Answer: A. 45
Answer: (A) 45
The first five multiples of 3 are 3, 6, 9, 12, 15. Sum \(= 3 + 6 + 9 + 12 + 15 = 45\).
(A.P. with \(a = 3\), \(d = 3\), \(n = 5\): \(S_5 = \dfrac{5}{2}[2(3) + 4(3)] = \dfrac{5}{2}(18) = 45\).)
Best answer
Answer: A. 45
Answer: (A) 45
The first five multiples of 3 are 3, 6, 9, 12, 15. Sum \(= 3 + 6 + 9 + 12 + 15 = 45\).
(A.P. with \(a = 3\), \(d = 3\), \(n = 5\): \(S_5 = \dfrac{5}{2}[2(3) + 4(3)] = \dfrac{5}{2}(18) = 45\).)
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