The sum of all three digit numbers which leave a remainder of 3, when

The sum of all three digit numbers which leave a remainder of 3, when divided by 7 is?

First three digit number which satisfies this can be found by looking at 100
100 when divided by 7 gives 2 remainder
=> 101 when divided by 7 will give 3 remainder

Last three digit number which satisfies this can be found by looking at 999
999 when divided by 7 gives 5 remainder
=> 997 will be the last three digit number which will give 3 remainder when divided by 7

=> Series will be
101, 108, 115, ...., 997

Arithmetic series with first term as 101, Last term as 997, Common difference d = 7 and number of terms, n as
(Last Term - First term)/d + 1 = \(\frac{997 - 101}{7}\) + 1 = 128 + 1 = 129

=> Sum = n * (First Term + Last Term)/2 = 129 * (101 + 997)/2 = 129 * 549 = 70821

So, Answer will be D
Hope it helps!

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