If an arithmetic progression starts with $a$ and ends with $b$, and has $k$ terms, then we must be able to write $b = a+(k-1)t$ for some $t\gt 0$ (and integer, given the conditions).
That means that $b-a$ must be a multiple of $k-1$. Conversely, if $b-a$ is a multiple of $k-1$, $b-a = (k-1)t$, then the arithmetic progression
$$a, a+t, a+2t,\ldots,a+(k-1)t = b$$
starts with $a$, ends with $b$, and has exactly $k$ terms.
So the number of arithmetic progressions that start with $a$ and end with $b$ and have at least $n$ terms is equal to the number of integer divisors of $b-a$ that are greater than $n-1$.
Can we count those more easily instead? In other words, given a number $K$, can we count how many divisors greater than a given $N$ it has?