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

The sum of natural numbers between 100 and 200 that are divisible by 4 is 104 + 108 + 112 + ….. + 196.

The above series is an A.P. with a = 104, d = 108 - 104 = 4 and l = 196.

Let 196 be nth term then,

⇒ 196 = 104 + 4(n - 1)
⇒ 196 - 104 = 4n - 4
⇒ 92 = 4n - 4
⇒ 4n = 92 + 4
⇒ 4n = 96
⇒ n = 24.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₂₄ = $\dfrac{24}{2}[2 \times 104 + 4(24 - 1)]$
⇒ S₂₄ = 12[208 + 92]
⇒ S₂₄ = 12 × 300
⇒ S₂₄ = 3600.

Hence, the sum of all two digits natural numbers between 100 and 200 which are divisible by 4 is 3600.