Question

If positive square root of $a^{\frac{1}{a}}.(2a{)}^{\frac{1}{2a}}.(4a{)}^{\frac{1}{4a}}.(8a{)}^{\frac{1}{8a}}\dots \infty$ is $\frac{8}{27},$ then the value of $a$ is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

$a^{\frac{1}{a}}.(2a{)}^{\frac{1}{2a}}.(4a{)}^{\frac{1}{4a}}.(8a{)}^{\frac{1}{8a}}\dots \infty$
$=a^{(\frac{1}{a}+\frac{1}{2a}+\frac{1}{4a}+\dots \infty )}\cdot 2^{(\frac{1}{2a}+\frac{2}{4a}+\frac{3}{8a}+\dots \infty )}$

Now, $\frac{1}{a}(1+\frac{1}{2}+\frac{1}{2^2}+\dots \infty )=\frac{1}{a}\cdot \frac{1}{1-\frac{1}{2}}=\frac{2}{a}$
and
$\frac{1}{2a}+\frac{2}{4a}+\frac{3}{8a}+\dots \infty$
$=\frac{1}{2a}(1+2\cdot \frac{1}{2}+3\cdot \frac{1}{4}+\dots \infty )$
It is forming AGP
$a=1,d=1,r=\frac{1}{2}$
For $|r|<1,n\to \infty$
$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$
$=\frac{1}{2a}[\frac{1}{1-\frac{1}{2}}+\frac{1\cdot \frac{1}{2}}{{(1-\frac{1}{2})}^2}]$
$=\frac{1}{2a}\cdot 4$
$=\frac{2}{a}$

$\therefore a^{\frac{1}{a}}\cdot 2^{\frac{1}{a}}=\frac{8}{27}={\left(\frac{1}{3}\right)}^3\cdot 2^3$
$\Rightarrow a=\frac{1}{3}$