设等差数列{an}的前n项和为Sn ,且S15>0,a8+a90,a8+a9

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设等差数列{an}的前n项和为Sn ,且S15>0,a8+a90,a8+a9
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设等差数列{an}的前n项和为Sn ,且S15>0,a8+a90,a8+a9
设等差数列{an}的前n项和为Sn ,且S15>0,a8+a9<0
设等差数列{an}的前n项和为Sn,且S15>0,a8+a9<0,则使得an十Sn/n<0的最小的n为?答案11

设等差数列{an}的前n项和为Sn ,且S15>0,a8+a90,a8+a9
设第一项为：a1 ,公差为：d
1、S15>0可得到a1>-7d
2、a8+a9

设公差为d，则S15=15(a1+7d)>0=>a8=a1+7d>0。而a8+a9<0
=>2a1+15d<0 => a1+7.5d<0
∴0.5d<-(a1+7d)<0，即d<0
要使an+Sn/n=an+(a1+an)/2=2a1+3(n-1)d/2<0
即要3(n-1)d/2<-2a1 <=> n-1>-4a1/3d
而-7d ...

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设公差为d，则S15=15(a1+7d)>0=>a8=a1+7d>0。而a8+a9<0
=>2a1+15d<0 => a1+7.5d<0
∴0.5d<-(a1+7d)<0，即d<0
要使an+Sn/n=an+(a1+an)/2=2a1+3(n-1)d/2<0
即要3(n-1)d/2<-2a1 <=> n-1>-4a1/3d
而-7d 30d<-4a1<28d
=>28/3<-4a1/3d<10
∴n>-4a1/3d+1>28/3+1=10+1/3
即n可取的最小值为11

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