Question

Sum of natural numbers between 101 and 999 which are divisible by both 5 and 2

Answer 1

First term, a = 110 (first number which is divisible by 2,5)
difference d = 10 
an = 990  (last number which is divisible by 2,5)


Formula is , an = a + (n - 1)d
        
                    990 = 110 + 10n -10

                    880 + 10 = 10n

                     890 = 10n 

                      n = 89.


Hope this helps!