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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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If the mean of $n$ observation $ 1^2, 2^2, 3^2.....n^2$ is $ \large\frac{46n}{11}$ then $n$ is equal to

$\begin {array} {1 1} (A)\;11 & \quad (B)\;12 \\ (C)\;23 & \quad (D)\;22 \end {array}$

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1 Answer

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Given :
$ \large\frac{1^2+2^2+3^2+.....n^2}{n} $ $ = \large\frac{46n}{11}$
$ \Rightarrow \large\frac{ (n+1)(2n+1)}{6}$ $ = \large\frac{46n}{11}$
$ 11[ (n+1)(2n+1)]=6(46n)$
$ 11[ 2n^2+n+2n+1]=276n$
$ 22n^2+11n+22n+11=276n$
$ 22n^2+33n-276n+11=0$
$ 22n^2-243n+11=0$
$(n-11)(22n-1)=0$
$n=11\: \: \: n=\large\frac{1}{22}$
$ \therefore n \neq \large\frac{1}{22}$
$n = 11$
Ans : (A)
answered Jan 30, 2014 by thanvigandhi_1
 

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