.已知等差数列{an}公差d不等于0,它的前n项和为sn,若s5=70,且a2,a7,a22成等比数列(1)求{an}的
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.已知等差数列{an}公差d不等于0,它的前n项和为sn,若s5=70,且a2,a7,a22成等比数列(1)求{an}的通项公式(2数列{sn分之1}的前n项和为Tn,求证6分之1≤Tn<8分之3
![.已知等差数列{an}公差d不等于0,它的前n项和为sn,若s5=70,且a2,a7,a22成等比数列(1)求{an}的](/uploads/image/z/14985881-17-1.jpg?t=.%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E5%85%AC%E5%B7%AEd%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E5%AE%83%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E8%8B%A5s5%3D70%2C%E4%B8%94a2%2Ca7%2Ca22%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%281%29%E6%B1%82%7Ban%7D%E7%9A%84)
(1)
an = a1+(n-1)d
S5=70
(a1+2d)5 = 70
a1+2d = 14 (1)
a2,a7,a22成等比数列
a2.a22= (a7)^2
(a1+d)(a1+21d)=(a1+6d)^2
(14-d)(14+19d)=(14+4d)^2 ( from (1) :a1+2d=14)
196+252d-19d^2= 196+112d+16d^2
35d^2-140d=0
d^2-4d=0
d=4
a1=6
an = 6+4(n-1) = 4n+2
(2)
Sn = 2n(n+2)
1/Sn = (1/4)[1/n -1/(n+2)]
Tn = 1/S1+1/S2+...+1/Sn
= (1/4)[ 1+ 1/2 - 1/(n+1) -1/(n+2) ]
< (1/4) ( 1+ 1/2)
= 3/8
Tn ≥ T1 = 1/6
ie
1/6≤Tn < 3/8
an = a1+(n-1)d
S5=70
(a1+2d)5 = 70
a1+2d = 14 (1)
a2,a7,a22成等比数列
a2.a22= (a7)^2
(a1+d)(a1+21d)=(a1+6d)^2
(14-d)(14+19d)=(14+4d)^2 ( from (1) :a1+2d=14)
196+252d-19d^2= 196+112d+16d^2
35d^2-140d=0
d^2-4d=0
d=4
a1=6
an = 6+4(n-1) = 4n+2
(2)
Sn = 2n(n+2)
1/Sn = (1/4)[1/n -1/(n+2)]
Tn = 1/S1+1/S2+...+1/Sn
= (1/4)[ 1+ 1/2 - 1/(n+1) -1/(n+2) ]
< (1/4) ( 1+ 1/2)
= 3/8
Tn ≥ T1 = 1/6
ie
1/6≤Tn < 3/8
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