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Maths!!請幫幫忙!

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2. In a sequence, the sum of first n terms is S(n )=2n^2+3n for all values of n. Find the 100th term. 4. Find the value(s) of x such that x+9, x-6 and 4 are in geometric sequence. 8. The third term of an arithmetic sequence is 25 and the tenth term is -3. Find the first term and the common difference. 更新: 可否詳細解釋第4題為何(x-6)/(x+9) = 4/(x-6)?

最佳解答:

2. Sum of the first 100 terms, S(100) = 2(100)2 + 3(100) = 20300 Sum of the first 99 terms, S(99) = 2(99)2 + 3(99) = 19899 The 100th term = S(100) - S(99) = 20300 - 19899 = 401 ====== 4. Consider the common ratio: (x-6)/(x+9) = 4/(x-6) (x-6)2 = 4(x+9) x2-12x+36 = 4x+36 x2-16 = 0 x(x-16) = 0 x = 0 ooro x = 16 ====== 8. Let a and d be the first term and the common difference. The 3th term: a + 2d = 25 ...... (1) The 1oth term: a + 9d = -3 ...... (2) (2)-(1): 7d = -28 d = -4 Put d=-4 into (1): a + 2(-4) = 25 a = 33 Ans: The first term is 33, and the common difference is -4. ====== 2008-09-08 01:23:34 補充： In Q.4: Consider the first two terms, common ratio = (x-6)/(x+9) Consider the last two terms, common ratio = 4/(x-6)

 

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