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QJEEM11Ma21e3: JEE Main 2011 Mathematics - Class 11 JEEM11Ma21e3 Solution: If the mean deviation about the median of the numbers a,…

If the mean deviation about the median of the numbers a, 2a,........., 50a is 50, then |a| equals

Multiple Choice Options

  • A 4
  • B 5
  • C 2
  • D 3

Step-by-Step Verified Solution

NOTE :

If total no of terms are even then median

=12= {1 \over 2} [ n2{n \over 2} th term +(n2+1)+ \left( {{n \over 2} + 1} \right) th term]

Here total terms =50,= 50, which is even

∴\therefore    \,\,\, Median =12= {1 \over 2} [ 502{{50} \over 2} th term +(502+1)+ \left( {{{50} \over 2} + 1} \right) th term]

=12= {1 \over 2} [ 2525 th term ++ 2626 th term ]

=12= {1 \over 2} [ 25a25a ++ 26a26a ]

=25.5a=25.5a

Mean deviation (M.D.) about the median

=∑i=150∣xi−Median∣N=50= {{\sum\limits_{i = 1}^{50} {\left| {{x_i} - Median} \right|} } \over N} = 50 (given)

∴\therefore 150[2×∣a∣×(0.5+1.5+2.5+....24.5)]=50{1 \over {50}}\left[ {2 \times \left| a \right| \times \left( {0.5 + 1.5 + 2.5 + ....24.5} \right)} \right] = 50

⇒2∣a∣×252×25=2500\Rightarrow 2\left| a \right| \times {{25} \over 2} \times 25 = 2500

⇒∣a∣=250025×25=4\Rightarrow \left| a \right| = {{2500} \over {25 \times 25}} = 4

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