The mean deviation of the numbers a, a+d, a+2 | vedclass

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(D) The given numbers are $a, a+d, a+2d, \ldots, a+2nd$. This is an arithmetic progression with $N = 2n+1$ terms. The mean $\bar{x}$ is the middle ter

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

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(D) The given numbers are $a, a+d, a+2d, \ldots, a+2nd$. This is an arithmetic progression with $N = 2n+1$ terms. The mean $\bar{x}$ is the middle term: $\bar{x} = a + nd$. The mean deviation about the mean is given by $MD = \frac{1}{N} \sum_{i=0}^{2n} |x_i - \bar{x}|$. $MD = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |k-n||d|$. Let $j = k-n$. As $k$ goes from $0$ to $2n$,$j$ goes from $-n$ to $n$. $MD = \frac{|d|}{2n+1} \sum_{j=-n}^{n} |j| = \frac{|d|}{2n+1} \left( 2 \sum_{j=1}^{n} j \right) = \frac{|d|}{2n+1} \cdot 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)|d|}{2n+1}$.

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

The mean deviation of the numbers $a, a+d, a+2d, \ldots, a+2nd$ from their mean is equal to

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