Let the six numbers a_1, a_2, a_3, a_4, a_5, a_6 be in A.P.

Question

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

Accepted Answer

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

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

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

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

Solutions · 1