Consider the set S = {1, 2, 3, ..., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?

- 3
- 4
- 6
- 7

**Answer**

Let the number of elements in a progression be n, then

1000 = 1 + (n-1)d

(n-1)d = 999 = 27 * 37

Possible values = 3, 37, 9, 111, 27, 333, 999

**The correct option is D.**