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

<p>${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$ are in AP.</p> <p>Let</p> <p>${a_1} = a$</p> <p>${a_2} = a + d$</p> <p>${a_3} = a + 2d$</p> <p>${a_4} = a + 3d$</p> <p>${a_5} = a + 4d$</p> <p>${a_6} = a + 5d$</p> <p>Now Mean of ${a_1},{a_2},{a_3},{a_4},{a_5}$ and ${a_6}$ is</p> <p>$= {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$</p> <p>$\Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}$</p> <p>$\Rightarrow 2a + 5d = 19$ ...... (1)</p> <p>Also, given,</p> <p>${a_1} + {a_3} = 10$</p> <p>$\Rightarrow a + a + 2d = 10$</p> <p>$\Rightarrow 2a + 2d = 10$</p> <p>$\Rightarrow a + d = 5$ ..... (2)</p> <p>From equation (1) and (2), we get $a = 2$ and $d = 3$</p> <p>$\therefore$ AP is 2, 5, 8, 11, 14, 17</p> <p>Now, Variance $({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$</p> <p>$$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$$</p> <p>$= {{105} \over 4}$</p> <p>$\therefore$ $8{\sigma ^2} = 8 \times {{105} \over 4} = 210$</p>