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Question: If the mean deviation about the median of the numbers \(a,2a,3a,............49a,50a\) is \(50\) , th...

If the mean deviation about the median of the numbers a,2a,3a,............49a,50aa,2a,3a,............49a,50a is 5050 , then |a| is equal to?
A) 22
B) 33
C) 44
D) 55

Explanation

Solution

We have to find mean deviation about median of all given numbers. So, we will first find the median of all numbers and then we will find the mean deviation about the median of all numbers using the same method like we find the mean of nn given numbers.

Complete step-by-step answer:
We can see that all the numbers are already arranged in ascending order and the number of terms in the set is 50 (which is even).
The median is the mid element of the increasing order set of a given set of numbers.
But when the number of elements in the set are even then the median will be average of (n2)th{\left( {\dfrac{n}{2}} \right)^{th}} element and (n2+1)th{\left( {\dfrac{n}{2} + 1} \right)^{th}} element of increasing order set of given set.
So the median of given will be (n2)thterm+(n2+1)thterm2\dfrac{{{{\left( {\dfrac{n}{2}} \right)}^{th}}term + {{\left( {\dfrac{n}{2} + 1} \right)}^{th}}term}}{2}, that is,
25a+26a2=25.5a\dfrac{{25a + 26a}}{2} = 25.5a
Now, the deviation of first number about median will be a25.5a=24.5a\left| {a - 25.5a} \right| = 24.5a
For second number, it will be 2a25.5a=23.5a\left| {2a - 25.5a} \right| = 23.5a
Similarly, we can find it for all terms.
Mean deviation of all deviations about median will be (sum of all deviations) divided by (num of terms), that is,
24.5a+23.5a+...............+23.5a+24.5a50\dfrac{{24.5a + 23.5a + ............... + 23.5a + 24.5a}}{{50}}
On simplifying this, we get,
(2×a×(24.5+23.5+..........+1.5+0.5))50\dfrac{{\left( {2 \times \left| a \right| \times (24.5 + 23.5 + .......... + 1.5 + 0.5)} \right)}}{{50}} --------(1)
So number of terms in series 0.5+1.5+2.5+.......+24.50.5 + 1.5 + 2.5 + ....... + 24.5 , can be found using nth{n^{th}} terms formula which is
nthterm=a+(n1)d{n^{th}}term = a + (n - 1)d
Where aa is first term and dd is common difference of series
24.5=1.5+(n1)×124.5 = 1.5 + (n - 1) \times 1
On solving,
n1=23n - 1 = 23
So number of terms in series 0.5+1.5+2.5+.......+24.50.5 + 1.5 + 2.5 + ....... + 24.5 is,
n=24n = 24
Now a+(a+d)+........+(a+(n1)d)=n2[2a+(n1)d]a + \left( {a + d} \right) + ........ + \left( {a + (n - 1)d} \right) = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] (sum of series formula where aa is first term and dd is common difference of series and nn is number of terms in series), using this in equation 11 , we get,
(2×a×252×(0.5+24.5))50\dfrac{{\left( {2 \times \left| a \right| \times \dfrac{{25}}{2} \times (0.5 + 24.5)} \right)}}{{50}}
Which simplifies to,
25×a×2550\dfrac{{25 \times \left| a \right| \times 25}}{{50}}
Which further simplifies to
25×a2\dfrac{{25 \times \left| a \right|}}{2}
Now, this mean deviation is given to us as 5050 in question,
So, we get
25a2=50\dfrac{{25\left| a \right|}}{2} = 50
On solving this, we get,
a=4\left| a \right| = 4

So the correct option is (C).

Note: We noted that all terms were already arranged in ascending order otherwise we would first arrange them in ascending order. Otherwise, the median of the set not in increasing order found with the applied formula of median for even number of terms would be incorrect.