Let the six numbers $a_1, a_2, ..., a_6$ be in $\mathrm{A}.\mathrm{P}.$ 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

Let $a_1=a$ and common difference be $d$.

Now,

$a_1+a_3=10$

$\Rightarrow 2a+2d=10$

$\Rightarrow a+d=5 ...\left(\mathrm{i}\right)$

Now,

Given mean $\overline{x}=\frac{19}{2}$

$\Rightarrow \frac{a+\left(a+d\right)+\left(a+2d\right)+...+\left(a+5d\right)}{6}=\frac{19}{2}$

$\Rightarrow \frac{6a+15d}{6}=\frac{19}{2}$

$\Rightarrow 2a+5d=19 ...\left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$a=2, d=3$

So, $\mathrm{A}.\mathrm{P}$ is $2, 5, 8, 11, 14, 17$.

Now variance is given by,

${\sigma }^2=\frac{1}{6}\sum _{i=1}^6{x}_i-{\left(\overline{x}\right)}^2$

$\Rightarrow {\sigma }^2=\frac{4+25+64+121+196+289}{6}-{\left(\frac{19}{2}\right)}^2$

$\Rightarrow {\sigma }^2=\frac{699}{6}-\frac{361}{4}$

$\Rightarrow {\sigma }^2=\frac{105}{4}$

$\Rightarrow 8{\sigma }^2=210$