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 $S_{n}=1683$