Find the sum of all natural numbers lying between 100 and 500, which are divisible by 8.

The numbers lying between 100 and 500 which are divisible by 8 are

104, 112, 120, 128, 136,…, 496

a₂ – a₁ = 112 – 104 = 8

a₃ – a₂ = 120 – 112 = 8

∵ a₃ – a₂ = a₂ – a₁ = 8

Therefore, the series is in AP

Here, a = 120, d = 8 and a_n = 496

We know that,

a_n = a + (n – 1)d

⇒ 496 = 104 + (n – 1)8

⇒ 496 – 104 = (n – 1)8

⇒ 392 = (n – 1)8

⇒ 49 = (n – 1)

⇒ n = 50

Now, we have to find the sum of this AP

⇒ S₅₀ = 25[208 + 49 × 8]

⇒ S₅₀ = 25[600]

⇒ S₅₀ = 15000

Hence, the sum of all numbers lying between 100 and 500 and divisible by 8 is 15000.