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Geometric series word problem

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data

    An infinite geometric progression is such that the sum of all the terms after the nth is equal to twice the nth term. Show that the sum to infinity of the whole progression is three times the first term.

    2. Relevant equations

    [itex] S_{n} = \frac{a(1-r^n)}{1-r}\\
    S_{\infty} = \frac{a}{1-r}\\
    [/itex]
    Where a is equal to the first term of the geometric series and r is equal to the common ratio.

    3. The attempt at a solution

    The sum of all the terms after the nth =
    [itex]\frac{a}{1-r} - \frac{a(1-r^n)}{1-r}
    [/itex]

    Twice the nth term =
    [itex]2ar^{n-1}\\
    [/itex]

    So
    [itex]\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\
    [/itex]
    I need to show that
    [itex]\frac{a}{1-r} = 3a\\
    [/itex]

    [itex]\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\
    a = 2ar^{n-1}(1-r) + a(1-r^n)\\
    2ar^{n-1}=3ar^{n}\\
    \frac{2a}{r}=3a
    [/itex]

    But
    [itex]
    \frac{2a}{r} \ne \frac{a}{1-r}
    [/itex]
    So either I've made an error, or much less likely, the author of the book has.
     
  2. jcsd
  3. Sep 21, 2014 #2

    LCKurtz

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    Science Advisor
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    Assuming ##a \ne 0## that last equation tells you that ##r = 2/3##. Use that to write down your series again and see if the result isn't true.
     
  4. Sep 22, 2014 #3
    Thanks for your reply, it seems I lost sight of the bigger picture.
     
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