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

Given:

Natural numbers divisible by 3.

To do:

We have to find the sum of all natural numbers between 1 and 100, which are divisible by 3.

Solution:

Natural numbers between 1 and 100 which are divisible by 3 are \( 3,6,9, \ldots, 99 \).

The sequence is in A.P.

Here,

\( a=3 \) and \( d=6-3=3 \) \( l=99 \)

We know that,

$l=a+(n-1) d$

$\Rightarrow 99=3+(n-1) \times 3$

$\Rightarrow 99=3+3 n-3$

$\Rightarrow 99=3 n$

$\Rightarrow n=\frac{99}{3}=33$

$\therefore n=33$

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

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

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

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

$=33 \times 51$

$=1683$

The sum of all natural numbers between 1 and 100 divisible by 3 is $1683$.