(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/b11的值为( )(2)已知等比数列{an},首项为81,数列{bn}满足bn=log3(an),其前n项和为Sn.①证明{bn}为等差数列;

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(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/b11的值为( )(2)已知等比数列{an},首项为81,数列{bn}满足bn=log3(an),其前n项和为Sn.①证明{bn}为等差数列;

(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/b11的值为( )(2)已知等比数列{an},首项为81,数列{bn}满足bn=log3(an),其前n项和为Sn.①证明{bn}为等差数列;
(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/b11的值为( )
(2)已知等比数列{an},首项为81,数列{bn}满足bn=log3(an),其前n项和为Sn.①证明{bn}为等差数列;②若S11≠S12,且S11最大,求{bn}的公差d的范围.

(1)若两等差数列{an},{bn}的前n项和分别为An,Bn,满足An/Bn=(7n+1)/(4n+27),则a11/b11的值为( )(2)已知等比数列{an},首项为81,数列{bn}满足bn=log3(an),其前n项和为Sn.①证明{bn}为等差数列;
1,S21=(a1+a21)*21/2
a1+a21=a1+a1+20d=2(a1+10d)=2a11
所以S21=2a11*21/2=21*a11
同理T21=21*b11
所以a11/b11=S21/T21
=(7*21+1)/(4*21+27)
=4/3
2,(1)等比数列{an},首项为81
设an=a1*q^(n-1)=81*q^(n-1)
数列{bn}满足bn=log3为底an
∴bn=log3为底[81*q^(n-1)]=log3为底81+log3为底q^(n-1)
=4+(n-1)log3为底q
=log3为底q*n+4-log3为底q
∵log3为底q为常数,∴bn为以4为首项,log3为底q为公差的等差数列
(2)若S11≠S12,且S11最大,
∴b11>0,b12<0
于是bn=n*d+4-d
∴10d+4>0,11d+4<0
∴-2/5<d<-4/11

1
a11/b11
=2a11/2b11
=(a1+a21)/(b1+b21)
=[21*(a1+a21)/2]/[21*(b1+b21)/2]
=A21/B21=(7*21+1)/(4*21+27)=148/111
2
(1)
bn-b(n-1)
=log3 an-log3 a(n-1)
=log3 [an/a(n-...

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1
a11/b11
=2a11/2b11
=(a1+a21)/(b1+b21)
=[21*(a1+a21)/2]/[21*(b1+b21)/2]
=A21/B21=(7*21+1)/(4*21+27)=148/111
2
(1)
bn-b(n-1)
=log3 an-log3 a(n-1)
=log3 [an/a(n-1)]
设等比数列公比为q
则bn-b(n-1)=log3 q为定值
∴{bn}为等差数列
(2)
b1=log3 a1=4
S12=S11+a12
∵S11最大,且S11≠S12
∴a12<0
即4+11d<0
d<-4/11

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