Question

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

• A. - 13/2
• B. 11
• C. 5
• D. none of these

Answer 1

Answer: (C)

$n=\frac{1}{n}\left\{1^2+2^2+3^2+...+n^2\right\}$
$\Rightarrow 11=\frac{1}{n}\left\{\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right\}$
$\Rightarrow 2n^2+3n+1=66$
$\Rightarrow 2n^2+3n-65=0$
$\Rightarrow n=\frac{-3\pm \sqrt{9+520}}{4}$
$\Rightarrow n=\frac{-3\pm 23}{4}=5,-\frac{13}{2};$
but $n\le 0$ therefore, n = 5.