.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

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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

.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
(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