Given numbers are,

1, 1 + d, 1 + 2d . . . . . 1 + 100d

$$ \therefore $$   Total 101 number are present.

$$ \therefore $$   n = 101

$$ \therefore $$   mean $$\left( {\overline x } \right)$$ = $${{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}$$

=   $${1 \over {101}} \times {{101} \over 2}$$ [1 + (1 + 100d)]

=   1 + 50d

$$ \therefore $$   Mean deviation from mean

=  $${1 \over {101}}$$ $$\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.$$

          $$\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]$$

=   $${{2\left| d \right|} \over {101}}$$ ( 1 + 2 + 3 + . . . . . + 50)

=   $${{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}$$

=   $${{2550} \over {101}}\left| d \right|$$

From question,

$${{2550} \over {101}}\left| d \right|$$ = 255

$$ \Rightarrow $$   $$\left| d \right|$$ = 10.1