Question
Find the sum of,
All $2$ - digit natural numbers divisible by $4$.
All $2$ - digit natural numbers divisible by $4$.
✓
Answer
We can see it forms an A.P. as the common difference is $4$ and the first term is $4$ . To find no. of terms $n$,
We know that
$a_n=a+(n-1) d$
$96=12+(n-1) 4$
$84=(n-1) 4$
$21=n-1$
$22=n$
Now,
First term $(a) = 12$
Number of terms $( n ) =22$
Common difference $( d )=4$
Now, using the formula for the sum of $n$ terms, we get
$S_{22}=\frac{22}{2}\{2(12)+(22-1) 4\}$
$S_{22}=11\{24+84\}$
$S_{22}=1188$
Hence, The sum of $22$ terms is $1188$ which are divisible by $4$ .
We know that
$a_n=a+(n-1) d$
$96=12+(n-1) 4$
$84=(n-1) 4$
$21=n-1$
$22=n$
Now,
First term $(a) = 12$
Number of terms $( n ) =22$
Common difference $( d )=4$
Now, using the formula for the sum of $n$ terms, we get
$S_{22}=\frac{22}{2}\{2(12)+(22-1) 4\}$
$S_{22}=11\{24+84\}$
$S_{22}=1188$
Hence, The sum of $22$ terms is $1188$ which are divisible by $4$ .
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