#### Answer

For the positions of the two middle elements in a list with an even number
$n$ of elements:
$\left[\frac{n}{2} \text { and } \frac{n}{2}+1\right]$ and their average $\left[\frac{n+1}{2}\right]$

#### Work Step by Step

We do the same with lists that consist of an even number of elements $n .$
We know that even numbers are divisible by $2 .$ So there isn't just one
distinct middle element, but two. The positions of the two middle elements
are the required number of comparisons to find them. The average of these
two numbers is their sum divided by 2.
Example
$\quad\bullet$ For $n=16$ the positions of the two middle elements are 8 and $9 .$
$\quad$$\quad$ Average $=8.5$
$\quad$For $n=40$ the positions of the two middle elements are 20 and $21 .$
$\quad$$\quad$Average $=20.5$
$\quad$$\bullet$ For $n=56$ the positions of the two middle elements are 28 and $29 .$
$\quad$$\quad$Average $=28.5$
The values for the two middle elements can be found using the expressions:
$\frac{n}{2}$ and $\frac{n}{2}+1$
Let's find an expression for the average of the two values for the middle
elements:
$\frac{\frac{n}{2}+\frac{n}{2}+1}{2}=\frac{\frac{2 \cdot n}{2}+1}{2}=\frac{n+1}{2}$