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Math Help - Geometric Progression rejection

  1. #1
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    [SOLVED] Geometric Progression rejection

    A geometric progression has first term a and common ratio r. The sum of the first three terms is 4.88 and the sum to infinity is 10.

    Find the values of a and r.

    S_3=4.88=\frac{a(1-r^3)}{1-r} ------(1)
     S_\infty=10=\frac{a}{1-r} ------(2)

    (2)-(1): \frac{a}{1-r}-\frac{a(1-r^3)}{1-r}=10-4.88

     \frac{ar^3}{1-r}=5.12

    ar^3=5.12-5.12r

    ar^3+5.12r-5.12=0 -----(3)

    From (2), a=10-10r -----(4)

    Sub (4) into (3): (10-10r)r^3+5.12r-5.12=0

    10r^3-10r^4+5.12r-5.12=0

    r=-0.4+0.6928, -0.4-0.6928, 1, 0.8

    =0.2928, -1.093, 1, 0.8

    I understand why i have to reject -1.093 and 1. But i don't understand why 0.2928 is rejected too
    Last edited by Punch; May 8th 2011 at 06:22 AM.
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  2. #2
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    Notice that \displaystyle S_3 = \frac{a(1 - r^3)}{1 - r} = \frac{a}{1 - r}(1 - r^3) = S_{\infty}(1 - r^3). Solve for \displaystyle r.
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