Question

If the variance of first $n$ even natural numbers is $133$ , then the value of $n$ is equal to

• A. $19$
• B. $24$
• C. $21$
• D. $20$

Answer 1

Answer: (D)

$Var\left(2,4,\mathrm{6.........2}n\right)=133$
$\Rightarrow 4Var\left(1,2,\mathrm{3.........}n\right)=133$
$\Rightarrow Var\left(1,2,\mathrm{3.........}n\right)=\frac{133}{4}$
$\Rightarrow \frac{1^2+2^2+.......n^2}{n}-{\left(\frac{1+2+3+.......n}{n}\right)}^2=\frac{133}{4}$
$\Rightarrow \frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{4}=\frac{133}{4}$
$\Rightarrow \frac{\left(n+1\right)}{2}\left[\frac{2n+1}{3}-\frac{n+1}{2}\right]=\frac{133}{4}$
$\Rightarrow \frac{\left(n+1\right)}{2}\left[\frac{n-1}{6}\right]=\frac{133}{4}$
$\Rightarrow n^2-1=399\Rightarrow n=20$