Question 174 of 208MediumMCQ
If the mean deviation of the numbers 1,1+d,1+2d,…,1+100d from their mean is 255, then d is equal to: Solutions · 1
Official solution
Mean \[\,\,\,\,\left( {\overline x } \right) = {{Sum\,\,of\,\,numbers} \over n}\]
\[ = {{{n \over 2}\left( {a + l} \right)} \over n}\]
\[ = {1 \over 2}\left( {1 + l + 100d} \right)\]
\[ = 1 + 50\,d.\]
Mean deviation (M.D) \[ = {1 \over n}\sum\limits_{i = 1}^{101} {\left| {{x_i} - \overline x } \right|} \]
\[ = {1 \over {101}}\][ 50d + 49d + 48d + .......d + 0 + ..... + 50d]
\[ = {1 \over {101}}.2d\left( {1 + 2 + .... + 50} \right)\]
\[ = {1 \over {101}}.2d.{{50 \times 51} \over 2}\]
\[ = {{50 \times 51\,d} \over {101}}\]
Given that M.D = 255
\[\therefore\,\,\,\] \[{{50 \times 51\,d} \over {101}} = 255\]
\[ \Rightarrow \,\,\,d = {{101 \times 255} \over {51 \times 50}}\]
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