3. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$. Then the number of such sequences is:
(A) 2;
(B) 3;
(C) 4;
(D) 5.
3. (C)

Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. Then, according to the problem, we have
$$
n a+\frac{n(n-1)}{2} d=97^{2},
$$

which can be rewritten as
$$
[2 a+(n-1) d] n=2 \times 97^{2} .
$$

Since $n$ is a natural number not less than 3, and 97 is a prime number, the value of $n$ can only be one of the following: 97, 2 $\times 97, 97^{2}, 2 \times 97^{2}$.

If $d>0$, then from $*$ we know $2 \times 97^{2} \geqslant n(n-1) d \geqslant n(n-1)$. Therefore, $n$ can only be 97. * becomes $a+48 d=97$, and in this case, $*$ has two solutions:
$$
\left\{\begin{array} { l } 
{ n = 9 7 , } \\
{ d = 1 , } \\
{ a = 4 9 ; }
\end{array} \quad \left\{\begin{array}{l}
n=97, \\
d=2, \\
a=1 .
\end{array}\right.\right.
$$

If $d=0$, then $*$ becomes $a n=97^{2}$, and in this case, $*$ has two solutions:
$$
\left\{\begin{array} { l } 
{ n = 9 7 , } \\
{ d = 0 , } \\
{ a = 9 7 ; }
\end{array} \quad \left\{\begin{array}{l}
n=97^{2}, \\
d=0, \\
a=1 .
\end{array}\right.\right.
$$

Therefore, there are a total of 4 arithmetic sequences that meet the conditions of the problem.