Let $9={\mathrm{x}}_1<{\mathrm{x}}_2<\dots ..,<{\mathrm{x}}_7$ be in an A.P. with common difference $d$. If the standard deviation of ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots .,{\mathrm{x}}_7$ is 4 and the mean is $\overline{x},$ $\text{ then }\overline{\mathrm{x}}+{\mathrm{x}}_6$ is equal to :

${\mathrm{x}}_1=9, {\mathrm{x}}_2=9+\mathrm{d}, {\mathrm{x}}_3=9+2\mathrm{d}, {\mathrm{x}}_4=9+3\mathrm{d}, {\mathrm{x}}_5=9+4\mathrm{d}, {\mathrm{x}}_6=9+5\mathrm{d}, {\mathrm{x}}_7=9+6\mathrm{d}$
Mean $\overline{\mathrm{x}}=\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4+{\mathrm{x}}_5+{\mathrm{x}}_6+{\mathrm{x}}_7}{7}$
$=\frac{63+21\mathrm{d}}{7}=9+3\mathrm{d}$
Variance,$=\sum _{\mathrm{i}=1}^7 \frac{{\left(\overline{\mathrm{x}}-{\mathrm{x}}_{\mathrm{i}}\right)}^2}{\mathrm{N}}$
$=\frac{2\left(9{\mathrm{d}}^2+4{\mathrm{d}}^2+{\mathrm{d}}^2\right)}{7}=4{\mathrm{d}}^2$
Given standard deviation $\mathrm{\sigma }=4$
Variance ${\mathrm{\sigma }}^2=16$
$4{\mathrm{d}}^2=16\Rightarrow \mathrm{d}=2$
Mean $\overline{x}=9+3\left(2\right)=15$
${\mathrm{x}}_6=9+5\mathrm{d}=19\overline{x}+{\mathrm{x}}_6=15+19=34$