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

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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)

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