Find the sum of all 3-digit natural numbers which are divisible by 13.

Given:

3-digit natural numbers which are divisible by 13.

To do:

We have to find the sum of all 3-digit natural numbers which are divisible by 13.

Solution:

3 -digit numbers which are divisible by 13 are \( 104,117,130,143, \ldots 988 \)
Here,

First term \( (a)=104 \)
Common difference \( (d)=117-104=13 \)

Last term \( (l)=988 \)

Let \( n \) be the numbers of terms

We know that,

$a_{n}=a+(n-1) d$
$\Rightarrow 988=104+(n-1) \times 13$
$\Rightarrow 988=104+13 n-13$
$\Rightarrow 13 n=988-104+13=897$

$n=\frac{897}{13}=69$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{69}{2}[2 \times 104+(69-1) \times 13]$
\( =\frac{69}{2}[208+68 \times 13] \)
\( =\frac{69}{2}[208+884] \)
\( =\frac{69}{2} \times 1092 \)

\( =69 \times 546=37674 \)

The sum of all 3-digit natural numbers which are divisible by 13 is $37674$.

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Updated on: 10-Oct-2022

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