Find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that
$$
f(x)+f(t)=f(y)+f(z)
$$
for all rational numbers $x<y<z<t$ that form an arithmetic progression.
Answer: any linear function $f$. These work.
Here is one approach: for any $a$ and $d>0$
$$
\begin{aligned}
f(a)+f(a+3 d) & =f(a+d)+f(a+2 d) \\
f(a-d)+f(a+2 d) & =f(a)+f(a+d)
\end{aligned}
$$
which imply
$$
f(a-d)+f(a+3 d)=2 f(a+d) .
$$

Thus we conclude that for arbitrary $x$ and $y$ we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right)
$$
thus $f$ satisfies Jensen functional equation over $\mathbb{Q}$, so linear.
The solution can be made to avoid appealing to Jensen's functional equation; here is a presentation of such a solution based on the official ones. Let $d>0$ be a positive integer, and let $n$ be an integer. Consider the two equations
$$
\begin{array}{l}
f\left(\frac{2 n-1}{2 d}\right)+f\left(\frac{2 n+2}{2 d}\right)=f\left(\frac{2 n}{2 d}\right)+f\left(\frac{2 n+1}{2 d}\right) \\
f\left(\frac{2 n-2}{2 d}\right)+f\left(\frac{2 n+1}{2 d}\right)=f\left(\frac{2 n-1}{2 d}\right)+f\left(\frac{2 n}{2 d}\right)
\end{array}
$$

Summing them and simplifying implies that
$$
f\left(\frac{n-1}{d}\right)+f\left(\frac{n+1}{d}\right)=2 f\left(\frac{n}{d}\right)
$$
or equivalently
$$
f\left(\frac{n}{d}\right)-f\left(\frac{n-1}{d}\right)=f\left(\frac{n+1}{d}\right)-f\left(\frac{n}{d}\right) .
$$

This implies that on the set of rational numbers with denominator dividing $d$, the function $f$ is linear.

In particular, we should have $f\left(\frac{n}{d}\right)=f(0)+\frac{n}{d} f(1)$ since $\frac{n}{d}, 0,1$ have denominators dividing $d$. This is the same as saying $f(q)=f(0)+q(f(1)-f(0))$ for any $q \in \mathbb{Q}$, which is what we wanted to prove.