An even number of consecutive integers results in what type of average?

When considering an even number of consecutive integers, the average can be analyzed through the properties of these numbers. Let's denote the first integer in the consecutive sequence as ( n ). The sequence would then include ( n, n+1, n+2, \ldots, n+k ), where ( k ) is odd, resulting in an even total count of integers.

To find the average of these integers, you add them up and divide by the number of integers. The sum of these consecutive integers can be simplified to ( \text{Sum} = n + (n+1) + (n+2) + ... + (n+k) ), which results in ( \text{Sum} = k \cdot n + \frac{k(k+1)}{2} ).

Now, since there is an even number of integers, the average is given by:

[ \text{Average} = \frac{\text{Sum}}{\text{Number of Integers}} = \frac{k \cdot n + \frac{k(k+1)}{2}}{k} ]

This simplifies to:

[ \text{Average} = n + \frac{k + 1}{2}