Question 172 of 208MediumMCQ
If the mean deviation about the median of the numbers a,2a,…,50a is 50, then ∣a∣ equals Solutions · 1
Official solution
NOTE :
If total no of terms are even then median
\[ = {1 \over 2}\] [ \[{n \over 2}\]th term \[ + \left( {{n \over 2} + 1} \right)\] th term]
Here total terms =50, which is even
\[\therefore\] \[\,\,\,\] Median \[ = {1 \over 2}\] [ \[{{50} \over 2}\] th term \[ + \left( {{{50} \over 2} + 1} \right)\] th term]
\[ = {1 \over 2}\] [ 25 th term + 26 th term ]
\[ = {1 \over 2}\] [ 25a + 26a ]
=25.5a
Mean deviation (M.D.) about the median
\[ = {{\sum\limits_{i = 1}^{50} {\left| {{x_i} - Median} \right|} } \over N} = 50\] (given)
\[\therefore\] \[{1 \over {50}}\left[ {2 \times \left| a \right| \times \left( {0.5 + 1.5 + 2.5 + ....24.5} \right)} \right] = 50\]
\[ \Rightarrow 2\left| a \right| \times {{25} \over 2} \times 25 = 2500\]
\[ \Rightarrow \left| a \right| = {{2500} \over {25 \times 25}} = 4\] Answer the question to unlock