An arithmetic progression whose sum of $n$ terms, for any integer $n$: $n(3 n+1)$. Which is this progression?
I. solution. According to the problem:

$$
\frac{n}{2}\left[2 a_{1}+n d-d\right]=n(3 n+1)
$$

from which

$$
n(d-6)+\left(2 a_{1}-d-2\right)=0
$$

This equation can only be 0 for all values of $n$ if

$$
d-6=0 \text { and } 2 a_{1}-d-2=0 \text {. }
$$

From these two equations:

$$
d=6, a_{1}=4
$$

Thus, the arithmetic sequence is:

$$
4,10,16, \ldots,(6 n-2)
$$

II. solution. If we substitute 1, 2, 3 for $n$, then

$$
s_{1}=4, s_{2}=14, s_{3}=30, \ldots
$$

Thus,

$$
a_{1}=4, a_{2}=s_{2}-s_{1}=10, a_{3}=s_{3}-s_{2}=16, \text { etc. }
$$

So the arithmetic sequence is

$4,10,16, \ldots$

(Miksa Pintér, Budapest.)

Number of solutions: 52.