trysoso
幼苗
共回答了18个问题采纳率:94.4% 举报
am,a(m+2),a(m+1)成等差数列
即
2a(m+2)=am+a(m+1)
两边同除以a1
2q^(m+1) =q^(m-1) +q^m .(1)
若Sm,S(m+2),S(m+1)成等差数列
2S(m+2)=Sm+S(m+1)
两边同除以a1/(1-q)
2[1-q^(m+2)] =1-q^m +1-q^(m+1)
2q^(m+2)] =q^m +q^(m+1)
两边同除以q
2q^(m+1) =q^(m-1) +q^m 即(1)
===>Sm,S(m+2),S(m+1)成等差数列
1年前
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