bxzbthf的部落格

標題:

S6 Maths. Sequences&Series 一題

發問:

• 香港酒店約$5-600,有咩好介紹-- 10分
• 刑法問題---希望知識+先進幫忙@1@
• 過失傷害的後續處理...請幫幫我!!謝謝!!@1@
• 補領身份証一問@1@
• 台中縣清水國小附設的安親班
• 護理係....（急２０點）@1@
• 找全球最高人口的五個地方
• 急需！！南部旅遊景點～台南、高雄@1@
• 電腦配備與顯卡階級
• 關於美職的球場!!

 

此文章來自奇摩知識+如有不便請留言告知

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

最佳解答:

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

其他解答:

還是要去 http://aaashops。com 品質不錯，老婆很喜歡。 厺及叛匕匄勴8758B59A7FA1EEA7