Question

If variance of first $n$ natural numbers is $10$ and variance of first $m$ even natural numbers is $16$, then $m+n$ is equal to

Answer 1

For $n$ natural number variance is given by
${\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2\frac{\sum x_i^2}{n}=\frac{1^2+2^2+3^2+....n\text{terms}}{n}\Rightarrow \frac{\sum x_i^2}{n}=\frac{n(n+1)(2n+1)}{6n}\frac{\sum x_i}{n}=\frac{1+2+3+....n\text{terms}}{n}\Rightarrow \frac{\sum x_i}{n}=\frac{n(n+1)}{2n}{\sigma }^2=\frac{n^2-1}{12}=10\Rightarrow n=11$

Variance of $(2,4,\mathrm{6...})$
$=4\times$ variance of $(1,2,3,\mathrm{4...})$
$=4\times \frac{m^2-1}{12}\Rightarrow \frac{m^2-1}{3}=16\Rightarrow m=7$

$\therefore n+m=11+7=18$