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Math Help - Geometric Progression

  1. #1
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    Geometric Progression

    A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1).

    I manage to get up to 2(r^5 - 1) = r^15 - r^5.

    Please help.
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  2. #2
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by azuki View Post
    A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1).



    I manage to get up to 2(r^5 - 1) = r^15 - r^5.



    Please help.
    Sum of a geometric series: S_n=\frac{a(r^n-1)}{r-1}

    S_5=\frac{a(r^5-1)}{r-1}

    S_{15}=\frac{a(r^{15}-1)}{r-1}

    The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies:

    S_5 = 2S_{15} - 2S_5

    3S_5 = 2S_{15}

    \frac{3a(r^5-1)}{r-1}=\frac{2a(r^{15}-1)}{r-1}

    3a(r^5-1)=2a(r^{15}-1)

    3(r^5-1)=2(r^{15}-1)

    3(r^5-1)=2[(r^5)^3-1] (note: the expression in the brackets is now a difference of two CUBES, so we can use: a^3-b^3=(a-b)(a^2+ab+b^2)

    So (r^5)^3-1=(r^5-1)(r^{10}+r^5+1)

    3(r^5-1)=2(r^5-1)(r^{10}+r^5+1)

    3=2(r^{10}+r^5+1)

    \frac{3}{2}=r^{10}+r^5+1

    0=r^{10}+r^5-\frac{1}{2}

    0=2r^{10}+2r^5-1

    0=2(r^5)^2+2r^5-1

    Let X=r^5

    0=2X^2+2X-1

    Using the quadratic formula ( x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}),

    X=\frac{-2\pm\sqrt{4-4\times2\times-1}}{4}

    X=\frac{-2\pm\sqrt{12}}{4}

    X=\frac{-2\pm2\sqrt{3}}{4}

    X=\frac{-1\pm\sqrt{3}}{2}

    Putting X=r^5 back in:

    r^5=\frac{-1\pm\sqrt{3}}{2}

    But the question said that r is a positive number, so r^5 must ALSO be positive, therefore r^5=\frac{-1+\sqrt{3}}{2}

    Proving that r^5=\frac{1}{2}(\sqrt{3}-1)
    Last edited by mr fantastic; January 20th 2009 at 03:22 AM. Reason: Added quote tags
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  3. #3
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    Why sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies:

    S_5 = 2S_15 – 2S_5
    Last edited by azuki; January 20th 2009 at 06:50 AM.
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