Let 9

Question

(B) The terms are $x_1, x_2, \ldots, x_7$ in $A.P.$ with $x_1 = 9$ and common difference $d > 0$.These can be written as $9, 9+d, 9+2d, \ldots, 9+6d$.

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(B) The terms are $x_1, x_2, \ldots, x_7$ in $A.P.$ with $x_1 = 9$ and common difference $d > 0$.These can be written as $9, 9+d, 9+2d, \ldots, 9+6d$.The mean $\overline{x} = \frac{1}{7} \sum_{i=0}^{6} (9+id) = 9 + \frac{d}{7} 	imes \frac{6 	imes 7}{2} = 9 + 3d$.The variance $\sigma^2$ is given by $\frac{1}{n} \sum x_i^2 - (\overline{x})^2$.Since standard deviation $\sigma = 4$,$\sigma^2 = 16$.Using the property of variance for an $A.P.$,$\sigma^2 = d^2 \frac{n^2-1}{12}$.Here $n=7$,so $\sigma^2 = d^2 \frac{49-1}{12} = d^2 \frac{48}{12} = 4d^2$.Given $4d^2 = 16$,we get $d^2 = 4$,so $d = 2$.Now,$\overline{x} = 9 + 3(2) = 15$.$x_6 = 9 + 5d = 9 + 5(2) = 19$.Therefore,$\overline{x} + x_6 = 15 + 19 = 34$.

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2, \ldots, x_7$ is $4$ and the mean is $\overline{x}$,then $\overline{x} + x_6$ is equal to:

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