Question

If the variance of first $\mathrm{n}$ natural numbers is $10$ and variance of first $\mathrm{m}$ even natural numbers is $16$, $\mathrm{m}+\mathrm{n}$ is equal to

Answer 1

Finding the value of $\mathbf{m}\mathbf{+}\mathbf{n}$:

Step 1: Finding the value of $\mathbf{n}$.

The variance of $\mathrm{n}$ natural numbers is given by

${\mathrm{\sigma }}^2=\frac{\sum _{{\mathrm{xi}}^2}}{\mathrm{n}}-{\left(\frac{\sum _{\mathrm{xi}}}{\mathrm{n}}\right)}^2$

According to the question,

$$\begin{gathered} \frac{\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{{\left(\mathrm{n}+1\right)}^2}{2}=10 \\ \Rightarrow \frac{({\mathrm{n}}^2-1)}{12}=10 \\ \Rightarrow {\mathrm{n}}^2-1=120 \\ \Rightarrow {\mathrm{n}}^2=121 \\ \Rightarrow \mathrm{n}=11 \end{gathered}$$

Step 2: Finding the value of $\mathbf{m}$.

Variance of$\left(2,4,6\dots \right)$ $=$$4\times$variance of $\left(1,2,3,4\dots \right)=$$4\times \frac{({\mathrm{m}}^2-1)}{12}$

$$\begin{gathered} \frac{\left({\mathrm{m}}^2-1\right)}{3}=16 \\ \Rightarrow {\mathrm{m}}^2-1=48 \\ \Rightarrow {\mathrm{m}}^2=49 \\ \Rightarrow \mathrm{m}=7 \end{gathered}$$

Now, by adding the value of $\mathrm{m}$ and $\mathrm{n}$, we get

$\begin{array}{rcl}\mathrm{n}+\mathrm{m}& =& 11+7\\ & =& 18\end{array}$

Hence, the required answer is $\mathbf{18}$.