Three - digit natural numbers which are divisible by 13 are 104, 117, 130, …, 988.

Sum of these numbers forms an arithmetic series 104 + 117 + 130 + … + 988.

Here, first term = a = 104

Common difference = d = 13

We first find the number of terms in the series.

Here, last term = l = 988

∴ 988 = a + (n - 1)d

⇒ 988 = 104 + (n - 1)13

⇒ 988 - 104 = 13n - 13

⇒ 884 = 13n - 13

⇒ 884 + 13 = 13n

⇒ 13n = 897

⇒ n = 69

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 69 terms of this arithmetic series is given by:

∴ S₆₉ = [2(104) + (69 - 1)(13)]

= (69/2) × [208 + 884]

= (69/2) × 1092

= 69 × 546

= 3767