Question

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

Answer 1

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
$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ =\left(\frac{100}{2}\right)\times [2\times 10+\left(100-1\right)\times 10]\left[\because a=10,d=10\mathrm{and}n=100\right] \\ =50\times \left[20+990\right]=50\times 1010=50500 \end{gathered}$$

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