Let's find arithmetic sequences of natural numbers

$$
\begin{array}{llllll}
a_{1}, & a_{2}, & a_{3}, & \ldots, & a_{k}, & \ldots \\
b_{1}, & b_{2}, & b_{3}, & \ldots, & b_{k}, & \ldots \\
c_{1}, & c_{2}, & c_{3}, & \ldots, & c_{k}, & \ldots \\
d_{1}, & d_{2}, & d_{3}, & \ldots, & d_{k}, & \ldots
\end{array}
$$

such that in the sequence obtained by merging them

$$
a_{1}, b_{1}, c_{1}, d_{1}, a_{2}, b_{2}, c_{2}, d_{2}, \ldots, a_{k}, b_{k}, c_{k}, d_{k}, \ldots
$$

any term after $a_{1}$ is the sum of the term immediately preceding it and the last digit of that term.
In the (2) sequence, we provide the value of $a_{1}$, and it is a natural number. Based on the stated property, we can determine every subsequent term of the sequence step by step. Therefore, we need to examine how to provide $a_{1}$ so that the sequence derived from it, under (2), has subsequences under (1) that are arithmetic sequences. According to the formation rule, in the (2) sequence, every term from the second onward is even, so $a_{1}$ must also be an even number; otherwise, the sequence $a_{1}, a_{2}, \ldots, a_{k}, \ldots$ could not be an arithmetic sequence. We will show that $a_{1}$ can be any even number.

If the last digit of $a_{1}$ is 0, then the terms of the (2) sequence are equal, and the difference of the arithmetic sequences under (1) is 0.

If the last digit of $a_{1}$ is 2, 4, 6, or 8, then one of these will be the last digit of every subsequent term in the (2) sequence. Let's select any term from the (2) sequence and examine how much larger the term four places to the right is. If the last digit of the selected number is 2, then $2+2=4$, $4+4=8$, $8+8=6$, so the increments are 2, 4, 8, and 6, which sum to 20. We get the same increments, just in a different order, if the last digit of the selected term is 4, 8, or 6, so the difference between the terms four places apart in the (2) sequence is always 20, which precisely means that the sequences under (1) are arithmetic sequences.