Proving that the mean and mode are equal for an arithmetic progression

Ask your students to prove this as an exercise, when learning how to prove things by cases. This theorem is used here, when proving that polite numbers cannot be of the form \(2^n.\)

Problem: Prove that the mean equals the mode for an arithmetic progression.

Solution:

$$a + (a + d) + (a + 2d) + \ldots + [a + (n - 1)d]$$

The sum of an arithmetic progression is

$$S_n = \dfrac{n}{2}[2a + (n - 1)d]$$

The mean is thus

$$\begin{align} & \dfrac{1}{2}[2a + (n - 1)d] \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align}$$

If \(n\) is odd, then the mode is the middle term, which is

$$\begin{align} & a + \left(\dfrac{n - 1}{2}\right)d \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align}$$

If \(n\) is even, then the mode is the average of the two middle terms, which is

$$\begin{align} & \dfrac{\left[a + \left(\dfrac{n}{2}\right)d\right] + \left[a + \left(\dfrac{n}{2} - 1\right)d\right]}{2} \\[0.5em] =&\,a + \dfrac{dn - d}{2} \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align}$$