设Sn是等差数列{an}的前n项和,已知1/3S3与1/4s4的等比中项为1/5S5,1/3S3与1/4s4的等差中项为1,求等差数列{an}的通项
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设Sn是等差数列{an}的前n项和,已知1/3S3与1/4s4的等比中项为1/5S5,1/3S3与1/4s4的等差中项为1,求等差数列{an}的通项
设Sn是等差数列{an}的前n项和,已知1/3S3与1/4s4的等比中项为1/5S5,1/3S3与1/4s4的等差中项为1,求等差数列{an}的通项
设Sn是等差数列{an}的前n项和,已知1/3S3与1/4s4的等比中项为1/5S5,1/3S3与1/4s4的等差中项为1,求等差数列{an}的通项
根据是等差数列,
把S4,S3,S5都用Sn=na1+n(n-i)d/2表示出来.
两个条件对应两个方程,
根据等差数列
与等比数列的性质
化简后
得到两个方程2a1+5/2d=23a1+5d=0
然后结得an=-12/5n+32/5
et4
Sn=na1 + [n(n-1)d]/2
1/3S3=1/3{3a1 + [3×(3-1)]d/2} = 1/3(3a1 + 3d)= a1 + d
1/4S4=1/4{4a1 + [4×(4-1)]d/2} = 1/4(4a1 + 6d)= a1 + 3d/2
1/5S5=1/5{5a1 + [5×(5-1)]d/2} = 1/5(5a1 + 10d)= a1 + 2d<...
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Sn=na1 + [n(n-1)d]/2
1/3S3=1/3{3a1 + [3×(3-1)]d/2} = 1/3(3a1 + 3d)= a1 + d
1/4S4=1/4{4a1 + [4×(4-1)]d/2} = 1/4(4a1 + 6d)= a1 + 3d/2
1/5S5=1/5{5a1 + [5×(5-1)]d/2} = 1/5(5a1 + 10d)= a1 + 2d
1/3S3与1/4S4的等比中项为1/5S5,
(a1 + 2d)^2 =(a1 + d)×(a1 + 3d/2)
3a1d/2 + 5d^2/2 = 0 ......(1)
1/3S3与1/4S4的等差中项为1,
(a1 + d)+(a1 + 3d/2)=2×1
a1 + 5d/4 = 1......(2)
(1),(2)两式联立,解得:d=0或d=-12/5
当d=0时,代入(2)中,a1=1 ,an=1
当d=-12/5时,代入(2)中,a1=4 ,an=32/5 - 12n/5赞同14| 评论
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