www.1862.net > 已知数列{An}为等比数列,A1×A9=64,A3+A7=20,求A...

已知数列{An}为等比数列,A1×A9=64,A3+A7=20,求A...

a1×a9=64,所以a3×a7=64,又a3+a7=20,解得a3=4,a7=16;或a3=16,a7=4.当a3=4,a7=16时,a11=64.当a3=16,a7=4时,a11=1

因为是等比数列,因此 a3*a7 = a1*a9 = 64 , 又 a3+a7 = 20 ,因此解得 a3 = 4 ,a7 = 16 或 a3 = 16 ,a7 = 4 , 由 a3、a7、a11 成等比得 a11 = (a7)^2/a3 = 64 或 1 。

解: 由等比中项性质得:a7²=a3·a11 a3=1,a7=5 a11=a7²/a3=5²/1=25 a11的值为25 本题考查的是等比中项性质,不需要求出首项和公比。

a3+a5+a7=42 计算过程 因为a1=3,a1+a3+a5=21, 所以a1+a3+a5=a1(1+q^2+q^4)=21 1+q^2+q^4=7 q^4+q^2-6=0 (q^2+3)(q^2-2)=0 q^2=2,q^2=-3(无解) q^2=2 然后计算a3+a5+a7=a1(q^2+q^4+q^6) a1q^2(1+q^2+q^4) =21q^2 =42 最终:a3+a5+a7=42 等比数...

(1)设等差数列{an}的公差为d,则5a1+5×42×d=20(a1+2d)2=a1?(a1+6d),解得d=1a1=2,∴an=2+n-1=n+1.(2)由(1)得,1anan+1=1(n+1)(n+2)=1n+1?1n+2,则Tn=(12?13)+(13?14)+…+(1n+1?1n+2)=12?1n+2=n2(n+2),∴Tn≤λan+1对一切n∈N*恒成...

let bn =an +1 bn = b1.q^(n-1) ; bn>0 b3 = a3+1 =8 b1.q^2 =8 (1) a7=127 b7 = 127+1=128 b1.q^6 = 128 (2) (2)/(1) q^4 = 16 q=2 from (1) b1.q^2 =8 b1= 2 bn = 2^n a5 = b5 -1 =32 -1 =31

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(1)依题意,{an}为等差数列设其公差为d;{bn}为等比数列,设其公比为q,则可知q>0∵a3+a7=10∴2a5=10即a5=5又a1=1∴a5-a1=4d=4解得d=1故an=a1+(n-1)d=n由已知b3=a4=4∴q2=b3b1=4即q=2∴bn=b1qn-1=2n-1∴an=n,bn=2n-1(2)∵cn=an?bn=n?2n-1∴Tn=...

设数列{an}的公差为d(d≠0),由a32=a1a7得(a1+2d)2=a1(a1+6d)?a1=2d,故q=a3a1=a1+2da1=2a1a1=2,故选 C.

a1+a3+a5+a7+a9=10 (1) a1+2+a3+2+a5+2+a7+2+a9+2=10+10 a2+a4+a6+a8+a10=20 (2) 由(1)+(2)得 a1+a3+a5+a7+a9+a2+a4+a6+a8+a10=10 +20 S10=30

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