Question

find the sum of three digit numbers which are divisible by 11

Answer 1

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

The first 3-digit number divisible by 11 is 110.

After 110, to find the next 3-digit number divisible by 11, we have to add 11 to 110.
So, the 2nd 3-digit number divisible by 11 is 121.

110,121,132.........990.

In the AP 110,121,132....990, we have

First-term a = 110

common difference d = 121- 110

                                     = 11

Last term l = 990.


We know that number of terms in an AP n = [(l - a)/d] + 1

                                                                        = [(990 - 110)/11] + 1

                                                                        = [880/11] + 1

                                                                         = 80 + 1

                                                                         = 81.


The number of 3-digit numbers divisible by 11 is 81.


We know that sum of n terms of an AP is = n/2(a + l)

a = 110, d = 11, l = 990, n = 81

= 81/2(110 + 990)

= 81/2(1100)

= 81 * 550

= 44550.


Therefore the sum of all three-digit numbers divisible by 11 = 44550.



Hope this helps!  --------------------- Gud Luck