Question

The sum of all natural numbers from $200$ to $300$, which are divisible by $6$ is:

Answer 1

Step 1: The arithmetic series is $204,210,216,\dots ..,294$

First term, $a=204$

Difference, $D=210–204=6$

Last term, $a_n=294$

By the formula of $n^{th}$ term of Arithmetic progression, we have;

$\begin{array}{c}{a}_n=294\\ a+(n-1)d=294\\ 204+(n-1)6=294\\ 204+6n-6=294\\ 198+6n=294\\ 6n=294-198\\ 6n=96\end{array}$

Therefore, $n=16$

Step 2: Substitute the value of $n$

Thus, there are $16$ terms in A.P., which we need to add.

$\begin{array}{r}{S}_n=\frac{n}{2}[a+an]\end{array}$

$\begin{array}{r}{S}_{16}=\frac{16}{2}[204+294]\end{array}$

$S_{16}=8\left[498\right]$

$S_{16}=3984$

Hence, the sum of all natural numbers from $200$ to $300$, which are divisible by $6$ is $3984$.