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(A) The given numbers are $a, 2a, 3a, \dots, 50a$. The number of terms $n = 50$.Since $n$ is even,the median $M$ is the average of the $25^{th}$ and $

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$4$

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(A) The given numbers are $a, 2a, 3a, \dots, 50a$. The number of terms $n = 50$.Since $n$ is even,the median $M$ is the average of the $25^{th}$ and $26^{th}$ terms:$M = \frac{25a + 26a}{2} = 25.5a$.The mean deviation about the median is given by $M.D. = \frac{1}{n} \sum_{i=1}^{n} |x_i - M|$.$M.D. = \frac{1}{50} \sum_{i=1}^{50} |ia - 25.5a| = \frac{|a|}{50} \sum_{i=1}^{50} |i - 25.5|$.Expanding the sum: $\sum_{i=1}^{50} |i - 25.5| = |1 - 25.5| + |2 - 25.5| + \dots + |50 - 25.5|$.$= 24.5 + 23.5 + \dots + 0.5 + 0.5 + \dots + 23.5 + 24.5$.$= 2 	imes (0.5 + 1.5 + \dots + 24.5)$.This is an arithmetic progression with $25$ terms where the first term $a_1 = 0.5$ and last term $l = 24.5$.Sum $= 2 	imes \left[ \frac{25}{2} (0.5 + 24.5) ight] = 25 	imes 25 = 625$.Given $M.D. = 50$,so $\frac{|a|}{50} 	imes 625 = 50$.$|a| 	imes 12.5 = 50 \Rightarrow |a| = \frac{50}{12.5} = 4$.

$x_i$	$2$	$4$	$6$	$8$	$10$
$f_i$	$1$	$2$	$3$	$2$	$1$

Marks obtained	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Number of students	$10$	$8$	$12$	$9$	$11$

Marks obtained	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
No of students	$2$	$3$	$8$	$14$	$8$	$3$	$2$

If the mean deviation about the median of the numbers $a, 2a, 3a, \dots, 50a$ is $50$,then what is the value of $|a|$?

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