等比数列{an}中,前n项和7n满足S1,s3,S2成等差数列.求…①该数列的公比q.②群a1-a3=3+求Sn
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![等比数列{an}中,前n项和7n满足S1,s3,S2成等差数列.求…①该数列的公比q.②群a1-a3=3+求Sn](/uploads/image/z/3649035-3-5.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%AD%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C7n%E6%BB%A1%E8%B6%B3S1%2Cs3%2CS2%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.%E6%B1%82%E2%80%A6%E2%91%A0%E8%AF%A5%E6%95%B0%E5%88%97%E7%9A%84%E5%85%AC%E6%AF%94q.%E2%91%A1%E7%BE%A4a1-a3%EF%BC%9D3%2B%E6%B1%82Sn)
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等比数列{an}中,前n项和7n满足S1,s3,S2成等差数列.求…①该数列的公比q.②群a1-a3=3+求Sn
等比数列{an}中,前n项和7n满足S1,s3,S2成等差数列.求…①该数列的公比q.②群a1-a3=3+求Sn
等比数列{an}中,前n项和7n满足S1,s3,S2成等差数列.求…①该数列的公比q.②群a1-a3=3+求Sn
S1,s3,S2成等差数列,
∴2a1(1+q+q^2)=a1(1+1+q),
∴2q^2+q=0,q=-1/2.
②a1-a3=a1(1-1/4)=3,a1=4.
∴Sn=4[1-(-1/2)^n]/(1+1/2)=(8/3)[1-(-1/2)^n].
s1=a1,s2=a1+a1*q,s3=a1+a1*q+a1*q*q
由S1,s3,S2成等差数列得
a1+a1+a1*q=2(a1+a1*q+a1*q*q)得q=-1/2
a1-a3=3得a1=4
所以Sn=a1(1-q^n)/(1-q)=4*[1-(-1/2)^n]/(1+1/2)=8/3*[1-(-1/2)^n]
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