Question

If $\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{4x^3}+\frac{1}{8x^4}+\frac{1}{16x^5}+\dots \dots ..\infty =\frac{1}{64}$, then the value of $x$ is

• A. $\frac{125}{2}$
• B. $\frac{127}{2}$
• C. $\frac{129}{2}$
• D. $\frac{131}{2}$

Answer 1

Answer: (C)

$\frac{\frac{1}{x}}{1-\frac{1}{2x}}=\frac{1}{64}\Rightarrow \frac{\frac{1}{x}}{\frac{2x-1}{2x}}=\frac{1}{64}$
$\frac{2}{2x-1}=\frac{1}{64}\Rightarrow 128=2x-1\Rightarrow x=\frac{129}{2}$