Let the six numbers a_1, a_2, a_3, a_4, a_5, | vedclass

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(B) Given $a_1, a_2, a_3, a_4, a_5, a_6$ are in $A.P.$ with common difference $d$.$a_1 + a_3 = a_1 + (a_1 + 2d) = 2a_1 + 2d = 10 \Rightarrow a_1 + d =

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$210$

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(B) Given $a_1, a_2, a_3, a_4, a_5, a_6$ are in $A.P.$ with common difference $d$.$a_1 + a_3 = a_1 + (a_1 + 2d) = 2a_1 + 2d = 10 \Rightarrow a_1 + d = 5$.The mean of the six numbers is $\frac{a_1 + a_2 + a_3 + a_4 + a_5 + a_6}{6} = \frac{19}{2}$.Sum of the numbers $= 6 	imes \frac{19}{2} = 57$.Using the sum formula for $A.P.$,$S_6 = \frac{6}{2}(2a_1 + 5d) = 3(2a_1 + 5d) = 57 \Rightarrow 2a_1 + 5d = 19$.Solving $a_1 + d = 5$ and $2a_1 + 5d = 19$,we get $d = 3$ and $a_1 = 2$.The numbers are $2, 5, 8, 11, 14, 17$.Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2}{6} - (\frac{19}{2})^2$.$\sigma^2 = \frac{4 + 25 + 64 + 121 + 196 + 289}{6} - \frac{361}{4} = \frac{699}{6} - \frac{361}{4} = 116.5 - 90.25 = 26.25 = \frac{105}{4}$.Therefore,$8 \sigma^2 = 8 	imes \frac{105}{4} = 210$.

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 $\frac{19}{2}$ and their variance is $\sigma^2$,then $8 \sigma^2$ is equal to

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