Question

From the 1^st 12 natural numbers how many Arithmetic Progressions of 4 terms can be formed such that the common difference is a factor of the 4^th term?

Answer 1

If a be the 1^st term and d be the common difference of the A.P. the 4^th term of the series will be a + 3d.

If a + 3d is divisible by d then a should be divisible by d. hence the cases are:

d = 1, a = 1, 2, 3, 4, 5, 6, 7, 8, 9 

d = 2, a = 2, 4, 6 

d = 3, a = 3 

So the required answer is 9 + 3 + 1 = 13