Question

4. Find the sum of all the natural numbers between $200$ and $300$ which are divisible by $4$.

Answer 1

Step 1: Note the given data

Given the series is the natural numbers between $200$ and $300$ which are divisible by $4$.

The first digit in the natural number is $a=204$ and the last term is $296$, which are divisible by $4$

Therefore the numbers are $204,208,212,........,296$

Let the last term $t_n=296$

Step 2: Find the common difference

The general form for finding the common difference $d={(n+1)}^{th}term-n^{th}term$

Common difference

$$\begin{gathered} d=208-204 \\ =4 \end{gathered}$$

Similarly, $d=208-204=212-208=4$

So, the series is in A.P.

Step 3: Find the $n^{th}$ term of A.P.

The formula for finding $n^{th}$term of A.P. $=a+(n-1)d$

where $a$ is the first term and $d$ is common difference.

The value of $n^{th}$term of A.P $=204+(n-1)\times 4$

$$\begin{gathered} =204+4n-4 \\ =200+4n \end{gathered}$$

Step 4: Equating both the $n^{th}$ value of A.P.

$200+4n=296$

$$\begin{gathered} \Rightarrow 4n=296-200 \\ \Rightarrow n=\frac{96}{4} \\ \Rightarrow n=24 \end{gathered}$$

Step 5: Find the required sum of the given series

The formula of the sum of the series of A.P. $=\frac{n}{2}(t_1+t_n)$

Where $t_1$ is the first term, $t_n$ is last term of an AP.

Sum of the given series

$$\begin{gathered} =\frac{24}{2}(204+296) \\ =\frac{24}{2}\times 500 \\ =12\times 500 \\ =6000 \end{gathered}$$

Hence, the sum of the given series is $6000$.