JEE Main 2023 (1 February Evening Shift)

Question

Let $9=x_{1} < x_{2} < \ldots < x_{7}$ be in an A.P. with common difference d. If the standard deviation of $x_{1}, x_{2}..., x_{7}$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_{6}$ is equal to :

Accepted Answer

$34$

Answer

$2\!\left(9+\dfrac{8}{\sqrt{7}}\right)$

Answer

$25$

Answer

$18\!\left(1+\dfrac{1}{\sqrt{3}}\right)$

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