Question
The sum of all three digit numbers, each of which on divide by 5 leaves remainder 3, is -
Answer: Option D $$\eqalign{
& {\bf{Series :}} \cr
& 103 + 108\, + ........ + \,998 \cr
& a{\text{ }} = {\text{ }}103 \cr
& d{\text{ }} = {\text{ }}5 \cr
& {\text{Last term = 998}} \cr
& {\text{Number of term }} \cr
& = \frac{{998 - 103}}{5} + 1 \cr
& = \frac{{895}}{5} + 1 \cr
& = 180 \cr
& {\text{Sum of n terms}} \cr
& = \frac{n}{2}\left[ {2a{\text{ }} + \left( {n - 1} \right)d} \right] \cr
& = \frac{{180}}{2}\left[ {2 \times 103 + \left( {180 - 1} \right)5} \right] \cr
& = 99090 \cr} $$
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