2. For the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$, each term starting from the third is equal to the sum of the previous two. Find the intersection of the sets $A=\left\{a_{n} \mid n \in \mathbb{N}\right\}$ and $B=\left\{b_{n} \mid n \in \mathbb{N}\right\}$, if $a_{1}=1, a_{2}=2, b_{1}=2$ and $b_{2}=1$.
Solution. Let $a_{3}=b_{3}=3$, then $\{1,2,3\} \subseteq A \cap B$.

We will prove, using the principle of mathematical induction, that

$$
a_{n-1}s$ we get

$$
b_{s}<b_{s+1}<a_{s+1}<a_{s+2}<b_{s+3}<a_{s+3}<\ldots<b_{k-1}<a_{k-1}<b_{k}
$$

which again leads to a contradiction;

3) For $k=s$ from (1) it follows $b_{k}<a_{k}$, so it is not possible $a_{k}=b_{k}$ again.

Therefore, apart from the first three, no other numbers occur in both sequences simultaneously, so $\{1,2,3\}=A \cap B$.

3 A. Let $k, n \in \mathbb{N}$. Find all arithmetic progressions such that the ratio of the sum of the first $n$ terms to the sum of the next $k n$ terms does not depend on $n$.