QJEEM23Maa8ce: JEE Main 2023 Mathematics - Class 11 JEEM23Maa8ce Solution: Let the six numbers , be in A.P. and . If the mean of these six …

${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$ 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 $({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$

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

$= {{105} \over 4}$

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