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How many different arithmetic progressions with integral terms can be formed with first term $a$, last term $b$, and at least $n$ terms?

$a \lt b$, and $a,b,n \in \mathbb{N}$ and they are always chosen in a manner which gives a valid answer.

How to solve this one? Please do explain your answer.

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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?

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    I don't know any way besides listing the factors and checking. But once you have the factorization that is not hard. For example, if you have $12=2^23^1$, you know it has $(2+1)(1+1)=6$ factors. To list them, they are of the form something from $\{1,2,4\}$ times something from $\{1,3\}$2011-11-03