已知等差数列an的前n项和为Sn,S4/S8=1/2,那么S8/S12=?

问题描述:

已知等差数列an的前n项和为Sn,S4/S8=1/2,那么S8/S12=?

答案是=2/3麻烦过程了,谢谢sn=na1+nd(n-1)/2,s4=4a1+6d ,s8=8a1+28d,s4/s8=1/2,s8=2s4即(4a1+6d)*2=8a1+28d,d=0,所以sn=na1,
S8/S12=8a1/12a1=2/3谢谢!