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Math Help - [SOLVED] Geometric Progression

  1. #1
    aod
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    [SOLVED] Geometric Progression

    Random question I came across when studying, couldn't work out how to do it.

    Question:
    The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

    Thanks.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by aod View Post
    Random question I came across when studying, couldn't work out how to do it.

    Question:
    The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

    Thanks.
    You know that: \frac{5x - 1}{x^2 - 1} = \frac{-3}{5x - 1}

    (5x - 1)^2 = -3x^2 + 3

    25x^2 - 10x + 1 = -3x^2 + 3

    28x^2 - 10x - 2 = 0

    Solve for x, i'm lazy

    (When you have x, you will have the terms and from there it will be easy to find the common ratio)
    Last edited by janvdl; November 11th 2007 at 01:07 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aod View Post
    Random question I came across when studying, couldn't work out how to do it.

    Question:
    The first three terms of a gemoetric sequence are x^2 - 1, 5x - 1 and -3. Find all possible values of a, the first term and r, the common ratio.

    Thanks.
    recall that in a geometric progression, the nth term, a_n, is given by:

    a_n = ar^{n-1} for n \in \mathbb{N} = 1,2,3,4,...

    thus, since the first term is x^2 - 1 it means that a = x^2 - 1

    also, the second term is:

    a_2 = ar = \left( x^2 - 1 \right)r = 5x - 1

    \Rightarrow r = \frac {5x - 1}{x^2 - 1} ...................(1)

    the third term is:

    a_3 = ar^2 = \left( x^2 - 1 \right)r^2 = -3

    \Rightarrow r = \sqrt{\frac {-3}{x^2 - 1}} ..................(2)

    Now, equate (1) and (2), we get:

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

    now solve that to et a range of values for x, and then you can find the possible values for a and r


    EDIT: Ah, Jan's way is easier
    Last edited by Jhevon; November 11th 2007 at 01:36 AM.
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  4. #4
    aod
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    Ah makes sense now, thanks for help
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