Question

Find the sum of all 3-digit natural numbers, which are multiples of 11.

Answer 1

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [2a + (n - 1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

all 3-digit natural numbers, which are multiples of 11.

We know that the first 3 digit number multiple of 11 will be 110. 

Last 3 digit number multiple of 11 will be 990.

So here,

First term (a) = 110

Last term (l) = 990

Common difference (d) = 11

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

`a_n = a + (n - 1)d`

So for the last term

990 = 110 + (n -1) 11

990 = 110 + 11n - 11

990 = 99 + 11n

891 = 11n

81 = n

Now, using the formula for the sum of n terms, we get

`S_n = 81/2 [2(110) + (81 - 1)11]`

`S_n = 81/2 [220 + 80 xx 11]`

`S_n = 81/2 xx 1100`

`S_n = 81 xx 550`

`S_n = 44550`

Therefore, the sum of all the 3 digit multiples of 11 is 44550.