Question

Let $9 =$ $x_1​<x_2​<…<x_7​$ be in an A.P. with common difference d. If the standard deviation of $x_1​,x_2​ …,x_7$​ 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)

34

Answer 2

The correct answer is (C) : $34$
$9=x_1​<x_2​<……<x_7​$
$9,9+d,9+2d,…….9+6d$
$0,d,2d,…….6d$
$x_{new} ​=\frac {21d​}{7}=3d$

$16=\frac 17​(0^2+1^2+…….+6^2)d^2−9d^2$

$16=\frac 17​(\frac {6×7×13}{6}​)d^2−9d^2$
$16 = 13d^2 - 9d^2$
$16=4d^2$
$d^2=4$
$d=2$
Now,
$\bar x+x_6​=6+9+10+9$
$\bar x+x_6= 34$