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Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5.
Ans: 103, 108..........998
a + (n-1)d
=
998
⇒
103 + (n-1)5
n
180
S_{180} = 180/2 [103 + 998]
= 90 x 1101
S_{180} = 99090
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