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設等差數列{an}前三項分別為a-d,a,a+d,則由題意得:a−d+a+a+d=−3(a−d)a(a+d)=8,解得:a=−1d=−3或a=−1d=3.當a=-1,d=-3時,首項a1=a-d=-1-(-3)=2,∴等差數列{an}的通項公式為an=2-3(n-1)=5-3n...
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