3. Let the first term of an arithmetic sequence be $a(a \neq 0)$, and the second term be $b$. Then the necessary and sufficient condition for this sequence to have a term equal to 0 is ( ).
(A) $a-b$ is a positive integer (B) $a+b$ is a positive integer
(C) $\frac{b}{a-b}$ is a positive integer (D) $\frac{a}{a-b}$ is a positive integer
3.(D).

Since the common difference $d=b-a \neq 0$, otherwise it would be impossible for any term to be 0. Thus, let $a_{n}=a+(n-1)(b-a)=0$, we get $n-1=$ $\frac{a}{a-b}$, which implies that $\frac{a}{a-b}$ is a natural number.
Conversely, if $\frac{a}{a-b}$ is a natural number, then the $\frac{a}{a-b}+1$-th term is 0.