Find the sum of all three digit numbers which are multiples of Ie digi

1.	Find the sum of all three digit numbers which are multiples of Ie digit numbers which are multiples of 11
Answer» 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 81Put all values in Sn Sn = n/2[2a+(n-1)d] = 81/2[2110+(81-1)(11) = 81/2[220+8011} = 81/2[220+880) = 81/21100 = 81550 = 44550

Find the sum of all three digit numbers which are multiples of Ie digit numbers which are multiples of 11

Answer»

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 81Put 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