 = \frac{a - a r^6}{1 - r} and S_3 = \frac{a - a r^3}{1 - r}. It follows from \frac{S_6}{S_3} = 9 that \frac{1 - r^6}{1 - r^3} = 9 \Rightarrow \frac{(1 - r^3)(1 + r^3)}{1 - r^3} = 9 (nb: r \neq 1).

    Therfore 1 + r^3 = 9 ....
    Last edited by mr fantastic; January 20th 2009 at 03:01 AM. Reason: Transcription errror. Powers wrong by 1 - wine stain on napkin.
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  3. #3
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    If you were to denote the first 6 terms of the geometric sequence as a, ar, \ldots , ar^5, then:

    \frac{a+ar+ar^2+\cdots+ar^5}{a+ar+ar^2} = 9

    1 + \frac{ar^3+ar^4+ar^5}{a+ar+ar^2} = 9

    1 + \frac{ar^3(1+r+r^2)}{a(1+r+r^2)} = 9

    \ldots
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