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Thread: geometric progression

  1. #1
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    geometric progression

    can anybody help me for this question?..i alrdy used the sum to infinity formula..but may be my first term and common ratio r wrong...so i cannot get the correct answer, the answer given is 1

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  2. #2
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    Re: geometric progression

    $x_1=\sqrt[3]{4x_1}$
    $x_1^3-4x_1=0$

    $x_1=0,\pm 2$
    Obviously, $x_1=2$

    For the denominator:
    $x_2=\sqrt{2x_2}$
    $x_2^2-2x_2=0$
    $x_2=0,2$
    Obviously $x_2=2$

    So you have $\dfrac{x_1}{x_2}= \dfrac{2}{2}=1$
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  3. #3
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    Re: geometric progression

    hi, thanks for answering,..but i cant understand what u wrote, especially these two lines...can u explain a bit?
    $x_1^3-4x_1=0$

    $x_1=0,\pm 2$


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  4. #4
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    Re: geometric progression

    Let $y = \sqrt[3]{4\sqrt[3]{4\sqrt[3]{4 \, ... }}} \implies y=\sqrt[3]{4y} \implies y^3 = 4y \implies y=2$ since $y>0$

    Let $x= \sqrt{2\sqrt{2\sqrt{2 \, ... }}} \implies x=\sqrt{2x} \implies x^2 = 2x \implies x=2$ since $x>0$

    $\dfrac{y}{x} = \dfrac{2}{2} = 1$
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