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 $\frac{50 	imes 51 	imes d}{101} = 255$.$d = \frac{255 	imes 101}{50 	imes 51} = \frac{5 	imes 101}{50} = \frac{101}{10} = 10.1$.

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
No of Girls	$6$	$8$	$14$	$16$	$4$	$2$

${x_i}$	$2$	$5$	$6$	$8$	$10$	$12$
${f_i}$	$2$	$8$	$10$	$7$	$8$	$5$

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:

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