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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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The geometric mean of $ 2.262.....2^n$ is

$\begin {array} {1 1} (A)\;2^{\large\frac{n}{2}} & \quad (B)\;n^{\large\frac{(n+1)}{2}} \\ (C)\;2^{\large\frac{n(n+1)}{2}} & \quad (D)\;2^{\large\frac{(n+1)}{2}} \end {array}$

 

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1 Answer

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GM = $ (2.2^2.....2^n)^{\large\frac{1}{n}}$
$ = 2^{\large\frac{1+2+....+n}{n}}$
$ = 2^{\large\frac{n(n+1)}{2n}}$
$ = 2^{\large\frac{n+1}{2}}$
Ans : (D)
answered Jan 29, 2014 by thanvigandhi_1
 

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