The mean of the squares of the first \( n \) natural numbers is given by: \[ \text{Mean} = \frac{1}{n} \sum_{k=1}^{n} k^2 \] The sum of squares of the first \( n \) natural numbers is: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] Given that the mean is 11, we set up the equation: \[ \frac{1}{n} \times \frac{n(n+1)(2n+1)}{6} = 11 \] \[ \frac{(n+1)(2n+1)}{6} = 11 \] \[ (n+1)(2n+1) = 66 \] Now, solving for \( n \): Expanding: \[ 2n^2 + n + 2n + 1 = 66 \] \[ 2n^2 + 3n - 65 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 3 \), and \( c = -65 \): \[ n = \frac{-3 \pm \sqrt{9 + 520}}{4} \] \[ n = \frac{-3 \pm \sqrt{529}}{4} \] \[ n = \frac{-3 \pm 23}{4} \] Solving for positive \( n \): \[ n = \frac{20}{4} = 5 \]