Question

find the sum of all 3 digit numbers which leave the remainder 2 when divisible by 5​

Answer 1

3 digit numbers which leave the remainder 2 when divided by 5 are

102, 107, 112,......, 997.

It forms an A.P

In this A.P

a (first term)= 102

d (Common difference)= 5

I(last term ) = 997

an = a + (n – 1) d

997= 102 + (n – 1) × 5

5 (n – 1) = 997 – 102 = 895

n-1= 885/5

(n – 1) = 179

n = 179 +1

n = 180

Sum of all three digit numbers which leaves remainder 2 when divided by 5

Sn = n/2 [a+l]

Sn= 180/2 [ 102+ 997]

Sn= 90 × 1099

Sn= 98910

98910 is required answer.