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

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(B) The given series $1, 1+d, 1+2d, \dots, 1+100d$ is an $A.P.$ with $n = 101$ terms.The mean of this series is $\bar{x} = \frac{\sum_{r=0}^{100} (1+r

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$10.1$

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(B) The given series $1, 1+d, 1+2d, \dots, 1+100d$ is an $A.P.$ with $n = 101$ terms.The mean of this series is $\bar{x} = \frac{\sum_{r=0}^{100} (1+rd)}{101} = \frac{101 + d \frac{100 	imes 101}{2}}{101} = 1 + 50d$.The mean deviation from the mean is given by $\frac{1}{101} \sum_{r=0}^{100} |(1+rd) - (1+50d)| = \frac{1}{101} \sum_{r=0}^{100} |(r-50)d|$.Assuming $d > 0$,this becomes $\frac{d}{101} [\sum_{r=0}^{50} (50-r) + \sum_{r=51}^{100} (r-50)]$.$= \frac{d}{101} [ (50+49+\dots+0) + (1+2+\dots+50) ] = \frac{d}{101} [ 2 	imes \frac{50 	imes 51}{2} ] = \frac{50 	imes 51 	imes d}{101}$.Given the mean deviation is $255$,we have $\frac{50 	imes 51 	imes d}{101} = 255$.$d = \frac{255 	imes 101}{50 	imes 51} = \frac{5 	imes 101}{50} = \frac{101}{10} = 10.1$.

Height in $cm$	Number of boys
$95-105$	$9$
$105-115$	$13$
$115-125$	$26$
$125-135$	$30$
$135-145$	$12$
$145-155$	$10$

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

If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is $255$,then $d$ is equal to:

Similar Questions

Find the mean deviation about the mean for the following data:
Height in $cm$ Number of boys
$95-105$ $9$
$105-115$ $13$
$115-125$ $26$
$125-135$ $30$
$135-145$ $12$
$145-155$ $10$

If the mean of $4, 7, 2, 8, 6$ and $a$ is $7$,find the mean deviation about the median of these observations.

If $10$ is the mean of the data $2, 3, 5, 18, 17, 15, 13, x, 9, 7$,then the mean deviation of this data about its mean is

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

The mean deviation about the mean for the following data is

Class Interval $0-4$ $4-8$ $8-12$ $12-16$

Frequency $4$ $3$ $2$ $1$

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