Let 9

Question

$9=x_1 < x_2 < \ldots \ldots < x_7$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{	ext {new }}=

Vedclass · Accepted Answer

$34$

Vedclass · Answer

$9=x_1 < x_2 < \ldots \ldots < x_7$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{	ext {new }}=\frac{21 d }{7}=3 d$

$16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots+6^2ight) d^2-9 d^2$

$=\frac{1}{7}\left(\frac{6 	imes 7 	imes 13}{6}ight) d ^2-9 d ^2$

$16=4 d^2$

$d^2=4$

$d=2$

$\bar{x}+x_6=6+9+10+9$

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

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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