Question

Find the sum of all natural numbers from $1$ to $200$ which are divisible by $4$.

Answer 1

We have to find sum $4+8+12+...+200$
Clearly, it is arithmetic progression.
Here, $a=4,d=4$
Let there are n terms and 200 is nth term. Then,
$a+(n-1)d=200$
$\Rightarrow 4+(n-1)4=200$
$\Rightarrow (n-1)4=196$
$\Rightarrow (n-1)=49$
$\Rightarrow n=50$
Therefore,
$S_{50}=\frac{50}{2}[4+200]$
$\Rightarrow S_{50}=25\times 204=5100$