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

Vedclass · Accepted Answer

$4$

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(B) The given numbers are $a, 2a, 3a, \dots, 50a$. The total number of terms is $n = 50$.Since $n$ is even,the median is the average of the $25^{th}$ and $26^{th}$ terms:$	ext{Median} = \frac{25a + 26a}{2} = 25.5a$.Mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - 	ext{Median}| = 50$.$\frac{1}{50} \sum_{i=1}^{50} |ia - 25.5a| = 50$.$\sum_{i=1}^{50} |i - 25.5| |a| = 2500$.$|a| (|25.5 - 1| + |25.5 - 2| + \dots + |25.5 - 50|) = 2500$.$|a| (24.5 + 23.5 + \dots + 0.5 + 0.5 + \dots + 24.5) = 2500$.$|a| 	imes 2 	imes (0.5 + 1.5 + \dots + 24.5) = 2500$.$|a| 	imes 2 	imes \frac{25}{2} (0.5 + 24.5) = 2500$.$|a| 	imes 25 	imes 25 = 2500$.$|a| 	imes 625 = 2500$.$|a| = 4$.

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

If the mean deviation about the median of the numbers $a, 2a, 3a, \dots, 50a$ is $50$,then $|a|$ equals:

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