Question

If the mean of $n$ observations $1^2,2^2,3^2\dots ..n^2$ is $\frac{46n}{11}$, then $n=$

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

Answer 1

The correct option is A

$11$

Explanation of the correct option.

Compute the required value.

Given: Mean of $n$ observations is $\frac{46n}{11}$.

Since mean $\left(\overline{x}\right)$ is given by,

$$\begin{gathered} \left(\overline{x}\right)=\frac{{\displaystyle \underset{i=1}{\overset{n}{\sum x_i}}}}{n} \\ \Rightarrow \frac{46n}{11}=\frac{1^2+2^2+3^2+......+n^2}{n} \\ \Rightarrow \frac{46n}{11}=\frac{n(n+1)(2n+1)}{6} \\ \Rightarrow 6\times 46n=11\left(2n^2+3n+1\right) \\ \Rightarrow 276n=22n^2+33n+11 \\ \Rightarrow 22n^2+243n+11=0 \\ \Rightarrow 22n^2+242n-n+11=0 \\ \Rightarrow 22n(n-11)-1(n-11)=0 \\ \Rightarrow \left(n-11\right)\left(22n-1\right)=0 \end{gathered}$$

Thus, the value of $n$ is $11$ or $\frac{1}{22}$.

Since $n$ can't be a fraction.

Therefore the value of $n$ is $11$.

Hence, option $\left(\mathrm{A}\right)$ is the correct option.