## Saturday, 26 November 2011

### Progressions-10 (CAT-2007)

Consider the set S = {2, 3, 4, ...., 2n + 1}, where n  is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of XY?
(1) 0
(2) 1
(3) n/2
(4) n+1/2n
(5) 2008
Solution follows here:

Solution:
Total numbers = last number – first number + 1 = (2n+1)-2+1 = 2n
Irrespective of the value of n, 2n+1 is an odd number =>
As the series is starting with even number and ending with odd number,
number of even numbers = number of of odd numbers = 2n/2 = n

Odd numbers = 3,5,7,.......2n+1
Sum of odd numbers = 3+5+7+.....+(2n+1)
It is an AP with initial number = a = 3 and common diff = d = 2
Applying sum to n terms in AP, Sn=n/2{2a+(n-1)d} =n/2{6+(n-1)2}
Average of odd numbers = X = [n/2{6+(n-1)2}]/n = {6+(n-1)2}/2     ---(1)

Even numbers = 2,4,6,.......2n
Sum of even numbers = 2+4+6+.....+2n
It is an AP with initial number = a = 2 and common diff = d = 2
Applying sum to n terms in AP, Sn=n/2[2a+(n-1)d] = n/2[4+(n-1)2]
Average of even numbers = Y = [n/2{4+(n-1)2}]/n
= {4+(n-1)2}/2        ---(2)
From (1) and (2), X – Y = {6+(n-1)2}/2 - {4+(n-1)2}/2 = (6-4)/2 = 1