Arithmetic Progressions in Set

An AP starting with 1 and ending with 1000 with common difference $d$ and $n$ elements satisfies: $1000 = 1 + (n - 1)d \implies (n - 1)d = 999 = 3^3 \times 37$. For at least 3 elements, $n \ge 3 \implies n - 1 \ge 2$. $d$ must be a factor of 999. Since $n - 1 = \frac{999}{d} \ge 2$, $d$ can be any divisor of 999 EXCEPT 999 itself (since $d = 999 \implies n - 1 = 1 \implies n = 2$, which gives only 2 elements). Total divisors of $999 = 3^3 \times 37^1$ is $(3+1)(1+1) = 8$. Excluding $d = 999$, we get $8 - 1 = 7$ valid APs.