Mathematics 2011 JAMB Past Questions - Question 21: Find the sum of the first 18 terms of the series 3,6,9,......36.

Correct Answer

B

Explanation

3,6,9,...,36a = 3 d = 3 I = 36, n = 18Ssubn = n/2 <2a+(n-1)dSsub18 = 18/2 <2x3+(18-1)3> = 9<6+(17x3) = 9<6+51> = 9 (57) = 513The given series is an arithmetic series with a common difference of 3. To find the sum of the first 18 terms, we can use the formula for the sum of an arithmetic series:Sn = (n/2)(2a + (n-1)d)Where Sn is the sum of the first n terms, a is the first term, and d is the common difference.In this case, n = 18, a = 3, and d = 3.Plugging in these values into the formula, we have:S18 = (18/2)(2(3) + (18-1)(3))Simplifying further:S18 = 9(6 + 17(3))S18 = 9(6 + 51)S18 = 9(57)S18 = 513Therefore, the sum of the first 18 terms of the series 3, 6, 9, ...... 36 is 513.