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等比数列A2=2,A2+A4+A6=14,A6=?

参考

等比数列中a2+a4+a6不具备像等差数列的这种性质。 等比数列中有类似的形式:a2·a4·a6=a4³

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

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...

由数列{an}是等比数列,a1=2,a4=14,可得 公比q=12,首项a1=2,∴an=(12)n?2,an+1=(12)n?1,∴anan+1 =(12)2n?3,又a1a2=2,∴数列{anan+1 }是2为首项,14公比为的等比数列,∴a1a2+a2a3+…+a5a6=2×[1?(14)5]1?14=2×43×(1?1210)=341128.故答案为34...

解:a1*q^3=a1^2*q^2 即a1=q 即q^2+q^4=5/16 ∴q^2=1/4 ∴q=1/2 即a5=q^5 =1/32 如有疑问,可追问!

∵数列{an}的奇数项成等差数列,偶数项成等比数列,公差与公比均为2,∴a3=a1+2,a5=a1+4,a7=a1+6,a4=2a2,a6=4a2,∵a2+a4=a1+a5,a4+a7=a6+a3∴a2+2a2=a1+4+a1,2a2+6+a1=4a2+2+a1∴a1=1,a2=2,∵am?am+1?am+2=am+am+1+am+2成立,∴由上面可以知数...

(Ⅰ) 设等比数列{an}的首项为a1>0,公比为q>0,∵a2?a4=a6,2a3+1a4=1a5,∴a1q?a1q3=a1q52a1q2+1a1q3=1a1q4,解得a1=q=12,∴an=12n.(Ⅱ)∵an=12n,∴Sn=12+122+…+12n=12×(1?12n)1?12=1?12n,Tn=12×122×…×12n=(12)n(n+1)2,若存在正整...

∵数列{an}是一个公差不为0等差数列,且a2=2,并且a3,a6,a12成等比数列,∴a62=a3?a12,∴(2+4d)2=(2+d)(2+10d),∵d≠0,∴d=1.∴an=2+(n-2)=n,∴1anan+1=1n-1n+1,∴1a1a2+1a2a3+1a3a4+…+1anan+1=1-12+12-13+…+1n-1n+1=1-1n+1=nn+1,故答案...

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