If the mean deviation of the numbers 1, 1 + d | vedclass

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(A) The given numbers are $1, 1+d, 1+2d, \dots, 1+100d$. This is an arithmetic progression with $n = 101$ terms.The mean $\bar{x}$ is given by:$\bar{x

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(A) The given numbers are $1, 1+d, 1+2d, \dots, 1+100d$. This is an arithmetic progression with $n = 101$ terms.The mean $\bar{x}$ is given by:$\bar{x} = \frac{1}{101} \sum_{k=0}^{100} (1 + kd) = \frac{1}{101} [101 + d 	imes \frac{100 	imes 101}{2}] = 1 + 50d$.The mean deviation from the mean is:$MD = \frac{1}{101} \sum_{k=0}^{100} |(1 + kd) - (1 + 50d)| = \frac{1}{101} \sum_{k=0}^{100} |(k - 50)d| = \frac{|d|}{101} \sum_{k=0}^{100} |k - 50|$.Calculating the sum: $\sum_{k=0}^{100} |k - 50| = |0-50| + |1-50| + \dots + |50-50| + \dots + |100-50| = 50 + 49 + \dots + 0 + 1 + \dots + 50 = 2 	imes \frac{50 	imes 51}{2} = 2550$.Thus,$MD = \frac{|d|}{101} 	imes 2550 = 255$.$|d| = \frac{255 	imes 101}{2550} = \frac{101}{10} = 10.1$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

If the mean deviation of the numbers $1, 1 + d, . . . , 1 + 100d$ from their mean is $255,$ then a value of $d$ is

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