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若等比数列An满足log3A1 log3A2......... log3A10=...

log3a1+log3a2+…+log3a10=log3(a1a10)+log3(a2a9)+…log3(a5a6)=5log3(a5a6)=10故答案为:10

由等比数列的性质得,a4a7=a5a6,因为a4a7+a5a6=18,所以a5a6=9,所以log3a1+log3a2+…+log3a10=log(a1a2a3…a10)3=log(a5a6)53=log(9)53=10,故选:A.

解: 由于{an}为等比数列 则:a5a6=a4a7=a3a8=a2a9=a1a10 又a5a6+a4a7=18 则: 2a5a6=18 a5a6=9 则: log3(a1)+log3(a2)+...+log3(a9)+log3(a10) =log3[a1*a2*a3*...*a10] =log3[(a1a10)*(a2a9)*...*(a5a6)] =log3[9*9*...*9] =log3[9^5] =log3[3^10...

∵a10a11=a9a12,∴a10a11+a9a12=2a10a11=486∴a10a11=486∴log3a1+log3a2+…log3a20=log3(a10a11)10=10log3(a10a11)=50.故答案为:50.

+log3a9+log3a10等于 等比数列{an}中a2a9=3,则log3a1+log3a2+…+log3a9+log3a10等于?... 等比数列{an}中a2a9=3,则log3a1+log3a2+…+log3a9+...

(1)设数列{an}的公差为d,则a1+2d=?42a1+9d=2,解之得a1=?8d=2,即an=-8+2(n-1)=2n-10(2)由an=log3bn,可得bn=3an=32n?10则Tn=b1?b2…bn=3-8?3-6…32n-10=3(?8+2n?1)n2=3n(?8+2n?10)2=3n2?9n∵Tn>1.∴3n2?9n>1∴n2-9n>0,∴n>9,

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