www.1862.net > 求解: 等比数列{An}中,A1+A4+A7=39,A2+A5+A8=...

求解: 等比数列{An}中,A1+A4+A7=39,A2+A5+A8=...

33=a2+a5+a8=(a1+d)+(a4+d)+(a7+d)=a1+a4a7+3d=39+3d 3d=-6 a3+a6+a9=a2+a5+a8+3d=33-6=27

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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...

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由等比数列的性质可得:a3+a4=(a1+a2)q2,∵a1+a2=30,a3+a4=60,∴q2=2,∴q6=(q2)3=8,则a7+a8=(a1+a2)q6=30×8=240.故答案为:240

a4a5a6=a5^3=8 a5=2 q^3=a5/a2=2 a2+a5+a8+a11=a2(1+q^3+q^6+a^9)=1*(1+2+4+8)=15

1.a1+a2=a1+a1q=10 2.a3+a4=a1*q*q+a1*q*q*q=40 2式除以1式,得q*q=4 a5+a6=(a3+a4)*q*q=160

log4a1+log4a2+...log4a8=log4(a1•a2•......•a7•a8) 因为a4•a5=a1•a8=a2•a7=a3•a6=32 所以有原式=log4(32的四次方)=log2的二次方(2的二十次方)=20/2=10

(1)设等比数列{an}的公比为q,由a5+a7=8(a2+a4),得a1q4(1+q2)=8a1q(1+q2),又∵a1=2,q≠0,1+q2>0,∴q=2,数列{an}的通项公式为an=2n,n∈N*,由题意有a1b1=(1-1)?21+1+2=2,∴b1=1,当n≥2时,anbn=(n-1)?2n+1-[(n-2)?2n+2]=n?2n,∴b...

(a3)^2+2a3a8+(a8)^2=36 (a3+a8)^2=36 因为an>0 a3+a8=6

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