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Thread: Geometric Progression rejection

  1. #1
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    [SOLVED] Geometric Progression rejection

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

    Find the values of $\displaystyle a$ and $\displaystyle r$.

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

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

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

    $\displaystyle ar^3=5.12-5.12r$

    $\displaystyle ar^3+5.12r-5.12=0$ -----(3)

    From (2), $\displaystyle a=10-10r$ -----(4)

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

    $\displaystyle 10r^3-10r^4+5.12r-5.12=0$

    $\displaystyle r=-0.4+0.6928, -0.4-0.6928, 1, 0.8$

    $\displaystyle =0.2928, -1.093, 1, 0.8$

    I understand why i have to reject $\displaystyle -1.093$ and $\displaystyle 1$. But i don't understand why $\displaystyle 0.2928$ 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 $\displaystyle \displaystyle S_3 = \frac{a(1 - r^3)}{1 - r} = \frac{a}{1 - r}(1 - r^3) = S_{\infty}(1 - r^3)$. Solve for $\displaystyle \displaystyle r$.
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