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Q.
If the arithmetic mean of $\log 1, \log 2, \log 4, \ldots \ldots . . \log 2^{ n -1}$ is $\log 32$ where all $\log$ arithms are on base 10, then $n$ equals
Sequences and Series
Solution:
$AM =\frac{\log 1+\log 2+\log 2^2+\log 2^3+\ldots \ldots+\log 2^{ n -1}}{ n }=\frac{\log \left(2^{1+2+3+\ldots \ldots+( n -1)}\right)}{ n }=\frac{\log 2^{\frac{( n -1) n }{2}}}{ n }$
$AM =\frac{1}{ n } \log 2^{\frac{( n -1) n }{2}}=\log 2^{\frac{( n -1)}{2}} $
$\therefore \log _{10} 2^{\frac{( n -1)}{2}}=\log 2^5$
$\therefore 2^{\frac{( n -1)}{2}}=2^5 \Rightarrow \frac{ n -1}{2}=5 \Rightarrow n =11 $