Let the six numbers $a_1,a_2,a_3,a_4,a_5,a_6$ be in A.P and $a_1+a_3=10$ . 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  $T_1=a_1=a$
Common difference =d
Also given  $a_1+a_3=10$
 $\Rightarrow a+a+2d=10\Rightarrow 2a+2d=\mathrm{10.........}\left(1\right)$
Mean of six numbers is  $\frac{19}{2}$
$$\begin{gathered} \Rightarrow \frac{a_1+a_2+a_3+a_4+a_5+a_6}{6}=\frac{19}{2} \\ \Rightarrow a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)=\frac{6\times 19}{2} \\ \Rightarrow 6a+15d=3\times 19 \\ \Rightarrow 2a+5d=\mathrm{19.........}\left(2\right) \end{gathered}$$
Eq. (1) & (2) 3d = 9  $\Rightarrow d=3$
Also  $2a+2\left(3\right)=10\Rightarrow 2a=4\Rightarrow a=2$
  six numbers are 2, 5, 8, 11, 14, 17
Wkt variance $={\sigma }^2=\frac{{\displaystyle \sum x{i}^2}}{n}-{\mu }^2$
$$\begin{gathered} =\frac{2^2+5^2+8^2+{11}^2+{14}^2+{17}^2}{6}-{\left(\frac{19}{2}\right)}^2 \\ =\frac{4+25+64+121+196+289}{6}-{\left(\frac{19}{2}\right)}^2 \\ =\frac{699}{6}-\frac{361}{4}\Rightarrow \frac{1398-1083}{12}=\frac{315}{12}=\frac{105}{4} \\ {\sigma }^2=\frac{105}{4}\Rightarrow 8{\sigma }^2=8\times \frac{105}{4}=2\times 105=210 \end{gathered}$$