Question

If the arithmetic mean of $\log 1,\log 2,\log 4,\dots \dots ..\log 2^{n-1}$ is $\log 32$ where all $\log$ arithms are on base 10, then $n$ equals

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

Answer: (B)

$AM=\frac{\log 1+\log 2+\log 2^2+\log 2^3+\dots \dots +\log 2^{n-1}}{n}=\frac{\log \left(2^{1+2+3+\dots \dots +(n-1)}\right)}{n}=\frac{\log 2^{\frac{(n-1)n}{2}}}{n}$
$AM=\frac{1}{n}\log 2^{\frac{(n-1)n}{2}}=\log 2^{\frac{(n-1)}{2}}$
$\therefore {\log }_{10}{2}^{\frac{(n-1)}{2}}=\log 2^5$
$\therefore 2^{\frac{(n-1)}{2}}=2^5\Rightarrow \frac{n-1}{2}=5\Rightarrow n=11$