The value of
Question:

The value of 91/3 . 91/9 . 91/27 … upto inf, is

(a) 1

(b) 3

(c) 9

(d) none of these

Solution:

$(\mathrm{b}) 3$

$9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots \infty$

$=9\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \infty\right)$

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

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

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

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