# 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 an = 988

We know that,

an = 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

S69 = 37674

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

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