Question

Consider the set \( S = \{2, 3, 4, \ldots, 2n+1\} \), 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 \( X - Y \)?

• A. 0
• B. 1
• C. \(\frac{1}{2}\)
• D. \(\frac{n+1}{2n}\)
• E. 2008

Hint:

For consecutive odd or even sequences, the average is simply the midpoint between the first and last term.

Answer 1

The Correct Option is B

The set \(S = \{2, 3, 4, \ldots, 2n+1\}\) contains both odd and even numbers. Odd integers in \(S\) are \(\{3, 5, 7, \ldots, 2n+1\}\). This is an arithmetic sequence with the first term \(a_1 = 3\) and the last term \(a_l = 2n+1\), with a common difference \(d = 2\). The number of terms \(k\) is found by solving \(a_k = 3 + (k-1) \cdot 2 = 2n+1\). Simplifying, \(2k - 2 = 2n - 2 \Rightarrow k = n\). The average \(X\) of odd integers is given by:

\(X = \frac{\text{sum of odd numbers}}{n} = \frac{\frac{n}{2}(3 + 2n + 1)}{n}=\frac{3 + 2n + 1}{2} = n + 2\) 

Even integers in \(S\) are \(\{2, 4, 6, \ldots, 2n\}\) with first term \(a_1 = 2\) and last term \(a_m = 2n\), common difference \(d = 2\). The number of terms \(m\) is calculated similarly: \(a_m = 2 + (m-1) \cdot 2 = 2n\), giving \(2m - 2 = 2n - 2 \Rightarrow m = n\). The average \(Y\) of even integers is:

\(Y = \frac{\text{sum of even numbers}}{n} = \frac{\frac{n}{2}(2 + 2n)}{n} = \frac{2 + 2n}{2} = n + 1\)

The difference \(X - Y\) is:

\(X - Y = (n + 2) - (n + 1) = 1\)