Find the sum of all the 3 digit natural numbers which are divisible by 13.

The three digit natural numbers which are divisible by 13 are

104, 117, 130,…, 988

a₂ – a₁ = 117 – 104 = 13

a₃ – a₂ = 130 – 117 = 13

∵ a₃ – a₂ = a₂ – a₁ = 13

Therefore, the series is in AP

Here, a = 104, d = 13 and a_n = 988

We know that,

a_n = a + (n – 1)d

⇒ 988 = 104 + (n – 1)13

⇒ 988 – 104 = (n – 1)13

⇒ 884 = (n – 1)13

⇒ 68 = (n – 1)

⇒ n = 69

Now, we have to find the sum of this AP

⇒ S₆₉ = 69[104 + 34 × 13]

⇒ S₆₉ = 69[546]

⇒ S₆₉ = 37674

Hence, the sum of three digit natural numbers which are divisible by 13 are 37674.