Problem: Find all functions $f \colon \QQ \to \QQ$ such that
\[ f(x)+f(t)=f(y)+f(z) \]
for all rational numbers $x<y<z<t$ that form an arithmetic progression.

Solution: Answer: any linear function $f$.
These work.

Here is one approach: for any $a$ and $d > 0$
\begin{align*}
  f(a) + f(a+3d) &= f(a+d) + f(a+2d) \\
  f(a-d) + f(a+2d) &= f(a) + f(a+d) \\
  \intertext{which imply}
  f(a-d) + f(a+3d) &= 2f(a+d).
\end{align*}
Thus we conclude that for arbitrary $x$ and $y$ we have
\[ f(x) + f(y) = 2f\left( \frac{x+y}{2} \right) \]
thus $f$ satisfies Jensen functional equation over $\QQ$, 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{align*}
  f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n+2}{2d} \right)
  &= f\left( \frac{2n}{2d} \right) + f\left( \frac{2n+1}{2d} \right) \\
  f\left( \frac{2n-2}{2d} \right) + f\left( \frac{2n+1}{2d} \right)
  &= f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n}{2d} \right)
\end{align*}
Summing them and simplifying implies that
\[ f\left( \frac{n-1}{d} \right) + f\left( \frac{n+1}{d} \right)
  = 2 f \left( \frac nd \right) \]
or equivalently
\[ f\left( \frac nd \right) - f\left( \frac{n-1}{d} \right)
  = f\left( \frac{n+1}{d} \right) - f\left( \frac nd \right). \]
This implies that on the set of rational numbers
with denominator dividing $d$, the function $f$ is linear.

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