Question

find the sum of all 3 digit numbers which are divisible by 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

Answer 2

Answer:

nth term =an =990

a= 110

an = a + (n -1)d

990= 110 + (n-1) 11

880/11 = n-1

81= n

Sn =n/2 { 2a + (n-1)d}

S81= 81/2 {2×110 + (81-1)11}

= 81/2 {220 +880}

=81/2 { 1100}

=81×550

=44550