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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 $1188$.
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