7. Let the first term and common difference of an arithmetic sequence be positive integers, the number of terms be a prime number no less than 3, and the sum of all terms be 2006. Then the number of such sequences is $\qquad$.


Ni, 7.15.
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$, then
$$
n a+\frac{1}{2} n(n-1) d=2006,
$$

i.e., $n[2 a+(n-1) d]=2 \times 2006=2 \times 2 \times 17 \times 59$.
Since $2 a+(n-1) d \geqslant 2+n-1=n+1>n$, hence $n=17,59$.
(1) If $n=17$, then $a+8 d=118$, there are 14 sets of positive integer solutions;
(2) If $n=59$, then $a+29 d=34$, there is 1 set of positive integer solutions. Therefore, there are a total of 15 sets of positive integer solutions.