Find the sum of the first 18 terms of the series 3, 6, 9,…, 36.
- 505
-
513
- 433
- 635
The correct answer is: B
Explanation
3, 6, 9,..., 36.a = 3, d = 3, i = 36, n = 18
Sn = \(\frac{n}{2}\) [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
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