Question

7. Find the sum of first $100$even natural numbers divisible by $5$.

Answer 1

Step 1. Finding the even numbers that are divisible by $5$.

The numbers that are divisible by $5$are

$5,10,15,20,25,30,35,40,45,50,....$

But the even natural numbers that are divisible by $5$are

$10,20,30,40,50.....$

Step 2. Finding the common difference.

The general formula of common differences is $d=n^{th}term-{\left(n-1\right)}^{th}term$

Here $t_1=10,t_2=20,t_3=30,..$

$d=t_2-t_1=t_3-t_2=10$

So the given series is an AP and the common difference is $10$.

Step 3. Finding the sum of first $100$even natural numbers divisible by $5$.

The general formula of first $n$term of an AP is $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

Here $a=10,d=10,n=100$

The sum is $=\frac{100}{2}\left(2·10+\left(100-1\right)10\right)$

$$\begin{gathered} =50\left(20+990\right) \\ =50\times 1010 \\ =50500 \end{gathered}$$

Hence, the sum of the first $100$even natural numbers divisible by $5$ is $50500$.