If the variance of first n even natural numbers is 133 , then the...

Var\left(2,4, 6......... 2n\right) = 133$<br />$⇒ 4Var\left(1,2, 3......... n\right) = 133$<br />$⇒ Var\left(1,2, 3......... n\right) = \dfrac{133}{4}$<br />$⇒ \dfrac{{1}^{2} + {2}^{2} +....... n^{2}}{n} - {\left(\dfrac{1 + 2 + 3 +....... n}{n}\right)}^{2} = \dfrac{133}{4}$<br />$⇒ \dfrac{\left(n + 1\right)\left(2n + 1\right)}{6} - \dfrac{{\left(n + 1\right)}^{2}}{4} = \dfrac{133}{4}$<br />$⇒ \dfrac{\left(n + 1\right)}{2}\left[\dfrac{2n + 1}{3} - \dfrac{n + 1}{2}\right] = \dfrac{133}{4}$<br />$⇒ \dfrac{\left(n + 1\right)}{2}\left[\dfrac{n - 1}{6}\right] = \dfrac{133}{4}$<br />$⇒ n^{2} - 1 = 399 ⇒ n = 20