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Math Help - Converging geometric sequence with the sum cubed

  1. #1
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    Exclamation Converging geometric sequence with the sum cubed

    The sum of the terms of an infinite geometric progression is 4 and the sum of the cubes of the terms is 192. Find the first three terms.

    Obviously you can't use the formula for the sum of a convergent series because you don't know the first term or the common ratio...

    I have no idea what else to use.
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  2. #2
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    Quote Originally Posted by KateMackay View Post
    The sum of the terms of an infinite geometric progression is 4 and....
    So:

    4\, =\, \frac{a}{1\, -\, r}

    Quote Originally Posted by KateMackay View Post
    ...the sum of the cubes of the terms is 192.
    So:

    192\, =\, \frac{a^3}{1\, -\, r^3}

    Note that:

    1\, -\, r^3\, =\, (1\, -\, r)(1\, +\, r\, +\, r^2)

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  3. #3
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    Okay, so what I've done is used the equation with 4 to solve for r and substitute that value into the equation with 192. Now there is way too much simplifying/factoring for my liking but I'll keep trying.
    Thanks!
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    Quote Originally Posted by KateMackay View Post
    Okay, so what I've done is used the equation with 4 to solve for r and substitute that value into the equation with 192. Now there is way too much simplifying/factoring for my liking but I'll keep trying.
    Thanks!
    here's a hint ...

    4 = \frac{a}{1-r}

    64 = \frac{a^3}{(1-r)^3}

    finally, note that ...

    \frac{192}{64} = 3
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