Question

Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

Answer 1

In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.

So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Last term (l) = 99

Common difference (d) = 3

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,

99 = 3 + (n - 1)3

99 = 3 + 3n - 3

99 = 3n

Further simplifying 

`n = 99/3`

n = 33

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

`S_n = 33/2 [2(3) + (33 - 1)3]`

`= 33/2 [6 + (32)3]`

`= 33/2 (6 + 96)`

`= (33(102))/2`

On further simplification, we get,

`S_n = 33(51)`

= 1683

Therefore, the sum of all the multiples of 3 lying between 1 and 100 is `S_n = 1683`