If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is 255, then $d =$
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The mean deviation of an arithmetic progression of $(2n+1)$ terms with common difference $d$ is $\frac{n(n+1)}{2n+1}|d|$. Here, $n=50$, so $\frac{50 \times 51}{101}|d| = 255 \implies |d| = 10.1$.