Consider the first few cases for $n$ with the entire $n$ numbers forming an arithmetic sequence $$(1, 2, 3, \ldots, n)$$ If $n = 3$ , there will be one ascending triplet (123). Let's only consider the ascending order for now.
If $n = 4$ , the first 3 numbers give 1 triplet, the addition of the 4 gives one more, for 2 in total.
If $n = 5$ , the first 4 numbers give 2 triplets, and the 5th number gives 2 more triplets (135 and 345).
Repeating a few more times, we can quickly see that if $n$ is even, the nth number will give $$\frac{n}{2} - 1$$ more triplets in addition to all the prior triplets from the first $n-1$ numbers.
If $n$ is odd, the $n$ th number will give $$\frac{n-1}{2}$$ more triplets.
Let $f(n)$ denote the total number of triplets for $n$ numbers. The above two statements are summarized as follows:
If $n$ is even, $$f(n) = f(n-1) + \frac{n}2 - 1$$ If $n$ is odd, $$f(n) = f(n-1) + \frac{n-1}2$$
Let's obtain the closed form for when $n$ is even: $$\text{f(n) = f(n-2) + n-2 f(n) = f(n-4) + (n-2) + (n-4) f(n) = i=1 n/2 n - 2i f(n even ) = n 2 - 2n 4}$$
Now obtain the closed form when $n$ is odd by using the previous result for when $n$ is even: $$\text{f(n) = f(n-1) + n-1 2 f(n) = (n-1) 2 - 2(n-1) 4 + n-1 2 f(n odd ) = (n-1) 2 4}$$
Note the ambiguous wording in the question! If the "arithmetic progression" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression.
Double the expression to account for the descending versions of each triple, to obtain: $$\text{f(n even ) = n 2 - 2n 2 f(n odd ) = (n-1) 2 2 f(n) = (n-1) 2 2}$$
~Lopkiloinm (corrected by integralarefun)