$${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$$ are in AP.

Let

$${a_1} = a$$

$${a_2} = a + d$$

$${a_3} = a + 2d$$

$${a_4} = a + 3d$$

$${a_5} = a + 4d$$

$${a_6} = a + 5d$$

Now Mean of $${a_1},{a_2},{a_3},{a_4},{a_5}$$ and $${a_6}$$ is

$$ = {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$$

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

$$ \Rightarrow 2a + 5d = 19$$ ...... (1)

Also, given,

$${a_1} + {a_3} = 10$$

$$ \Rightarrow a + a + 2d = 10$$

$$ \Rightarrow 2a + 2d = 10$$

$$ \Rightarrow a + d = 5$$ ..... (2)

From equation (1) and (2), we get $$a = 2$$ and $$d = 3$$

$$\therefore$$ AP is 2, 5, 8, 11, 14, 17

Now, Variance $$({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$$

$$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$$

$$ = {{105} \over 4}$$

$$\therefore$$ $$8{\sigma ^2} = 8 \times {{105} \over 4} = 210$$