The sum of all three digit numbers, each of which on divide by 5 leaves remainder 3, is -

Solution(By Examveda Team)

$$\eqalign{ & {\bf{Series :}} \cr & 103 + 108\, + ........ + \,998 \cr & a{\text{ }} = {\text{ }}103 \cr & d{\text{ }} = {\text{ }}5 \cr & {\text{Last term = 998}} \cr & {\text{Number of term }} \cr & = \frac{{998 - 103}}{5} + 1 \cr & = \frac{{895}}{5} + 1 \cr & = 180 \cr & {\text{Sum of n terms}} \cr & = \frac{n}{2}\left[ {2a{\text{ }} + \left( {n - 1} \right)d} \right] \cr & = \frac{{180}}{2}\left[ {2 \times 103 + \left( {180 - 1} \right)5} \right] \cr & = 99090 \cr} $$