Progressions

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?

  1. 3
  2. 4
  3. 6
  4. 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.