Let 9 = x _ 1 < x _ 2 < < x _ 7 be in an A.P. with common...

Given,

$9 = {x}_{1} < {x}_{2} < \ldots < {x}_{7}$ be in an A.P. with common difference $d$

And the standard deviation of ${x}_{1}, {x}_{2} \ldots , {x}_{7}$ is 4 and the mean is $\overset{¯}{x}$,

Now solving, $9 = {x}_{1} < {x}_{2} < \ldots \ldots < {x}_{7}$ which is an A.P, we get,

$9, 9 + d, 9 + 2d, \ldots \ldots \ldots . 9 + 6d$

Now subtracting $9$ from the series we get,

$0, d, 2d, . . . . . . . 6d$

So, mean will be $\overset{¯}{x}new = \dfrac{21d}{7} = 3d$

Now using the formula of variance we get,

${\sigma}^{2} = \mathop{∑}_{i = 1}^{n}{\left({x}_{i}\right)}^{2} - {\left(\overset{¯}{x}\right)}^{2}$

$⇒ 16 = \dfrac{1}{7}\left({0}^{2} + {1}^{2} + \ldots \ldots . . + {6}^{2}\right){d}^{2} - 9{d}^{2}$

$⇒ 16 = \dfrac{1}{7}\left(\dfrac{6 \times 7 \times 13}{6}\right){d}^{2} - 9{d}^{2}$

$⇒ 16 = 4{d}^{2}$

$⇒ {d}^{2} = 4$

$⇒ d = 2$

So, mean $\overset{¯}{x} = \dfrac{9 + 9 + d + 9 + 2d \ldots \ldots \ldots . 9 + 6d}{7} = 15$

Hence, $\overset{¯}{x} + {x}_{6} = 15 + 9 + 5d = 15 + 9 + 10 = 34$