If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16. Then n-m=______

$$\begin{gathered} {\sigma }^2=\frac{{\displaystyle \sum x_i^2}}{n}-{\left(\frac{{\displaystyle \sum x_i}}{n}\right)}^2=\frac{1^2+2^2+\mathrm{.......}+n^2}{n}-{\left(\frac{\frac{n(n+1)}{2}}{K}\right)}^2 \\ =\frac{n{(n+1)}^2(2n+1)}{6n}={\left(\frac{n+1}{2}\right)}^2\Rightarrow 10=\frac{n^2-1}{12}\Rightarrow n=11 \end{gathered}$$

Variance of (2, 4, 6, -----) = (Variance of 1, 2, 3, ----)

$$\begin{gathered} \Rightarrow \frac{4(m^2-1)}{12}=\frac{m^2-1}{3}\Rightarrow 16=\frac{m^2-1}{3}\Rightarrow m=7 \\ \therefore n-m=11-7=4 \end{gathered}$$