Find the sum of first 100 even natural numbers which are divisible by 5.

Question:

Find the sum of first 100 even natural numbers which are divisible by 5.

Solution:
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 − 10) = 10 and n = 100

The sum of n terms of an AP is given by

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\left(\frac{100}{2}\right) \times[2 \times 10+(100-1) \times 10] \quad[\because a=10, d=10$ and $n=100]$

$=50 \times[20+990]=50 \times 1010=50500$

Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.