Question

The value of ${\left[{\left(0.16\right)}^{{\log }_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\infty \right)}\right]}^{\frac{1}{2}}$is equal to

• A. $1$
• B. $0$
• C. $-$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct option

Step 1: Calculation of the GP series

The given GP series is $\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\infty$

The common difference $r=\frac{1}{3}$

The first term $a=\frac{1}{3}$

For infinite GP series we know that the sum is $\frac{a}{1-r}$

Therefore the sum of the given GP series is $=\frac{{\displaystyle \frac{1}{3}}}{1-\frac{1}{3}}=\frac{1}{2}$

$$\begin{gathered} \therefore {\log }_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\infty \right) \\ ={\log }_{2.5}\left(\frac{1}{2}\right) \end{gathered}$$

Step 2. Find the value of the given expression

We know that,

$$\begin{gathered} a^{{\log }_a\left(x\right)}=x \\ {a}^{{\log }_b\left(x\right)}=x^{{\log }_b\left(a\right)} \end{gathered}$$

$$\begin{gathered} \therefore {\left[{\left(0.16\right)}^{{\log }_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\infty \right)}\right]}^{\frac{1}{2}} \\ ={\left[{\left(\frac{4}{25}\right)}^{{\log }_{\frac{5}{2}}\left(\frac{1}{2}\right)}\right]}^{\frac{1}{2}} \\ ={\left[{\left(\frac{2}{5}\right)}^{2{\log }_{\frac{5}{2}}\left(\frac{1}{2}\right)}\right]}^{\frac{1}{2}} \\ ={\left[{\left(\frac{1}{2}\right)}^{2{\log }_{\frac{5}{2}}\left(\frac{2}{5}\right)}\right]}^{\frac{1}{2}} \\ ={\left[{\left(\frac{1}{2}\right)}^{-2{\log }_{\frac{5}{2}}\left(\frac{5}{2}\right)}\right]}^{\frac{1}{2}}\left[\log \left(a^n\right)=n\log \left(a\right)\right] \\ ={\left[{\left(\frac{1}{2}\right)}^{-2}\right]}^{\frac{1}{2}}\left[\because {\log }_a\left(a\right)=1\right] \\ =2 \end{gathered}$$

Hence, the correct option is$\mathrm{Option}\left(\mathrm{D}\right)$