Step 1: Write the data set.
The data is $1, 1+d, 1+2d, \ldots, 1+100d$, which is an AP with 101 terms, first term $a_0=1$ and common difference $d > 0$.
Step 2: Find the mean.
Mean $\bar{x} = \frac{\text{Sum}}{101}$. Sum $= 101 \cdot 1 + d(0+1+2+\cdots+100) = 101 + d \cdot \frac{100 \cdot 101}{2} = 101 + 5050d$. So $\bar{x} = 1 + 50d$.
Step 3: Compute deviations from mean.
$|x_k - \bar{x}| = |(1+kd) - (1+50d)| = |k-50|d$ for $k = 0, 1, 2, \ldots, 100$.
Step 4: Compute the mean deviation.
Mean deviation $= \frac{1}{101} \sum_{k=0}^{100} |k-50|d = \frac{d}{101} \sum_{k=0}^{100} |k-50|$.
$\sum_{k=0}^{100} |k-50| = 2(1+2+\cdots+50) = 2 \cdot \frac{50 \cdot 51}{2} = 2550$.
Step 5: Set the mean deviation equal to 255.
$\frac{d \cdot 2550}{101} = 255$. So $d = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.
Step 6: Match with options.
$d = 10.1$ is option (1).
\[ \boxed{d = 10.1} \]