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Sequence problem

  1. Feb 12, 2006 #1
    Hey guys:cool: , I need help for sum of a geometric sequence problem:

    The first and second terms of a geometric sequence have a sum of 15, while the second and third terms have a sum of 60. Use an algebraic method to find the three terms.

    This is what I have so far:

    a + b + c
    a + b = 15
    b + c = 60
    a = a
    r = b/a
    S2 = 15 = a(1-(b/a)^2)/1-(b/a)
    S3 = 60 = a(1-(b/a)^3)/1-(b/a)

    I then solved for a and b and got a = 3.75 and b = 11.25.

    After knowing a and b, I find the common ratio: 3. But my numbers do not work for S3 because I got 48.75:grumpy: (Which, interestly enough is 11.25 away from 60:uhh: )

    I'm really confused:confused: :cry:

    Thanks
     
  2. jcsd
  3. Feb 12, 2006 #2
    Ok I just figured it out, lets get rid of the a, b, and c stuff and just deal with a and r. well you have 3 terms that differ by a power of r you have
    {ar0, ar1, ar2}

    you know that

    a + ar = 15

    and that

    ar + ar2 = 60

    divide the second equation by 4 and you can then set the two equations equal to one another and you end up with a nice quadratic to find r.
     
  4. Feb 12, 2006 #3
    thanks man:smile:
    it's actually so simple i never thought about that:blushing:
    thanks again:smile:
     
  5. Feb 12, 2006 #4
    Yea it took me a bit to realize that too, and I felt so stupid because I almost immediately realized what the asnwers had to be but couldn't figure out how to derive them for a bit.
     
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