Question

Find the sum of all three digit natural numbers, which are multiples of 11.

Answer 1

The smallest and the largest three digit natural numbers, which are divisible by 11 are 110 and 990 respectively.

So, the sequence of three digit numbers which are divisible by 11 are 110, 121, 132, …, 990.

Clearly, it is an A.P. with first term, a = 110 and common difference, d = 11.

Let there be n terms in the sequence.

So, a_n = 990

`rArr a+(n-1)d=990`

`rArr 110+(n-1)11=990`

`rArr 110+11n=990`

`rArr 11n+99=990`

`rArr 11n=990-99`

`rArr 11n=891`

`rArr n=81`

Now, required sum `n/2[2a+(n-1)d]`

`=81/2[2xx110+(81-1)xx11]`

`=81/2[220+880]`

`=81/2xx1100`

`=44550`

Hence, the sum of all three digit numbers which are multiples of 11 is 44550.