Question

Find the sum of all three digit numbers which leaves remainder 2 when divided by 3

Answer 1

101 is the first and 998 is the last three digit number which leaves remainder 2 when divided by 3
So, first term=101 and last term=998
also, Required AP = 101, 104, 107.........998
then, common difference=3
Now, nth term= 998
or, a+(n-1)d = 998
or, 101+(n-1)3 = 998
or, n = 300
so, There are 300 terms.

Now, Sum = 300/2{a+l}
or, Sum = 300/2(101+998)
or, Sum = 300/2× 1099
or, Sum = 150× 1099
Hence, Sum = 164850

So, the correct answer is 164850.

Hope it helped
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