Question

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

Answer 1

For n natural number variance is given by
${\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2\frac{\sum x_i^2}{n}=\frac{1^2+2^2+3^2+....nterm}{n}=\frac{n(n+1)(2n+1)}{6n}$
$\frac{\sum x_i}{n}=\frac{1+2+3+....nterms}{n}=\frac{n(n+1)}{2n}$
${\sigma }^2=\frac{n^2-1}{12}=10$
$\Rightarrow n=11$
Variance of $(2,4,\mathrm{6...})=4\times$ variance of $(1,2,3,\mathrm{4...})=4\times \frac{m^2-1}{12}=\frac{m^2-1}{3}=16\Rightarrow m=7$
Therefore, n + m = 11 + 7 = 18