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(A) The given numbers are in an arithmetic progression: $1, 1+d, 1+2d, \dots, 1+100d$. Here,the number of terms is $N = 101$. The mean of these terms

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(A) The given numbers are in an arithmetic progression: $1, 1+d, 1+2d, \dots, 1+100d$. Here,the number of terms is $N = 101$. The mean of these terms is $\bar{x} = \frac{1}{101} \sum_{i=0}^{100} (1 + id) = 1 + \frac{d}{101} 	imes \frac{100 	imes 101}{2} = 1 + 50d$. The mean deviation about the mean is given by $M.D. = \frac{1}{N} \sum_{i=0}^{100} |(1+id) - (1+50d)| = \frac{1}{101} \sum_{i=0}^{100} |(i-50)d| = \frac{|d|}{101} \sum_{i=0}^{100} |i-50|$. Calculating the sum: $\sum_{i=0}^{100} |i-50| = (50+49+\dots+1) + 0 + (1+2+\dots+50) = 2 	imes \frac{50 	imes 51}{2} = 2550$. Thus,$M.D. = \frac{|d|}{101} 	imes 2550 = 255$. $|d| = \frac{255 	imes 101}{2550} = \frac{101}{10} = 10.1$.

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

$x_i$	$10$	$30$	$50$	$70$	$90$
$f_i$	$4$	$24$	$28$	$16$	$8$

$x_i$	$15$	$21$	$27$	$30$	$35$
$f_i$	$3$	$5$	$6$	$7$	$8$

$x$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

If the mean deviation about the mean of the numbers $1, 1 + d, 1 + 2d, \dots, 1 + 100d$ is $255$,then what is the value of $|d|$?

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