QJEEM16Maadfb: JEE Main 2016 Mathematics - Class 11 JEEM16Maadfb Solution: If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d…

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