Question

Find the sum of all three digit numbers which leave the remainder 3 when divided by 5. 

Answer 1

The sequence is 103, 108, 113, …., 998 

The sequence is an A.P with a = 103, d = 5, t_n = 998 

Answer 2

Minimum 3 digits number that gives remainder 3 when divided by 5 is 103.and maximum 3 digit number having same characteristics is 998.

So the series will constitute an AP series having first term 103 and last term 998 with common difference 5.

Let the number of terms be n then

103+(n-1)×5=998

=>n = 895/5+1=180

So the sum of the series

S= 180/2(103+998)=99090