Given AP: \(1,\,1+d,\,1+2d,\,\ldots,\,1+100d\) (total \(101\) terms). The mean is the middle term \(1+50d\).

Mean Deviation (about mean): For \(2n+1\) terms \(a,a+d,\ldots,a+2nd\), $$\text{MD}=\frac{d\,n(n+1)}{2n+1}.$$

Here \(2n+1=101 \Rightarrow n=50\). So $$\text{MD}=\frac{d\cdot 50\cdot 51}{101}=\frac{2550}{101}\,d.$$

Given \(\text{MD}=255\). Solve for \(d\): $$\frac{2550}{101}\,d=255 \;\Rightarrow\; d=\frac{255\times 101}{2550}=10.1.$$

Answer: \(d=10.1\).