Find the sum of all 3-digit natural numbers, which are multiples of 11.

Given:

3-digit natural numbers which are divisible by 11.

To do:

We have to find the sum of all 3-digit natural numbers which are divisible by 11.

Solution:

3 -digit numbers which are divisible by 11 are \( 110,121,132,143, \ldots 990 \)
Here,

First term \( (a)=110 \)
Common difference \( (d)=121-110=11 \)

Last term \( (l)=990 \)

Let \( n \) be the numbers of terms

We know that,

$a_{n}=a+(n-1) d$
$\Rightarrow 990=110+(n-1) \times 11$
$\Rightarrow 990=110+11 n-11$
$\Rightarrow 11 n=990-99=891$

$n=\frac{891}{11}=81$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{81}{2}[2 \times 110+(81-1) \times 11]$
\( =\frac{81}{2}[220+80 \times 11] \)
\( =\frac{81}{2}[220+880] \)
\( =\frac{81}{2} \times 1100 \)

\( =81 \times 550=44550 \)

The sum of all 3-digit natural numbers which are divisible by 11 is $44550$.

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Updated on: 10-Oct-2022

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