Question

Find the sum of all three digit numbers which are divisible by 7 leaves remainder 2

Answer 1

Minimum three digit number is 100 and maximum three digit number is 999. 
The first three digit number that leaves remainder 5 when divided by 7 is 103.
14 * 7 = 98 +5 = 103
The second three digit number that leaves remainder 5 when divided by 7 is 110.
15 * 7 = 105 +5 =110 
The third three digit number that leaves remainder 5 when divided by 7 is 117
and so on

The last three digit number that leaves remainder 5 when divided by 7 is 999
142 * 7 = 994 + 5 = 999

Therefore, we identify the sequence 
103,110,117.....999

use the formula of last term
Last term = first term + (n - 1) * common difference

you will get the answer 129 that is definitely E.