The mean deviation from the mean of the serie | vedclass

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(B) The given series is an arithmetic progression with $2n+1$ terms: $a, a+d, a+2d, \ldots, a+2nd$. The mean $m$ of the series is the middle term: $m

Vedclass · Accepted Answer

$\frac{n(n+1)d}{2n+1}$

Vedclass · Answer

(B) The given series is an arithmetic progression with $2n+1$ terms: $a, a+d, a+2d, \ldots, a+2nd$. The mean $m$ of the series is the middle term: $m = a + nd$. The mean deviation from the mean is given by $\frac{1}{2n+1} \sum_{i=0}^{2n} |x_i - m|$. Substituting the terms: $	ext{Mean Deviation} = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |(k-n)d|$. This sum is $\frac{d}{2n+1} [| -n | + | -(n-1) | + \ldots + | 0 | + \ldots + | n |]$. The sum inside the bracket is $2 	imes (1 + 2 + \ldots + n) = 2 	imes \frac{n(n+1)}{2} = n(n+1)$. Thus,the mean deviation is $\frac{n(n+1)d}{2n+1}$. Therefore,option $B$ is correct.

The mean deviation from the mean of the series $(a), (a+d), (a+2d), \ldots, (a+2nd)$ is

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