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

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,

$99=3+(n-1) 3$

$99=3+3 n-3$

$99=3 n$

Further simplifying,

$n=\frac{99}{3}$

$n=33$

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

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

We get,

$S_{n}=\frac{33}{2}[2(3)+(33-1) 3]$

$=\frac{33}{2}[6+(32) 3]$

$=\frac{33}{2}(6+96)$

$=\frac{33(102)}{2}$

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 $S_{n}=1683$