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在等比数列{An}中,若公比q>1,且A3A7=16,A4+A6=...

由题意可得:数列{an}为等比数列,所以a2015a2010=q5.因为数列{an}为等比数列,a3a4=12,所以a3a4=a1a6=12.因为a1+a6=8,公比q>1,解得a1=2,a6=6,所以q5=a6a1=3.故答案为:3

∵a2a8=a4a6=6,a4+a6=5,∴a4和a6为方程x2-5x+6=0的两根解得两个根为:2和3、∵q>1∴a6=3,a4=2∴q2=32∴a5a7=1q2=23故答案为:23

an = a1.q^(n-1) a3+a6=36 a1q^2.(1+q^3) =36 (1) a4+a7=18 a1q^3.(1+q^3) = 18 (2) (2)/(1) q= 1/2 from (1) a1q^2.(1+q^3) =36 (1/4)a1( 1+ 1/8) = 36 (9/32)a1 = 36 a1=128 an = 128. (1/2)^(n-1) an =1/2 128. (1/2)^(n-1) = 1/2 (1/2)^(n-8...

在等比数列{an}中,∵a4+a6=3,∴a42+2a4a6+a5a7=a42+2a4a6+a62=(a4+a6)2=32=9,故答案为:9.

(Ⅰ)设等比数列{an}的公比为q(q≠0),由a7=a1q6=1,得a1=q-6,从而a4=a1q3=q-3,a5=a1q4=q-2,a6=a1q5=q-1.…(3分)因为a4,a5+1,a6成等差数列,所以a4+a6=2(a5+1),即q-3+q-1=2(q-2+1),q-1(q-2+1)=2(q-2+1).所以q=12.故an=a1qn-...

解:∵a3+a7=34,a4×a6=64即:a1q^2+a1q^6=34 ①a1q^3*a1q^5=64 既 a1q^4=8 ②②式带入①式 即1/(q^2)+q^2=34/8=17/4又∵公比q<0∴解得 q=-1/2,a1=128∴an=a1×q^(n-1)=128×(-1/2)^(n-1)

(1)设公比为q,∵a3=2,a6=16,∴a1q2=2,a1q5=16,两式相比得q=2,所以a 1=12,an=12?2n?1=2n?2.…(4分)(2)∵{an}为等比数列,a3+a6=36,a4+a7=18,∴a1q2+a1q5=36a1q3+a1q6=18,解得a1=128,q=12,∴an=128×(12)n-1=28-n.∵an=12,∴28...

设首项为a1,公比为q a2+a3+a4=a1Xq+a2Xq+a3Xq=q(a1+a2+a3)=-3 此为1式 由a1+a2+a3=6,结合1式得到6Xq=-3 得到q=-1/2 a1+a2+a3=a1+a1Xq+a1XqXq=6 q=-1/2 得到a1=8 所以a1=8 a2=-4 a3=2 a4=-1 a5=1/2 a6=-1/4 a7=1/8 a3+a4+a5+a6+a7=11/8

(1)q^3=a7/a4=8/2=4 q=4^(1/3)=2^(2/3) a1=a4/q^3=2/4=1/2 an=a1*q^(n-1)=1/2*(2^(2/3))^(n-1)=2^(-1+2n/3-2/3)=2^(2n/3-5/3) (2) a2+a5=a2(1+q^3)=18 a3+a6=a3(1+q^3)=9 下式/上式得:q=a3/a2=1/2 a2+a5=a1*1/2+a1*(1/2)^4=18 a1=32 an=a1*q...

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