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Math Help - Goemetric seqeunce

  1. #1
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    Goemetric seqeunce question

    a,b and c are three consecutive numbers terms of a goemetric sequence and that verify
    a+b+c=19
    2a+b-c=5
    calculate a,b and c.
    Last edited by lebanon; May 10th 2012 at 04:50 PM.
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  2. #2
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    Re: Goemetric seqeunce

    a + ar + ar^2 = 19

    2a + ar - ar^2 = 5

    --------------------

    3a + 2ar = 24

    a(3+2r) = 24

    a = \frac{24}{3+2r}


    \frac{24}{3+2r} + \frac{24r}{3+2r} + \frac{24r^2}{3+2r} = \frac{19(3+2r)}{3+2r}

    24 + 24r + 24r^2 = 57 + 38r

    24r^2 - 14r - 33 = 0

    (12r + 11)(2r - 3) = 0

    r = -\frac{11}{12} or r = \frac{3}{2}

    two possible solutions for a, b, c ...

    \frac{144}{7} , -\frac{132}{7} , \frac{121}{7}

    4, 6, 9
    Thanks from lebanon
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