hzb53jl55v的部落格

標題:

nth term的maths問題

發問:

(a):16,13,10,7......求nth term (b):2,-6,18,-54.....求nth term (c):1/8,-6/13,9/5,-54/7......求nth term

最佳解答:

a) common difference = 13-16 = -3 nth term = 16-3(n-1) b) common ratio = (-6)/2 = -3 nth term = 2*(-3)^(n-1) c) T(3)-T(2) = 9/5-(-6/13) = 147/65 T(2)-T(1) = -6/13-1/8 = -61/104 ∴No common difference ∴It is not A.S. T(3)/T(2) = (9/5)/(-6/13) = -39/10 T(2)/T(1) = (-6/13)/(1/8) = -48/13 ∴No common ratio ∴It is not G.S.

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• 43%與40% vol Black Label既分別@1@
• 鋼之煉金術師 10分!!
• 腳的長度是由哪部份至哪部份？@1@

 

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其他解答:

a) a = 16 There exists a common difference d = 13-16 = 10-13 = 7-10 = -3 The series is an AP. Hence, the nth term T(n) = a+(n-1)d = 16-3(n-1) = 19 - 3n b) a = 2 There exists a common ratio r = (-6)/2 = 18/(-6) = (-54)/18 = -3 The series is a GP. Hence, the nth term T(n) = a r^(n-1) = 2*(-3)^(n-1) c) There is no common difference nor common ratio found, however, the series can be rewritten as: 2/16, -6/13, 18/10, -54/7 .... It can be observed that each term is a combination of the terms in (a) and (b). The numerator comes from (b) and the denumerator comes from (a). Hence, the nth term T(n) = [2*(-3)^(n-1)] / (19 - 3n)