Question

find the sum of all three digit number which are multiples of 11

Answer 1

The first three digit no. 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=(n-1)11

880/11=(n-1)

80=n-1

n=81

So the no. of terms are 81...Putting all values in Sn

Sn=n/2[2a+(n-1)d]
=81/2[2*110+(81-1)(11)

=81/2[220+80*11}

=81/2[220+880)

=81/2*1100

=81*550

=44550 Ans