Question

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

• A. -6.5
• B. 11
• C. 5
• D. none of these

Answer 1

Answer: (C)

5

Answer 2

n = \frac{1}{n} \left\\{1^{2}+2^{2}+3^{2} + ... +n^{2}\right\\} \Rightarrow 11 = \frac{1}{n} \left\\{ \frac{n\left(n+1\right)\left(2n+1\right)}{6}\right\\} $\Rightarrow 2n^{2} + 3n + 1 = 66$ $\Rightarrow 2n^{2} + 3n -65 = 0$ $\Rightarrow n = \frac{-3 \pm\sqrt{9+520}}{4}$ $\Rightarrow n = \frac{-3 \pm23}{4} = 5, - \frac{13}{2} ;$ but $n \le0$ therefore, n = 5.