If the mean of the squares of first n natural numbers be 11, then n is equal to?

Given: Mean of squares of first n natural numbers is 11.

Use the identity for sum of squares: $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$

Mean \(=\) (sum) \( \div n \Rightarrow \) $$\frac{1}{n}\sum_{k=1}^{n}k^2=\frac{(n+1)(2n+1)}{6}=11$$

Solve for \(n\): $$(n+1)(2n+1)=66 $$ $$\;\Rightarrow\; 2n^2+3n+1=66 \;\Rightarrow\; $$ $$n=\frac{-3\pm\sqrt{9+520}}{4}=\frac{-3\pm 23}{4}$$

Only positive integer solution: $$n=\frac{20}{4}=5$$

Answer: \(n=5\).