2011
Mathematics
67ca00ce0c643c71d77dee1f
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: 513
Explanation
3, 6, 9,..., 36. a = 3, d = 3, i = 36, n = 18 Sn = n/2 [2a + (n - 1)d SS18 = 18/2 [2 x 3 + (18 - 1)3] = 9[6 + (17 x 3)] = 9 [6 + 51] = 9(57) = 513
3, 6, 9,..., 36. a = 3, d = 3, i = 36, n = 18 Sn = n/2 [2a + (n - 1)d SS18 = 18/2 [2 x 3 + (18 - 1)3] = 9[6 + (17 x 3)] = 9 [6 + 51] = 9(57) = 513THIS WEEK's
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