Question

The value of $9^{\frac{1}{3}}{.9}^{\frac{1}{9}}{.9}^{\frac{1}{27}}.....$ to $\infty$ is

• A. 1
• B. 3
• C. 9

Answer 1

The correct option is B

3

(b) 3

$9^{\frac{1}{3}}\times 9^{\frac{1}{9}}\times 9^{\frac{1}{27}}\times .....\infty =9^{(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....\infty )}$

Here, it is a G.P. with $a=\frac{1}{3}$ and $r=\frac{1}{3}$

$\therefore 9^{\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}=9^{\left(\frac{1}{2}\right)}=3$