Question

Find the sum of all natural numbers between 100 and 1000 which are divisible by 11

Answer 1

Here

a=110,a'(last term)=990 & d=121-110=11

we know

a`=a+(n-1)d

990=110(n-1)11

990÷110=(n-1)

9=n-1

n=10

we know

Sum=(n÷2){a+a'}

S=(10÷2)(110+990)

S=5(1100)

S=5500

Hence Solved