{an}是等差数列,S8=48,S12=108求Sn

问题描述:

{an}是等差数列,S8=48,S12=108求Sn

s8=8a1+28d=48,2a1+7d=12(1)
s12=12a1+66d=108,2a1+11d=18(2)
(1)-(2)解得:4d=6,d=1.5
把d=1.5代入(1)得a1=0.75
没分就给个好评吧!谢了!