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If $x=1+\frac{y}{2}+\left(\frac{y}{2}\right)^{2}+\left(\frac{y}{2}\right)^{3}+\ldots .$ where $|y| < 2$, what is $1 y$
equal to ?
Options:
equal to ?
Solution:
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Verified Answer
The correct answer is:
$\frac{2 x-2}{x} \quad$
$\mathrm{x}=1+\frac{\mathrm{y}}{2}+\left(\frac{\mathrm{y}}{2}\right)^{2}+\left(\frac{\mathrm{y}}{2}\right)^{3}+\ldots . .$
Here, $\frac{\mathrm{y}}{2} < 1$ and this a G. with first term $=1$ and common
ratio $=\frac{\mathrm{y}}{2}$ so,
$\Rightarrow \quad x=\frac{1}{1-\frac{y}{2}} \Rightarrow x=\frac{2}{2-y}$
$\Rightarrow 2 x-x y=2 \Rightarrow y=\frac{2 x-2}{x}$
Here, $\frac{\mathrm{y}}{2} < 1$ and this a G. with first term $=1$ and common
ratio $=\frac{\mathrm{y}}{2}$ so,
$\Rightarrow \quad x=\frac{1}{1-\frac{y}{2}} \Rightarrow x=\frac{2}{2-y}$
$\Rightarrow 2 x-x y=2 \Rightarrow y=\frac{2 x-2}{x}$
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