Question

The value of ${\left(0.05\right)}^{{\log }_{\sqrt{20}}(0.1+0.01+0.001+....)}$ is

• A. $81$
• B. $\frac{1}{81}$
• C. $20$
• D. $\frac{1}{20}$

Answer 1

The correct option is A $81$

$0.1+0.01+0.0001+.....\Rightarrow G.P$

$=\left(0.1\right)(1-0.1)S_{\infty }=-\frac{a}{1-r}$

$=\frac{0.1}{0.9}=\frac{1}{9}$

$\therefore \left(0.05\right){\log }_{\sqrt{20}}(0.1+0.01+....)$

$=\left(\frac{1}{20}\right)\log {\sqrt{20}}^{1/9}$

$=\left(\frac{1}{9}\right)\log {\sqrt{20}}^{1/20}$

$=\left(\frac{1}{20}\right)\log {\sqrt{20}}^{(\sqrt{20}{)}^{-2}}={\left(\frac{1}{20}\right)}^{-2}$

$=81$