Question

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 $\bar{x}$, then $\bar{x}+x_6$ is equal to :

• A. $18\left(1+\frac{1}{\sqrt{3}}\right)$
• B. $2\left(9+\frac{8}{\sqrt{7}}\right)$
• C. 34
• D. 25

Answer 1

Answer: (C)

$ 9=x_1< x_2< \ldots \ldots< x_7 $
$ 9,9+d, 9+2 d, \ldots \ldots .9+6 d $
$ 0, d, 2 d, \ldots \ldots .6 d$
$ \overline{ x }_{\text {new }}=\frac{21 d }{7}=3 d $
$ 16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots .+6^2\right) d ^2-9 d ^2$
$=\frac{1}{\not 7}\left(\frac{6 \times \not 7 \times 13}{\not 6}\right) d ^2-9 d ^2$
$ 16=4 d^2 $
$ d^2=4$
$ d=2$
$\bar{x}+x_6=6+9+10+9$