Question 153 of 208MediumMCQ
If the mean deviation of the numbers 1,1+d,…,1+100d from their mean is 255, then a value of d is:
Solutions · 1
Official solution
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
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