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. $\mathrm{C}$ Let the first term of the arithmetic sequence be $a_{1}$, and the common difference be $d$. From the given, we have:
$n a_{1}+\frac{1}{2} n(n-1) d=97^{2} \quad$ which means $\left[2 a_{1}+(n-1) d\right] n=2 \times 97^{2}$, and since $n \geqslant 3$. Therefore, $n$ can only be $97, 2 \times 97, 97^{2}, 2 \times 97^{2}$.
Also, because $a_{1} \geqslant 0, d \geqslant 0$, we have $2 \times 97^{2}=n(n-1) d+2 a_{1} n \geqslant n(n-1) d$. If $d>0$, and since $d$ is an integer, we know $d \geqslant 1$, thus $2 \times 97^{2} \geqslant n(n-1)$, which can only be true when $n=97$.
At this time, $\left\{\begin{array}{l}n=97 \\ d=1 \\ a_{1}=49\end{array}\right.$ or $\left\{\begin{array}{l}n=97 \\ d=2 \\ a_{1}=1\end{array}\right.$
If $d=0 \quad$ then $a_{1} n=97^{2} \quad$ at this time $\left\{\begin{array}{l}n=97 \\ d=0 \\ a_{1}=97\end{array} \quad\right.$ or $\left\{\begin{array}{l}n=97^{2} \\ d=0 \\ a_{1}=1\end{array}\right.$
Therefore, there are 4 arithmetic sequences that meet the conditions, so the answer is C.