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

Let $\overline{X}$ denote the mean of the given numbers.

Then,

  $\overline{X}=\frac{1+(1+d)+(1+2d)+\dots +(1+100d)}{101}$

$\Rightarrow \overline{X}=\frac{\left.\frac{101}{2}\{1+(1+100d\left)\right)\right\}}{101}=1+50d$

$\therefore$   Mean deviation $=\frac{1}{101}\left\{\sum _{r=0}^{100} \left|\right(1+rd)-(1+50d\left)\right|\right\}$

$\Rightarrow$ Mean deviation $=\frac{1}{101}\sum _{r=0}^{100} |r-50|$

$\Rightarrow$Mean deviation $=\frac{{\displaystyle d}}{{\displaystyle 101}}\times 2\sum _{r=1}^{50}r$

$\Rightarrow$Mean deviation $=\frac{2d}{101}\times \frac{50\times 51}{2}=\frac{50\times 51}{101}d.$

It is given that the mean deviation is $255.$

$\therefore 255=\frac{50\times 51}{101}d\Rightarrow d=10.1$