Question

Find the sum of all three digits natural number wch is divisible by 11

Answer 1

the first three digits natural number which is divisible by 11 is 110

so,

a(first term)=110
d(common difference)=11
an(last term)=990

an=a+(n-1)d
990=110+(n-1)*11
990=110+11n-11
990=99+11n
990-99=11n
891=11n
n=891/11
n=81
so number of term are 81... putting all value in Sn
Sn=n/2[2a+(n-1)d]
Sn=81/2[2*110+(81-1)*11]
Sn=81/2[220+80*11]
Sn=81/2[220+880]
Sn=81/2*1100
Sn=89100/2
Sn=44550. Ans.