设Sn是等比数列的前n项和,S3,S9,S6成等差数列 求证a2,a8,a5成等差数列
问题描述：
设Sn是等比数列的前n项和,S3,S9,S6成等差数列 求证a2,a8,a5成等差数列
答
首项a2S9=2a(q^9-1)/(q-1)S3+S6=a(a^3-1)/(q-1)+a(a^6-1)/(q-1)2S9=S3+S6显然a不等于02(q^9-1)=a^3-1+q^6-12q^9=q^3+q^62q^7=q+q^42aq^7=aq+aq^42a8=a2+a5所以a2,a8,a5成等差数列