Question

An为等差数列,A1=-9,S3=S7,那么使其前n项和Sn最小的n是

Answer 1

1、 a4+a5+a6+a7=a1+18d=0,a1=-9所以d=1/2Sn=n*a1+n*(n-1)*d/2=n^2/4-(37/4)*nn=37/2时,Sn取最小值2、 Sn/n=a1+(n-1)d所以 a1+2008d - (a1+2006d)= 2d =2 ,d=1.S2011=-20113、 a4+a6+a8+a10+a12 = 5* a8 =120 ,a8= 2...