(3) The sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ are both arithmetic sequences of positive integers, with common differences $d_{1}$ and $d_{2}$, respectively, and both are greater than zero. If the sets $A=\left\{a_{n} \mid n \in \mathbf{N}^{*}\right\}, B=\left\{b_{n} \mid n \in \mathbf{N}^{*}\right\}$, then the first terms of the two sequences being equal (i.e., $a_{1}=b_{1}=a$) is a ( ) for the elements of the set $C=A \cap B$ to also form a positive integer arithmetic sequence.
(A) necessary but not sufficient condition
(B) sufficient but not necessary condition
(C) necessary and sufficient condition
(D) neither sufficient nor necessary condition
(3) Let $\left[d_{1}, d_{2}\right]$ denote the least common multiple of $d_{1}$ and $d_{2}$. If $a_{1}=b_{1}=a$, then for any natural number $n$, we have $a+n\left[d_{1}, d_{2}\right] \in C$. For any $c_{n} \in C$, then
$$
c_{n}=a+\left(n_{1}-1\right) d_{1}=a+\left(n_{2}-1\right) d_{2},
$$

it follows that
$$
d_{2}\left|\left(n_{1}-1\right) d_{1}, d_{1}\right|\left(n_{2}-1\right) d_{2},
$$

so
$$
c_{n}=a+m\left[d_{1}, d_{2}\right],
$$

where $m$ is a natural number, thus the condition is sufficient.
However, the condition is not necessary. A counterexample is as follows: $A=\{1,2, \cdots, n, \cdots\}, B=\{2$, $3, \cdots, n+1, \cdots\}, C=A \cap B=B$, but $a_{1} \neq b_{1}$. Therefore, the answer is B.