**Question **

Consider a set of numbers from 1 to 751. How many Arithmetic Progressions can be formed from this set of numbers such that the first term is 1 and the last term is 751 and the AP has at least 3 terms.

**Solution **

Numbers from 1 to 751 such that the first term is one and the last term is 751. From the basic definition of AP,

a = 1, a + (n-1)d = 751.

Subtracting both, we get -> (n-1)d = 750.

Now, if we find the number of ways 750 can be written as a product of two numbers, we have our answer. (After excluding cases that have n<3)

750 can be written as -> 1 x 750, 2 x 375, 3 x 250, 5 x 150 … and so on.

1 x 750 is a case and so is 750 x 1, this is the same with all other ways of writing the product. Effectively, if we count the number of factors in the number, then we should have the answer.

750 = 5^{3} x 2^{1} x 3^{1}. => Number of factors = 4 x 2 x 2 = 16.

Now we need to exclude the case where n will be lesser than 3, this means that (n-1) should be greater than or equal to 2.

One case gets eliminated -> 1 x 750.

So totally there are 16 – 1 = 15 cases.

Answer = 15