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 $n$. $MD = \frac{|d|}{2n+1} \sum_{j=-n}^{n} |j| = \frac{|d|}{2n+1} \left( 2 \sum_{j=1}^{n} j \right) = \frac{|d|}{2n+1} \cdot 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)|d|}{2n+1}$.

$x_i$	$3$	$9$	$17$	$23$	$27$
$f_i$	$8$	$10$	$12$	$9$	$5$

$x$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

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

Similar Questions

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In a series of $3n$ observations,if $n$ observations are equal to $a$ and the remaining $2n$ observations are equal to $-2a$,then the mean deviation of the observations about their mean is:

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