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Math Help - Geometric Progressions

  1. #1
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    Geometric Progressions

    Hello everyone.

    I'm working on the following problem:

    A geometric progression containing 20 terms has 48 as its fifth term. The ratio of the sum of its first 6 terms to that of its first 3 terms is 9:1. Find the sum of its last 5 terms.

    Here's what I've done with it so far:

    G.P.

    n = 20

    T_5 = 48 => a * r^4 = 48 . . . . . . (i)

    S_6 : S_3 = 9 : 1

    S_6 / S_3 = 9 / 1

    S_6 = 9 * S_3

    [ a * (1 - r^5) ] / (1 - r) = { [ a * (1 - r^3) ] / (1 - r) } * 9

    1 - r^5 = 9 (1 - r^3) . . . . . . (ii)

    At this point I'm stuck... can't seem to figure out a way to get a and r.

    Once I have a and r the other part of the question is trivial.

    Any help will be greatly appreciated.

    Thanks,

    Shahz.
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  2. #2
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    A geometric progression containing 20 terms has 48 as its fifth term. The ratio of the sum of its first 6 terms to that of its first 3 terms is 9:1. Find the sum of its last 5 terms.
    \frac{a+ar+ar^2+ar^3+ar^4+ar^5}{a+ar+ar^2} = \frac{9}{1}

    1+r+r^2+r^3+r^4+r^5 = 9+9r+9r^2

    r^3+r^4+r^5 = 8(1+r+r^2)

    r^3(1+r+r^2) = 8(1+r+r^2)

    r = 2

    since ar^4 = 48

    a = 3
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{a+ar+ar^2+ar^3+ar^4+ar^5}{a+ar+ar^2} = \frac{9}{1}

    1+r+r^2+r^3+r^4+r^5 = 9+9r+9r^2

    r^3+r^4+r^5 = 8(1+r+r^2)

    r^3(1+r+r^2) = 8(1+r+r^2)

    r = 2

    since ar^4 = 48

    a = 3
    Gr8!! Thanks!!
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