If the mean deviation of the data 1, 1+d, 1+2 | vedclass

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(A) The given data is an arithmetic progression with $n = 101$ terms,where the first term $a = 1$ and the common difference is $d$. The mean $\bar{x}

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$10.1$

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(A) The given data is an arithmetic progression with $n = 101$ terms,where the first term $a = 1$ and the common difference is $d$. The mean $\bar{x} = \frac{1}{101} \sum_{i=0}^{100} (1 + id) = 1 + \frac{d}{101} 	imes \frac{100 	imes 101}{2} = 1 + 50d$. The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=0}^{100} |x_i - \bar{x}|$. $MD = \frac{1}{101} \sum_{i=0}^{100} |(1 + id) - (1 + 50d)| = \frac{1}{101} \sum_{i=0}^{100} |(i - 50)d| = \frac{d}{101} \sum_{i=0}^{100} |i - 50|$. The sum $\sum_{i=0}^{100} |i - 50| = |0-50| + |1-50| + \ldots + |50-50| + \ldots + |100-50| = 50 + 49 + \ldots + 0 + \ldots + 50 = 2 	imes \frac{50 	imes 51}{2} = 2550$. Given $MD = 255$,we have $255 = \frac{d}{101} 	imes 2550$. $255 = d 	imes \frac{2550}{101} \Rightarrow d = \frac{255 	imes 101}{2550} = \frac{101}{10} = 10.1$.

$x_i$	$5, 15, 25, 35, 45$
$f_i$	$8, 48, 56, 32, 16$

If the mean deviation of the data $1, 1+d, 1+2d, \ldots, 1+100d$ $(d > 0)$ from their mean is $255$,then $d$ is equal to:

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