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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 $1683$.
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