QJEEM09Maf2f5: JEE Main 2009 Mathematics - Class 11 JEEM09Maf2f5 Solution: If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 +…

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

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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}}

= 10.1

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