Let the six numbers a _ 1 , a _ 2 , . . . , a _ 6 be in A . P . and...

Let ${a}_{1} = a and common difference be$d$.</p><p>Now,</p><p>${a}_{1} + {a}_{3} = 10$</p><p>$⇒ 2a + 2d = 10$</p><p>$⇒ a + d = 5 . . . \left(i\right)$</p><p>Now,</p><p>Given mean$\overset{¯}{x} = \dfrac{19}{2}$</p><p>$⇒ \dfrac{a + \left(a + d\right) + \left(a + 2d\right) + . . . + \left(a + 5d\right)}{6} = \dfrac{19}{2}$</p><p>$⇒ \dfrac{6a + 15d}{6} = \dfrac{19}{2}$</p><p>$⇒ 2a + 5d = 19 . . . \left(ii\right)$</p><p>Solving \left(i\right) \text{ and } \left(ii\right)$, we get

$a = 2, d = 3$

So, $A . P$ is $2, 5, 8, 11, 14, 17$.

Now variance is given by,

${\sigma}^{2} = \dfrac{1}{6}\mathop{∑}_{i = 1}^{6}{x}_{i} - {\left(\overset{¯}{x}\right)}^{2}$

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

$⇒ {\sigma}^{2} = \dfrac{699}{6} - \dfrac{361}{4}$

$⇒ {\sigma}^{2} = \dfrac{105}{4}$

$⇒ 8{\sigma}^{2} = 210$