What is the mean deviation about the mean of | vedclass

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(B) The number of terms in the series is $N = 2n + 1$.The mean $\bar{x}$ is given by $\bar{x} = \frac{a + (a + d) + (a + 2d) + \dots + (a + 2nd)}{2n +

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

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(B) The number of terms in the series is $N = 2n + 1$.The mean $\bar{x}$ is given by $\bar{x} = \frac{a + (a + d) + (a + 2d) + \dots + (a + 2nd)}{2n + 1}$.Using the sum of an arithmetic progression,$\bar{x} = \frac{1}{2n + 1} \left[ \frac{2n + 1}{2} (a + a + 2nd) \right] = a + nd$.The mean deviation about the mean is $MD = \frac{1}{N} \sum_{i=1}^{N} |x_i - \bar{x}|$.$\sum |x_i - \bar{x}| = |a - (a + nd)| + |(a + d) - (a + nd)| + \dots + |(a + 2nd) - (a + nd)|$$= |-nd| + |(1 - n)d| + \dots + |0| + \dots + |nd| = 2|d| [n + (n - 1) + \dots + 1] = 2|d| \frac{n(n + 1)}{2} = n(n + 1)|d|$.Therefore,$MD = \frac{n(n + 1)|d|}{2n + 1}$.

$\text{Size}$	$20$	$21$	$22$	$23$	$24$
$\text{Frequency}$	$6$	$4$	$5$	$1$	$4$

$x$	$10$	$11$	$12$	$13$
$f$	$6$	$12$	$18$	$12$

What is the mean deviation about the mean of the series $a, a + d, a + 2d, \dots, a + 2nd$?

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