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S6 Maths. Sequences&Series 一題

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Given that the sum of the first 10 terms of an arithmetic series is 310, and the sumfrom the 6th term to the 13th term is 408, find (a) the first term and the common difference (b) the sum of the first 30 terms of the series. 更新: 求 : 完整步驟 + 答案

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first term = a nth term = a + (n-1)d Sum of n terms = {(a) + [a+(n-1)d]}(n)/2 sum of the first 10 terms of an arithmetic series is 310 gives (a + a + 9d)(10)/2 = 310 2a + 9d = 62 ................... (1) 6,7,8,9,10,11,12,13 6th term to 13th term are 8 terms 6th term is a + 5d 13th term is a + 12d [(a + 5d) + (a + 12d)](8)/2 = 408 (2a + 17d) = 102 ............ (2) (2) - (1) ===> 8d = 40 d = 5 sub into (1) ==> 2a + 9(5) = 62 2a = 62 - 45 = 17 a = 8.5 (b) the sum of the first 30 terms Sum = (a + a+29d)(30)/2 = [8.5 + 8.5 + (29)(5)](30)/2 = 2430

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