Browse Questions

# If the mean of $n$ observation $1^2, 2^2, 3^2.....n^2$ is $\large\frac{46n}{11}$ then $n$ is equal to

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

Can you answer this question?

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