Let a_1, a_2, dots, a_{101} be a group of rea | vedclass

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(B) The mean square deviation of a set of numbers is minimum when taken about the arithmetic mean. Given that the mean square deviation is minimum abo

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$a_{51}$

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(B) The mean square deviation of a set of numbers is minimum when taken about the arithmetic mean. Given that the mean square deviation is minimum about $a_{51}$,it implies that the mean of the group is $a_{51}$. Since the sequence $a_1, a_2, \dots, a_{101}$ is strictly decreasing $(a_i > a_{i+1})$,the median of the group is the middle term,which is $a_{51}$. For a symmetric distribution or a sequence where the mean,median,and mode coincide,the mode is also $a_{51}$.

Let $a_1, a_2, \dots, a_{101}$ be a group of real numbers such that $a_i > a_{i+1}$ for all values of $i$ and the mean square deviation of the group is minimum about the number $a_{51}$. Then the mode of the group will be:

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