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Thread: 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

    $\displaystyle a + ar + ar^2 = 19$

    $\displaystyle 2a + ar - ar^2 = 5$

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

    $\displaystyle 3a + 2ar = 24$

    $\displaystyle a(3+2r) = 24$

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


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

    $\displaystyle 24 + 24r + 24r^2 = 57 + 38r$

    $\displaystyle 24r^2 - 14r - 33 = 0$

    $\displaystyle (12r + 11)(2r - 3) = 0$

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

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

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

    $\displaystyle 4, 6, 9$
    Thanks from lebanon
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