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

Correct answer: (A) 10.1. Mean x-=1/1011+1+d+1+2d….1+100d =1/101 ×101/2 1+1+100d=1+50d Mean deviation from mean =1/1011-1+50d+1+d-1+50d….1+100d-1+50d =2d/101 1+2+3….+50 =2d/101 (50 50+1)/2 =2550/101d=255 ⇒d=10.1

Q41 / 90 · JEE Main 2016 (09 Apr Online) · Mathematics

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

Correct answer
A
$10.1$
B
$5.05$
C
$20.2$
D
$10$

Solution

Mean $\bar{x} = \frac{1}{101} [1 + (1 + d) + (1 + 2 d)] \dots . (1 + 100 d)$

$= \frac{1}{101} \times \frac{101}{2} [1 + (1 + 100 d)] = 1 + 50 d$

Mean deviation from mean

$= \frac{1}{101} \{|1 - (1 + 50 d)| + |(1 + d) - (1 + 50 d)| \dots . |[1 + 100 d] - (1 + 50 d)|\}$

$= \frac{2 |d|}{101} (1 + 2 + 3 \dots . + 50)$

$= \frac{2 |d|}{101} \frac{(50) (50 + 1)}{2}$

$= \frac{2550}{101} |d| = 255$

$\Rightarrow |d| = 10 .1$

Hence, the correct answer is option: 'A'.

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