Question

等差数列{an}的前n项和为Sn,已知S9＜0,S10＞0,则此等差数列的前n项和中,n是多少时取得最小值.

如题,详细过程.

Answer 1

Sn=n*(a1+an)/2
S9=9*(a1+a1+8d)/2<0,a1+4d<0
S10=10*(a1+a1+9d)/2>0,a1+4.5d>0
d>0,a1<0,-4d>a1>-4.5d
a1=-4d,an=-4d+(n-1)d=0,n=5
a1=-4.5d,an=-4.5d+(n-1)d=0,n=5.5
可知5<n<5.5时，即n=5时,Sn取得最小值.