Question

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

Answer 1

the first and last even number divisible by 5 are 10 and 1000 respectively
1000=10+(n-1)10
990=(n-1)10
99=n-1
n=100

s=n/2(a+l)
=100/2(10+1000)
=50(1010)
=50500

Answer 2

⇒ First even natural number divisible by 5 = 10.

⇒ Second even natural number divisible by 5 = 20.

Here,

First term a = 10.

Common difference d = 20 - 10 = 10.

Number of terms n = 100.

We know that Sum of n terms of an AP:

Sn = (n/2)[2a + (n - 1) * d]

⇒ (100/2)[2 * 10 + 99 * 10]

⇒ 50[20 + 990]

⇒ 50[1010]

⇒ 50500.

Therefore, the sum is 50500.

hope this helps!