1. ## 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. ## 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 $\dfrac{x_1}{x_2}= \dfrac{2}{2}=1$

3. ## 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. ## 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 $x>0$

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