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Math Help - The sum of the firsts two terms of a geometric progression..?

  1. #1
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    The sum of the firsts two terms of a geometric progression..?

    I hope this is the right section. If not please move.

    The sum of the firsts two terms of a geometric progression is six times the magnitude of the third term:

    i.e. S2 = 6*T3

    This is only possible for two values of the common ratio, r = 0.5 is one value.
    Calculate the other value correct to three decimal places.

    Any help with this would be appreciated. If you could give a step by step explanation on how its done it would be great.
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  2. #2
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    T_2=rT_1

    T_3=r^2T_1

    S_2=T_1+rT_1

    6r^2T_1=rT_1+T_1\Rightarrow\ T_1\left(6r^2-r-1)=0

    6r^2-r-1=0\Rightarrow\ (3r+1)(2r-1)=0
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  3. #3
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    90 *(-1/3) -30 *(-1/3) 10
    90 + (-30) = 60
    10 * 6 = 60
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  4. #4
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    Oh I see that it is a quadratic equation.

    I like to use the b+- formula.

    Thanks for the help.
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