The three-digit number range is from 100 to 999 (inclusive).

The first three-digit number completely divisible by 7 is 105 and the last three-digit number completely divisible by 7 is 994.

But, as we need the remainder of 3 after division, the first three-digit number will be 101 ( when divided by 7 leaves remainder 3).

So, the first number is 101 and the last number is 997 with common difference of 7. {101,108,115,.....997}

Note: 101 = 7*14+3 and 997 = 7*142 + 3 (Don't get confuse - common difference of '7' is between numbers which are divided by 7 leaves remainder of 3).

Number of terms in an A.P. : \(T_n\) = a+(n-1)d

=> 997 = 101 + (n-1)7
=> 997-101+7=7n
=> n = 129.

Sum of an A.P. when first and last term is known: \(\frac{n}{2}\)[first term + last term]
=> \(\frac{129 }{ 2}\) [ 101+997]
=> 70,821

Answer D
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Hi MathRevolution

Got the same answer and used the same approach. But took some time to figure out the last value 997.

Is there a faster way to identify.

I divided 999 by 7 and back-calculated to 997.

Thanks in Advance.