Question

find the sum of all natural numbers between 400 and 600 which are divisible by 11

Answer 1

Hey

Here is your answer,

407,418,429,.........,594

a=407
d=11
An = l= 594

An= a+(n-1)d
594-407/11 = n-1
187/11=n-1
N=17+1
N=18

Sn= n/2(a+l)
=9(407+594)
=9x1001
= 9009

Answer 2

see my answer the answer is 2179089