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

A.
505 
B.
513 
C.
433 
D.
635
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
S_{18} = (frac{18}{2}) [2 x 3 + (18 – 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513