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Math Help - Complicated Geometric Series Question.

  1. #1
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    Complicated Geometric Series Question.

    The question is:

    The sum of the first term and the second term of a geometric series is equal to twice the sum of the second term and third term of the series.
    a] given that the common ratio is positive, find the value of the common ratio.
    b] the sum to infinity of the series is 12. find, correct to two decimal places, the sum of the first eight terms of the series.

    I will be able to do the part' b] OK, but the first part is troubling and I find it complicated.

    I see that u1 + u2 = 2 (u2 + u3)
    So [ u1 =  ar^{-1} , u2 =  ar ] = 2 x [ u2 =  ar , u3 =  ar^2 ] ...{I think}

    What happens from here?
    Please could someone help?
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  2. #2
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    Quote Originally Posted by db5vry View Post
    The question is:

    The sum of the first term and the second term of a geometric series is equal to twice the sum of the second term and third term of the series.
    a] given that the common ratio is positive, find the value of the common ratio.
    b] the sum to infinity of the series is 12. find, correct to two decimal places, the sum of the first eight terms of the series.

    I will be able to do the part' b] OK, but the first part is troubling and I find it complicated.

    I see that u1 + u2 = 2 (u2 + u3)
    So [ u1 =  ar^{-1} , u2 =  ar ] = 2 x [ u2 =  ar , u3 =  ar^2 ] ...{I think}

    What happens from here?
    Please could someone help?
    A) U_1 = a by definition.

    <br />
a + ar = 2(ar + ar^2)<br />

    Factor out: a(1+r) = 2ar(1+r)

    As a(1+r) is a common factor you can cancel to give 2r = 1 and so r = \frac{1}{2}

    B)
    <br />
S_{infinity} = \frac{a}{1-r} which allows you to find a.

    The sum of n terms is S_n = \frac{a(1-r^n)}{1-r}
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