3. In how many ways can 4 numbers forming an increasing arithmetic progression be chosen from the first $n$ natural numbers $1,2, \ldots, n$?
Solution. The number of progressions with a difference of 1 is $n-3$ (the first term of the progression can take values from 1 to $n-3$ inclusive), the number of progressions with a difference of 2 is $n-$ $6, \ldots$, the number of progressions with a difference of $d$ is $n-3 d$. The difference $d$ satisfies the inequality $1+3 d \leq n$ (if the first term of the progression is 1, then its fourth term, $1+3 d$, does not exceed n). Therefore, the maximum value of the difference is $d_{\max }=\left[\frac{n-1}{3}\right]$ (square brackets denote the integer part of the number).

Therefore, the number of progressions satisfying the condition of the problem is:

$$
(n-3)+(n-6)+\cdots+(n-3 k)=\frac{(2 n-3 k-3) k}{2}, \text { where } k=d_{\max }
$$

Answer: $\frac{(2 n-3 k-3) k}{2}$, where $k=\left[\frac{n-1}{3}\right]$.