3A. Determine all arithmetic progressions with difference $d=2$, such that the ratio $\frac{S_{3 n}}{S_{n}}$ does not depend on $n$.
Solution. Let the first term of the arithmetic progression that satisfies the given condition be $a_{1}$. The sum of the first $k$ terms of this progression is

$$
S_{k}=\frac{k}{2}\left[2 a_{1}+(k-1) \cdot 2\right]=k\left(a_{1}+k-1\right)
$$

From the condition, it follows that for $n=1$ and $n=2$ we have $\frac{S_{3}}{S_{1}}=\frac{S_{6}}{S_{2}}$, from which we obtain the equation $\frac{3\left(a_{1}+2\right)}{a_{1}}=\frac{6\left(a_{1}+5\right)}{2\left(a_{1}+1\right)}$, whose only solution is $a_{1}=1$.

Thus, if any progression with a difference of 2 satisfies the given condition, then it has the first term $a_{1}=1$.

Conversely, if $a_{1}=1$, then

$$
\frac{S_{3 n}}{S_{n}}=\frac{3 n(1+3 n-1)}{n(1+n-1)}=\frac{9 n^{2}}{n^{2}}=9
$$

so the given ratio does not depend on $n$.

Thus, there is only one such progression, and it is $1,3,5, \ldots$.