Question

If the mean of '$n$' observations, $1^2,2^2,3^2,...,n^2$ is $\frac{46n}{11}$, find the value of '$n$'.

• A. $22$
• B. $23$
• C. $12$
• D. $11$

Answer 1

The correct option is D $11$

Observations are:

$1^2,2^2,3^2,.....,n^2$

$\therefore Mean=\frac{1^2+2^2+3^2+.....+n^2}{n}=\frac{46n}{11}$

$\Rightarrow \frac{n(n+1)(2n+1)}{6n}=\frac{46n}{11}$

$\Rightarrow 22n^2+33n+11=276n$

$\Rightarrow 22n^2-243n+11=0$

$\therefore n=11$

$(Ans\to D)$