Let the six numbers $a_1,a_2,a_3,a_4,a_5,a_6$ be in A.P and ${\mathrm{a}}_1+{\mathrm{a}}_3=10$. If the mean of these six numbers
is $\frac{19}{2}$ and their variance is ${\sigma }^2$ , then $8{\sigma }^2$ is equal to

Given $a_1,a_2,a_3,a_4,a_5,a_6$ are in A.P 
Let ${\mathrm{T}}_1={\mathrm{a}}_1=\mathrm{a}$
Common difference =d
Also given $a_1+a_3=10$
$\Rightarrow \mathrm{a}+\mathrm{a}+2\mathrm{d}=10\Rightarrow 2\mathrm{a}+2\mathrm{d}=10\dots \dots \dots \text{ (1) }$
Mean of six numbers is $\frac{19}{2}$
$\begin{array}{l}\Rightarrow \frac{{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+{\mathrm{a}}_4+{\mathrm{a}}_5+{\mathrm{a}}_6}{6}=\frac{19}{2}\\ \Rightarrow \mathrm{a}+(\mathrm{a}+\mathrm{d})+(\mathrm{a}+2\mathrm{d})+(\mathrm{a}+3\mathrm{d})+(\mathrm{a}+4\mathrm{d})+(\mathrm{a}+5\mathrm{d})=\frac{6\times 19}{2}\\ \Rightarrow 6\mathrm{a}+15\mathrm{d}=3\times 19\\ \Rightarrow 2\mathrm{a}+5\mathrm{d}=19\dots \dots \dots .\left(2\right)\end{array}$
$\text{ Eq. (1) \& (2) }3\mathrm{d}=9\Rightarrow \mathrm{d}=3$
$\text{ Also }2a+2\left(3\right)=10\Rightarrow 2a=4\Rightarrow a=2$
 six numbers are 2, 5, 8, 11, 14, 17
Wkt variance $={\mathrm{\sigma }}^2=\frac{\sum {\mathrm{xi}}^2}{\mathrm{n}}-{\mathrm{\mu }}^2$
$\begin{array}{l}=\frac{2^2+5^2+8^2+{11}^2+{14}^2+{17}^2}{6}-{\left(\frac{19}{2}\right)}^2\\ =\frac{4+25+64+121+196+289}{6}-{\left(\frac{19}{2}\right)}^2\\ =\frac{699}{6}-\frac{361}{4}\Rightarrow \frac{1398-1083}{12}=\frac{315}{12}=\frac{105}{4}\\ {\mathrm{\sigma }}^2=\frac{105}{4}\Rightarrow 8{\mathrm{\sigma }}^2=8\times \frac{105}{4}=2\times 105=210\end{array}$