1.已知递增的等比数列{an}中,a1a9=2304,a4+a6=120,求an的通项公式.2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q
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![1.已知递增的等比数列{an}中,a1a9=2304,a4+a6=120,求an的通项公式.2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q](/uploads/image/z/12488302-46-2.jpg?t=1.%E5%B7%B2%E7%9F%A5%E9%80%92%E5%A2%9E%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1a9%3D2304%2Ca4%2Ba6%3D120%2C%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.2.%E8%AE%BE%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%8B%A5S3%2BS6%3D2S9%2C%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E5%85%AC%E6%AF%94q)
1.已知递增的等比数列{an}中,a1a9=2304,a4+a6=120,求an的通项公式.2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q
1.已知递增的等比数列{an}中,a1a9=2304,a4+a6=120,求an的通项公式.
2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q
1.已知递增的等比数列{an}中,a1a9=2304,a4+a6=120,求an的通项公式.2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q
1.a1a9=2304,
即a1×(a1×q^8)=a1^2×q^8=(a1×q^4)^2=a5^2=2304,故a5=48
a4= a5/q,a6=a5×q,a4+a6=120,即48/q+48q=120,转化后得到方程
2q^2-5q+2=0,(2q-1)(q-2)=0,q=1/2或q=2,由于{an}是递增的等比数列,故q=2
a5=a1×q^4=a1×2^4=48,所以,a1=3,an=3×2^(n-1)
2.S3+S6=2S9
s3+s6=[a1(1-q^3)+a1(1-q^6)]/(1-q)= a1(2-q^3-q^6) /(1-q)
2s9=2a1(1-q^9)/(1-q)
故,a1(2-q^3-q^6)=2a1(1-q^9)
化简后,q^3(2q^6-q^3-1)=0,q≠0,则(2q^6-q^3-1)=0
设q^3=K,则
2K^2-K-1=0,(2K+1)(K-1)=0
K= -1/2或K=1
即q^3= -1/2 或q^3=1
则q= -3√ 1/2(这是开立方,我这不会表示,不是“3×√ ”)或q=1 (等比数列中,q≠1)
所以,q= -3√ 1/2