Find the sum of first 12 natural numbers each of which is a multiple of 7.

First 12 natural numbers that are a multiple of 7 are,

7, 14, 21, ……….., 12^th term.

The above sequence is an A.P. with a = 7 and common difference = 7.

$S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] = \dfrac{12}{2}(2 \times 7 + (12 - 1) \times 7) \\[1em] = 6(14 + 77) \\[1em] = 6 \times 91 \\[1em] = 546.$

Hence, sum of first 12 natural numbers that are divisible by 7 is 546.