The value of {(0.2)^{log_{sqrt{5}}left( frac{ | vedclass

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(D) The given series is an infinite geometric progression: $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty$.Here,the first term $a = \frac{1}

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$4$

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(D) The given series is an infinite geometric progression: $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty$.Here,the first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{1/4}{1-1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.Now,the expression becomes $(0.2)^{\log_{\sqrt{5}}(1/2)}$.Since $0.2 = \frac{1}{5} = 5^{-1}$ and $\sqrt{5} = 5^{1/2}$,we have:$(5^{-1})^{\log_{5^{1/2}}(2^{-1})} = (5^{-1})^{\frac{-1}{1/2} \log_{5} 2} = (5^{-1})^{-2 \log_{5} 2} = 5^{2 \log_{5} 2}$.Using the property $n \log_{b} a = \log_{b} a^n$,we get $5^{\log_{5} 2^2} = 2^2 = 4$.

The value of ${(0.2)^{\log_{\sqrt{5}}\left( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty \right)}}$ is:

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