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Math Help - Geometric Sequence Help

  1. #1
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    Geometric Sequence Help

    Having trouble finding a way of working through this problem:

    A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

    I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by fletch View Post
    Having trouble finding a way of working through this problem:

    A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

    I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.
    I am not sure what you are asking.
    You are given that a_1=1,~a_5=r^4,~\&~a_9=r^8.
    Also a_9-a_5=240.
    From that find r.
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  3. #3
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    Quote Originally Posted by fletch View Post
    Having trouble finding a way of working through this problem:

    A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

    I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.

    Any help would be appreciated.
    The geometric sequence is

    1,\ r,\ r^2,\ r^3,\ r^4,\ r^5,\ r^6,\ r^7,\ r^8,\ r^9....

    From your information

    \left(r^4\right)^2=r^4+240

    x^2-x-240=0

    (x-16)(x+15)=0

    x is not -15, so x is 16

    r is the 4th root of 16, which is \pm2.

    The 5th and 9th terms are both positive, regardless of whether r is 2 or -2

    The 8th term is the two solutions to r^7
    Last edited by Archie Meade; July 14th 2010 at 04:33 AM.
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