Question

Let $P=3^{1/3}.3^{2/9}.3^{3/27}...\infty$, then $P^{1/3}$ is equal to

• A. $3^{2/3}$
• B. $\sqrt{3}$
• C. $3^{1/3}$
• D. $3^{1/4}$

Answer 1

The correct option is D $3^{1/4}$

Above in an infinite A.G.S. with $a=1,d=1$ for A.P., $b=\frac{1}{3},r=\frac{1}{3}$ for G.P.
$\therefore S_{\infty }=\frac{ab}{1-r}+\frac{dbr}{(1-r{)}^2}=\frac{\frac{1}{3}}{1-\frac{1}{3}}+\frac{1.\frac{1}{3}.\frac{1}{3}}{{(1-\frac{1}{3})}^2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$
$\therefore P=3^S=3^{3/4}\therefore P^{1/3}=3^{1/4}$.