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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...

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...

a3^2-2a3a5+a5^2=(a3-a5)^2=36 ∵an是递增数列∴a5-a3=6,q>0 ∴q^4-q^2-6=0∴q=根下3 an=3^(n-1)/2

∵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...

设等比数列{an}的公比为q,∵a1+a6=11,a3?a4=329,23a2, a23, a4+49成等差数列,∴由等比数列的性质知a1a6=a3a4=329,∴a1,a6是方程x2?11x+329=0的两个根,解得a1=13,a6=323或a1=323,a6=13,当a1=13,a6=323时,13×q5=323,解得q=2...

a1+a2+a3+a4=1,a5+a6+a7+a8=2 两式相减得 4d+4d+4d+4d=1 d=1/16 a1+a1+d+a1+2d+a1+3d=1 ∴a1=5/32 5n/32+n(n-1)/32=15 解得n=20

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