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(B) The given numbers are $k, 2k, 3k, \dots, 1000k$. Here $n = 1000$ (even).Median $X_M = \frac{(\frac{n}{2})k + (\frac{n}{2} + 1)k}{2} = \frac{500k +

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$4$

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(B) The given numbers are $k, 2k, 3k, \dots, 1000k$. Here $n = 1000$ (even).Median $X_M = \frac{(\frac{n}{2})k + (\frac{n}{2} + 1)k}{2} = \frac{500k + 501k}{2} = 500.5k$.Mean deviation about median $= \frac{1}{n} \sum_{i=1}^{n} |x_i - X_M| = \frac{1}{1000} \sum_{i=1}^{1000} |ik - 500.5k| = \frac{k}{1000} \sum_{i=1}^{1000} |i - 500.5|$.This sum is $2 	imes (0.5 + 1.5 + 2.5 + \dots + 499.5) = 2 	imes \frac{500}{2} (0.5 + 499.5) = 500 	imes 500 = 250000$.Mean deviation $= \frac{k 	imes 250000}{1000} = 250k$.Given $250k = 500$, so $k = 2$.Therefore, $k^{2} = 2^{2} = 4$.

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

If the mean deviation about the median of the numbers $k, 2k, 3k, \dots, 1000k$ is $500$, then $k^{2}$ is equal to :

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