1.设数列{an}是等差数列,且a2=-6,a8=6,前n项和为Sn,当Sn最大时,则n=___.2.在等差数列{an}中,若a10=10,a19=100,Sn=0,则n=___.3.已知等差数列{an},an=2n-1,则S1+S2+S3+S4+S5=___.4.在等差数列{an}中,d不等于0,已知S100=100S10,则a100/a10=___.
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2008-12-15
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2007-09-10
在等差数列an中,已知公差d?0,前n项和为Sn,且满足a2×a3 = 45,a1 + a4 = 14。若bn = Sn÷(n + c)且bn为等差数列,求非零常数。 a1 + a4 = 14 (a2 - d) + (a3 + d) = 14 a2 + a3 = 14 a2×a3 = 45 联立、,并解得 a2 = 5,a3 = 9 d = 4 a1 = 5 - 4 = 1 S1 = a1 = 1 S2 = S1 + a2 = 6 S3 = S2 + a3 = 15 。 。。。。。 b1 = S1/(1 + c) = 1/(1 + c) b2 = S2/(2 + c) = 6/(2 + ...全部
在等差数列an中,已知公差d?0,前n项和为Sn,且满足a2×a3 = 45,a1 + a4 = 14。若bn = Sn÷(n + c)且bn为等差数列,求非零常数。 a1 + a4 = 14 (a2 - d) + (a3 + d) = 14 a2 + a3 = 14 a2×a3 = 45 联立、,并解得 a2 = 5,a3 = 9 d = 4 a1 = 5 - 4 = 1 S1 = a1 = 1 S2 = S1 + a2 = 6 S3 = S2 + a3 = 15 。 。。。。。 b1 = S1/(1 + c) = 1/(1 + c) b2 = S2/(2 + c) = 6/(2 + c) b3 = S3/(3 + c) = 15/(3 + c) 。 。。。。。 2×b2 = b1 + b3 2×6/(2 + c) = 1/(1 + c) + 15/(3 + c) 12(1 + c)(3 + c) = (2 + c)(3 + c) + 15(1 + c)(2 + c) 12c^2 + 48c + 36 = c^2 + 5c + 6 + 15c^2 + 45c + 30 4c^2 + 2c = 0 2c^2 + c = 0 c(2c + 1) = 0 c = -1/2 _____________________________ b1 = 1/(1 - 1/2) = 2 b2 = 6/(2 - 1/2) = 4 b3 = 15/(3 - 1/2) = 6 。 。。。。。 由此可见,{bn}是首项为2,公差也为2的等差数列。收起
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