Find the sum of all 2-digit natural numbers divisible by 4.

Given:

2-digit natural numbers divisible by 4.

To do:

We have to find the sum of all 2-digit natural numbers divisible by 4.

Solution:

Two digits natural numbers which are divisible by 4 are \( 12,16,20, \ldots, 96 \).

The sequence is in A.P.

Here,

\( a=12 \) and \( d=4 \) \( l=96 \)

We know that,

$l=a+(n-1) d$
$\Rightarrow 96=12+(n-1) \times 4$
$\Rightarrow 96=12+4 n-4$
$\Rightarrow 96=8+4 n$
$\Rightarrow 4 n=96-8$
$\Rightarrow n=\frac{88}{4}=22$
$\therefore n=22$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{22}{2}[2 \times 12+(22-1) \times 4]$
\( =11[24+21 \times 4] \)
\( =11[24+84] \)
\( =11 \times 108 \)

\( =1188 \)

The sum of all 2-digit natural numbers divisible by 4 is $1188$.