Find the sum of the first 18 terms of the series 3, 6, 9,…, 36.

Correct Answer: Option B

Explanation

3, 6, 9,…, 36.
a = 3, d = 3, i = 36, n = 18
S_n = (frac{n}{2}) [2a + (n – 1)d
S₁₈ = (frac{18}{2}) [2 x 3 + (18 – 1)3] = 9[6 + (17 x 3)] = 9 [6 + 51] = 9(57)
= 513