7. Given that the lengths of the three sides of a triangle are all integers less than 100, and form an arithmetic sequence, how many such triangles are there.
7. Let the three sides of the triangle be $x, y, z (x \leqslant y \leqslant z)$.

From the problem, we get $0 < x \leqslant y \leqslant z < x + y$.
Eliminating $z$ gives $0 \leqslant x < y < \frac{x}{2} + 50, y < 2x$. The number of integer points in the plane region that satisfies the above system of inequalities is the number of triangles we are looking for. Using the line $x = k (1 \leqslant k \leqslant 99$, and $k \in \mathbf{N})$ to cut the plane region, we have:
(1) When $1 \leqslant k \leqslant 33$, the integer points on the line $x = k$ are $(k, k), (k, k+1), \cdots, (k, 2k-1)$, a total of $k$ points. Here, the number of integer points that satisfy the problem is $\sum_{k=1}^{33} k = 561$;
(2) When $34 \leqslant k \leqslant 99$, if $k$ is even, the integer points on the line $x = k$ are $(k, k), (k, k+1), \cdots, \left(k, \frac{k}{2} + 50 - 1\right)$, a total of $50 - \frac{k}{2}$ points;

If $k$ is odd, the integer points on the line $x = k$ are $(k, k), (k, k+1), \cdots, \left(k, \frac{k-1}{2} + 50\right)$, a total of $50 - \frac{k-1}{2}$ points.
Therefore, the number of integer points for $34 \leqslant k \leqslant 99$ is $2(33 + 32 + \cdots + 1) = 1122$.
Combining (1) and (2), the number of triangles that satisfy the problem is $561 + 1122 = 1683$.