標題:
Maths
http://x5f.xanga.com/46fc527373435161478067/z121578884.jpg 更新: ALL WRONG
最佳解答:
22. Let the first term be a and the common ratio be r. a + ar = 54 a (1 + r) = 54 a = 54 / (1 + r) … (1) a / (1 – r) = 72 … (2) Substitute (1) into (2): [54 / (1 + r)] / (1 – r) = 72 54 / (1 + r) = 72(1 – r) 54 = 72 (1 – r) (1 + r) 3 = 4 (1 – r^2) 4 – 4r^2 = 3 4r^2 – 1 = 0 r^2 = 1/4 r = ±1/2 When r = 1/2, a = 54 / (1 + 1/2) a = 54 / (3/2) a = 36 When r = -1/2, a = 54 / (1 – (1/2)) a = 54 / (1/2) a = 108 ∴ The first term is 108 and the common ratio is -1/2 or the first term is 36 and the common ratio is 1/2. 26. (a) Since logx + log2, logx2 + log4, logx3 + log8 are in arithmetic sequence. The first term, a = log(x) + log2 = log(2x) The common difference, d = (logx2 + log4) – (logx + log2) = 2logx + 2log2 – logx – log2 = logx + log2 = log(2x) The general term = (log(2x)) + (n – 1) (logx + log2) = log(2x) + n(logx + log2) – log(2x) = n(logx + log 2) = nlog(2x) (b) ∵ the given sequence is an arithmetic sequence, ∴ the sum of the first ten terms = (10/2)[2(log(2x)) + (10 – 1)(log(2x))] = 5[2log(2x) + 9log(2x)] = 5(11log(2x)) = 55log(2x)
其他解答:
logx+2log*2log+log4*3log+log8
Maths
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發問:http://x5f.xanga.com/46fc527373435161478067/z121578884.jpg 更新: ALL WRONG
最佳解答:
22. Let the first term be a and the common ratio be r. a + ar = 54 a (1 + r) = 54 a = 54 / (1 + r) … (1) a / (1 – r) = 72 … (2) Substitute (1) into (2): [54 / (1 + r)] / (1 – r) = 72 54 / (1 + r) = 72(1 – r) 54 = 72 (1 – r) (1 + r) 3 = 4 (1 – r^2) 4 – 4r^2 = 3 4r^2 – 1 = 0 r^2 = 1/4 r = ±1/2 When r = 1/2, a = 54 / (1 + 1/2) a = 54 / (3/2) a = 36 When r = -1/2, a = 54 / (1 – (1/2)) a = 54 / (1/2) a = 108 ∴ The first term is 108 and the common ratio is -1/2 or the first term is 36 and the common ratio is 1/2. 26. (a) Since logx + log2, logx2 + log4, logx3 + log8 are in arithmetic sequence. The first term, a = log(x) + log2 = log(2x) The common difference, d = (logx2 + log4) – (logx + log2) = 2logx + 2log2 – logx – log2 = logx + log2 = log(2x) The general term = (log(2x)) + (n – 1) (logx + log2) = log(2x) + n(logx + log2) – log(2x) = n(logx + log 2) = nlog(2x) (b) ∵ the given sequence is an arithmetic sequence, ∴ the sum of the first ten terms = (10/2)[2(log(2x)) + (10 – 1)(log(2x))] = 5[2log(2x) + 9log(2x)] = 5(11log(2x)) = 55log(2x)
其他解答:
logx+2log*2log+log4*3log+log8
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