Question

find the sum of all three digit numbers which leave the remainder 3 when divided by5

Answer 1

AP for these numbers is, 103,108,113,.................,998

 a= 103, d = 5

Let there are n terms in this AP so an = 998

a+(n-1)d = 998
103 + (n-1)5 = 998
     (n-1) = 179
        so, n = 180

So, Sn = S180 = 180/2[a+l]
                        = 90[103+998]
                        = 90*1101
                        = 99090