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Math Help - Geometric sequence question.

  1. #1
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    Geometric sequence question.

    The first term of a finite geometric series is 6 and the last term is 4374. The sum of all terms is 6558. Find the common ratio and the number of terms in the series.

    So i get that T1= 6.So, a=6. Sn= 6558. how to find r?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Geometric sequence question.

    What is the formula for the sum of a geometric series?
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  3. #3
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    Re: Geometric sequence question.

    If t_1 = a, t_2 = ar, t_3 = ar^2, ..., t_n = ar^{n-1} and S_n = t_1 + t_2 + \dots + t_n, then, according to the standard formula for the sum of geometric progression, we get

    S_n = \frac{a(r^n-1)}{r-1}=\frac{ar^n-a}{r-1}=\frac{rt_n-t_1}{r-1}

    from where

    r=\frac{S_n-t_1}{S_n-t_n}=\frac{S_n-t_1}{S_{n-1}}.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Geometric sequence question.

    Not using any memorized formulas, just thinking about it:

    Given that 6r^n = 4374, divide through by 6 to get r^n = 729. Note that 729 = 27^2, or 9^3, or 3^6; so the answer must be that either r=27 and n=2, r=9 and n=3, or r=3 and n=6. See which one results in a series whose sum of terms 6 + 6r + 6r^2 + ...6r^n = 6558.
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