Question

find the sum of three digit numbers,which leaves the remainder 3 when divided by 7

Answer 1

Heya...


Your Query :- Sum of all three digit numbers which leaves remainder when divided by 7. 

Now, Here is the answer of your query.

The three digit numbers which leaves remainder 3 when divided by 7 are 101, 108, 115, 122, ..........,991, 997.

Clearly, we can see that this forms an AP with common difference, d = 7.
Also, a = 101 and nth term (last term) = 997.

By using formula.
aⁿ = a + (n-1)d 
997 = 101 + (n-1)7
896 = 7(n-1)
n -1 = 128
n = 129

Now, using 
Sⁿ = n/2{2a+ (n-1) d}
= 129/2 {2 X 101 + ( 129 -1) 7}
= 129/2 {202 + 128X 7}
= 129/2 {202 + 896)
$= \frac{129}{2} X1098 \\ = 129 X 549 \\ =70821$.

Hence, your query is solved and the answer is 70821.


HOPE IT HELPS