Consider the set

*S*= {2, 3, 4, ...., 2*n*+ 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*X*–*Y*?
(1) 0

(2) 1

(3)

*n*/2*(4)*

*n*+1/2

*n*

(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, S

_{n}=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, S

_{n}=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)

= {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****Answer (2)**
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