MATHEMATICS
JAMB 2009 - Question 18
Mathematics 2009 JAMB Past Questions - Question 18: The sum of the first n terms of the arithmetic progression 5,11,17,23,29,35 .... is
A:
B:
C:
D:
Correct Answer
B
Explanation
Sn=n/2{2a+(n-1)d}=n/2{10+(n-1)6}=1/2<10+6n-6>=n2<4+6n>=n(2+3n)To find the sum of the first n terms of an arithmetic progression, we can use the formula: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, the first term a = 5 and the common difference d = 6 (since each term increases by 6).Plugging these values into the formula, we get:Sn = (n/2)(2(5) + (n-1)(6))Simplifying further:Sn = (n/2)(10 + 6n - 6)Sn = (n/2)(6n + 4)Sn = 3n² + 2nTherefore, the sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35 ... is 3n² + 2n.
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