Problem 1264

1. Initial Setup and Assumptions:
We are given a sequence $a_1, a_2, a_3, \ldots$ of real numbers. We need to find all such sequences that satisfy the property: for any sequence $b_1, b_2, b_3, \ldots$ of positive integers such that $b_n \neq b_{n+1}$ and $b_n \mid b_{n+1}$ for all $n \geq 1$, the subsequence $a_{b_1}, a_{b_2}, a_{b_3}, \ldots$ forms an arithmetic progression.

2. Choosing Specific Sequences:
Consider the sequence $b_1 = 1, b_2 = p, b_3 = np$ for some $n > 1$ and a prime $p$. According to the problem, $a_1, a_p, a_{np}$ must form an arithmetic progression. Therefore, we have:
$$a_{np} = 2a_p - a_1$$
This equation must hold for any $n$ and any prime $p$.

3. Implication for General Terms:
Since $a_{np} = 2a_p - a_1$ must hold for any $n$, it implies that $a_{2p} = a_{3p} = \ldots = a_{np}$ for any prime $p$. This suggests that the terms $a_{kp}$ for $k \geq 2$ are all equal.

4. Further Analysis with Different Sequences:
Now, consider the sequences $b_1 = 1, b_2 = 2p, b_3 = 4p$ and $b_1 = p, b_2 = 2p, b_3 = 4p$. For these sequences, the subsequences $a_1, a_{2p}, a_{4p}$ and $a_p, a_{2p}, a_{4p}$ must also form arithmetic progressions. This gives us:
$$a_{4p} = 2a_{2p} - a_1 \quad \text{and} \quad a_{4p} = 2a_{2p} - a_p$$
Since both expressions for $a_{4p}$ must be equal, we have:
$$2a_{2p} - a_1 = 2a_{2p} - a_p$$
This simplifies to:
$$a_1 = a_p$$
for any prime $p$.

5. Conclusion:
Since $a_1 = a_p$ for any prime $p$, and we have already established that $a_{2p} = a_{3p} = \ldots = a_{np}$ for any $n$ and any prime $p$, it follows that all terms $a_i$ in the sequence must be equal. Therefore, the sequence $a_1, a_2, a_3, \ldots$ is constant.

The final answer is $\boxed{ a_i = c }$ for some constant $c$.