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

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

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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