S4+S8=2S12推出1+q^4=2q^8.解出q^4=1.即q^2=1
a13,a11,a7成等差数列即2*a11=a13+a7.即2*q^4=q^6+1.当然成立
已知Sn是等比数列{an}的前n项和,若S4+S8=2S12,求证：a13,a11,a7成等差数列.
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