If the mean deviation of number $$1, 1+d, 1+2d, \ldots, 1+100d$$ from their mean is $$255$$, then the $$d$$ is equal to

The data form an arithmetic progression with first term $$a=1$$ and common difference $$d$$. There are $$n = 101$$ terms: $$1,\;1+d,\;1+2d,\ldots,1+100d$$.

Step 1: Compute the mean of the 101 terms
For an AP the mean equals the average of the first and last terms: $$\text{Mean} = \frac{1 + (1+100d)}{2} = 1 + 50d.$$

Step 2: List the deviations from the mean
Because $$n$$ is odd, the mean coincides with the middle (51-st) term $$1+50d$$. Hence the 101 deviations are $$-50d,\,-49d,\ldots,-d,\,0,\,d,\ldots,49d,\,50d.$$ Their absolute values are $$50d,\,49d,\ldots,d,\,0,\,d,\ldots,49d,\,50d.$$

Step 3: Mean deviation (about the mean)
Mean deviation is the arithmetic mean of the absolute deviations: $$\text{M.D.} = \frac{1}{101}\Bigl[2\bigl(d+2d+\cdots+50d\bigr)\Bigr].$$ The factor 2 accounts for symmetry on either side of the mean.

The sum of the first 50 positive integers is $$1+2+\cdots+50 = \frac{50 \times 51}{2} = 1275.$$ Therefore, $$\text{M.D.} = \frac{1}{101}\bigl[2 \times 1275 \times d\bigr] = \frac{2550\,d}{101}.$$

Step 4: Use the given mean deviation
It is given that the mean deviation equals $$255$$: $$\frac{2550\,d}{101} = 255.$$ Solve for $$d$$: $$d = 255 \times \frac{101}{2550} = 0.1 \times 101 = 10.1.$$

Hence $$d = 10.1.$$

Option C which is: $$10.1$$