Bunuel wrote:
Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is a positive integer. If X is the average of the odd integers in set S and Y is the average of the even integers in set S, what is the value of (X − Y)?
(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2
In set S, there are 2n + 1 integers: n + 1 of them are odd and n of them are even. Recall that the sum of the first k positive odd integers is k^2, so the sum of the first n + 1 positive odd integers is (n + 1)^2. Likewise, the sum of first k positive even integers is k(k + 1), so the sum of the first n positive even integers is n(n + 1). Therefore, we have:
X = (1 + 3 + … + (2n + 1))/(n + 1) = (n + 1)^2/(n + 1) = n + 1
and
X = (2 + 4 + … + 2n))/n = n(n + 1)/n = n + 1
Therefore, the difference is X - Y = (n + 1) - (n + 1) = 0.
Alternate Solution:
The odd numbers in set S are 1, 3, 5, … , 2n + 1. Since these numbers form an evenly spaced sequence of integers, the average of these numbers is equal to the average of the first and last terms, which is ((2n + 1) + 1) / 2 = (2n + 2) / 2 = n + 1.
Similarly, the even numbers in set S, which are 2, 4, 6, … , 2n, also form an evenly spaced sequence of integers. Thus, the average of the even numbers in set S is (2n + 2) / 2 = n + 1.
We see that the difference is (n + 1) - (n + 1) = 0.
Second Alternate Solution:
We know that the number (2n + 1) must be odd. Therefore, the set {1, 2, 3, 4, …, 2n + 1} will begin and end with an odd integer. So let’s assume that set S = {1,2,3,4,5}. The average of the odd numbers is (1 + 3 + 5)/3 = 3. The average of the even numbers is (2 + 4)/2 = 3.
Since the two averages are equal, their difference will be 3 - 3 = 0. This difference of 0 will hold for any chosen positive integer value of n.
Answer: A
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