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

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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