Question

find the sum of all three digit natural number which are multiply of 11

Answer 1

If seperately ...three digit natural nos that are multiples of 11 is 
110,121,132,143,......,990
so here d= 11
FIRST term 110
last term 990
so 990=110+(n-1)11
or n=81
so total sum here is (81/2)*(2*110+(80*11))=44550
again multiple of 13 is 104,117,130,143,.....,988
so 988=104+(n-1)*13
or n=69
so here total sum 69/2 (2*104+(69-1)*13)=37674
so total = 44550+37674=82224

Answer 2

11=X

22=X+11

33=X+22

X+X+11+X+22

3x+33