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(C) Let the $100$ consecutive positive integers be $a_1, a_1+1, a_1+2, \ldots, a_1+99$.The mean $\bar{x}$ is given by:$\bar{x} = \frac{1}{100} \sum_{i

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$\mathbb{N}$

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(C) Let the $100$ consecutive positive integers be $a_1, a_1+1, a_1+2, \ldots, a_1+99$.The mean $\bar{x}$ is given by:$\bar{x} = \frac{1}{100} \sum_{i=0}^{99} (a_1+i) = a_1 + \frac{1}{100} 	imes \frac{99 	imes 100}{2} = a_1 + 49.5$.The mean deviation about the mean is:$MD = \frac{1}{100} \sum_{i=0}^{99} |(a_1+i) - (a_1+49.5)| = \frac{1}{100} \sum_{i=0}^{99} |i - 49.5|$.This sum is $|0-49.5| + |1-49.5| + \ldots + |99-49.5|$.$= 49.5 + 48.5 + \ldots + 0.5 + 0.5 + \ldots + 48.5 + 49.5$.$= 2 	imes (0.5 + 1.5 + \ldots + 49.5) = 2 	imes \frac{50}{2} (0.5 + 49.5) = 50 	imes 50 = 2500$.Thus,$MD = \frac{2500}{100} = 25$.Since the mean deviation is independent of $a_1$,the condition holds for all natural numbers $a_1 \in \mathbb{N}$.Therefore,$S = \mathbb{N}$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a_1, a_2, a_3, \ldots, a_{100}$ is $25$. Then $S$ is

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