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?​give the approach also

1) 4
2) 6
3) 7
4) 8

• 1000-1=999
  999=3^3*37
  no of factors of 999=(3+1)(1+1)=8
  but 999 is one of the factors so answer is 8-1=7
• 11 years agoHelpfull: Yes(1) No(1)