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设数列{An}是等差数列,数列{Bn}是等比数列,记数列{An}、{Bn}的前n项和分别为Sn、Tn...

由等比数列的性质可知,a4?a6=a52,又a4?a6=2a5,∴a52=2a5,∴a5=2.∴b5=2a5=4.则s9=9(b1+b9) 2 =9b5=36.故答案为:36.

解:设等差数列的等差为d,等比数列的等比是q则a3=b3a4-d=b4/q又∵a4=b4∴a4-d=a4/qa4-a4/q=d∵(S5-S3)/(T4-T2)=5∴(a5+a4)/(b4+b3)=(a4+d+a4)/(a4+b4/q)=(2a4+d)/(a4+a4/q)=5即(2a4+a4-a4/q)/(a4+a4/q)=5左边可以分子分母同时除以

答案是-5/13由S7-S5=4(T6-T4)可得a7+a6=4(b6+b5)=4(a6+a5),就可以得到首项公差的关系:a1=-25d/6由a5=b5,a6=b6得a5/a6=b5/b6=1/q,将a1=-25d/6代入上面的关系式就可以求得等比数列的公比q=-5(a7+a5)/(b7+b5) =2a6/(b7+b5)=2b6/(b7+b5)=2q/1+q^2=-10/26=-5/13

设等差数列的公差为d,等比数列的公比是q∵a3=b3,a4=b4,∴a4-d=a4q,a4-a4q=d,∵S5-S3T4-T2=5,∴a5+a4b4+b3=2a4+da4+a4q=5,即2a4+a4-a4qa4+a4q=5,左边可以分子分母同时除以a4,得:3-1q1+1q=5,左边分数上下同时乘以q,得:3q-1q+1=5,解得q=-3,根据等差中项可知,a5+a3=2a4,∴a5+a3b5+b3=2a4b4q+b4q=2a4-3a4-a43=2-103=-35.

a3=b3,a4=b4q=b4/b3=a4/a3q=(a1十3d)/(a1十2d)(s5-s3)/(t4-t2)=55a1十10d-3a1-3d=10a1十25d4a1十9d=0a1=-9d/4q=-3a5十a3/b5十b3=2a4/a3十9a3=a4/6a3=(-9d/4十3d)/(-27d/2十12d)=3d/4÷(-3d/2)=-1/2

设等差数列的等差为d,等比数列的等比是q,由a3=b3,得a4-d=b4q,又∵a4=b4,∴a4-a4q=d,∵S5-S3T4-T2=7,∴a5+a4b4+b3=a4+d+a4a4+a4q=7,即3a4-a4qa4+a4q=7,即q=-2.∴a5b3+b6=a5a3+4a4=a4+32a4a4-32a4+4a4=57.故选:C.

解:∵an是sn与2的等差中项∴2an=sn+2 (*)令n=1,得2a1=s1+2=a1+2∴a1=2由(*)得:2a(n+1)=s(n+1)+2两式相减,得:2a(n+1)-2an=a(n+1)即a(n+1)=2ana(n+1)/an=2∴{an}是以首项a1=2,公比q=2的等比数列∴an=22^(n-1)=2^n点p(

抄错题了吧?{an}为等差数列,a1=1,a2=1,那么d=0,an=1{bn}为等比数列,b1=1,b2=1,那么q=1,bn=1然后这明显和已知条件不符

∵{an}是等差数列数列,{bn}是各项都为正数的等比数列∴b1+b2=a2∵a1=b1=1即1+q=1+d ∴q=d又∵2b3=a1+a4 ∴2q=2+3d结合q=d得q=2或q=-1/2∵bn各项均为正数∴q=-1/2(舍去) q=d=2∴an=1+2(n-1)=2n-1bn=2^(n-1)令Tn=an/bn=(2n-1)/(2^(n-1))应该是会用到错位相减法,太麻烦了,不算了

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