Question

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

• A. $11.1$
• B. $20.0$
• C. $10.1$
• D. $20.2$

Answer 1

Answer: (C)

$Mean\left(\bar{x}\right)=\frac{sumofquantities}{n}=\frac{\frac{n}{2}\left(a+l\right)}{n}=\frac{1}{2}\left[1+1+100d\right]=1+50d$
$M.D.=\frac{1}{n}\sum \left|x_i-\bar{x}\right|\Rightarrow 255=\frac{1}{101}\left[50d+49d+48d+....+d+0+d+.....+50d\right]=\frac{2d}{101}\left[\frac{50\times 51}{2}\right]$
$\Rightarrow d=\frac{255\times 101}{50\times 51}=10.1$