If the mean deviation of the numbers $1$,

$1 + d, ...., 1 + 100d$ from their mean is $255$, then $d$is equal to

$\begin{array}{l}\overline{x}=\frac{1}{101}[1+\mid (1+d)+(1+2d)+\dots ..+(1+100d\left)\right]\\ =\frac{1}{101}\times \frac{101}{2}[1+(1+100d\left)\right]=1+50d\end{array}$

$\therefore$   Mean deviation from mean 

$\begin{array}{l}=\frac{1}{101}\left[\right|1-(1+50d)|+\mid 1+d-\\ (1+50d)|+\dots ..+|1+100d-(1+50d)\mid ]\\ =\frac{2\left|d\right|}{101}\{1+2+\dots .+50]\end{array}$

$=\frac{2\left|d\right|}{101}\frac{50\left(51\right)}{2}=\frac{2550}{101}\left|d\right|$

Now, $\frac{2550}{101}\left|d\right|=255\Rightarrow \left|d\right|=10.1$ 

Thus, we may take $d = 10.1$.