Let 7 be the first term and the common difference of an arithmetic sequence both be non-negative integers, the number of terms is no less than 3, and the sum of all terms is $97^{2}$. How many such sequences are there?
Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. According to the problem, we have $n a + \frac{1}{2} n(n-1) d = 97^2$, which means
$$
[2 a + (n-1) d] n = 2 \times 97^2
$$

Since $n$ is a positive integer no less than 3, and 97 is a prime number, the value of $n$ can only be one of $97$, $2 \times 97$, $97^2$, or $2 \times 97^2$. If $d > 0$, then from (1) we know $2 \times 97^2 \geqslant n(n-1) d \geqslant n(n-1)$, so $n$ can only be $97$. Thus, (1) becomes $a + 48 d = 97$, from which we can get $n=97, d=1, a=49$ or $n=97, d=2, a=1$; If $d=0$, then from (1) we get $n a = 97^2$, in which case $n=97, a=97$ or $n=97^2, a=1$. Therefore, there are 4 arithmetic sequences that meet the conditions.