Question

2. Find the sum of all natural numbers between $100$ and $200$ which are divisible by $4$.

Answer 1

Step 1: Note the given data

Given the series is the natural numbers between $100$ and $200$ which are divisible by $4$

The first digit in the natural number is $a=104$ and the last term is $196$, which are divisible by $4$

Therefore the numbers are $104,108,112,116,........,196$

Let the last term $t_n=196$

Step 2: Find the common difference

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

Common difference

$$\begin{gathered} d=108-104 \\ =4 \end{gathered}$$

Similarly, $d=108-104=112-108=4$

So, the series are in A.P.

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

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

The value of $n^{th}$term of A.P $=104+(n-1)\times 4$

$$\begin{gathered} =104+4n-4 \\ =100+4n \end{gathered}$$

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

$100+4n=196$

$$\begin{gathered} \Rightarrow 4n=196-100 \\ \Rightarrow 4n=96 \end{gathered}$$

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

Step 5: Find the required sum of the given series

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

Sum of the given series

$$\begin{gathered} =\frac{24}{2}(104+196) \\ =12\times 300 \\ =3600 \end{gathered}$$

Hence, the sum of the given series is $3600$