递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?

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递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?递减等差数列{an}的前n项和sn满足:

递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?
递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?

递减等差数列{an}的前n项和sn满足:s5=s10,则要sn最大,n等于多少?
设an=a1+bn
因为s10=s5则a6+a7+a8+a9+a10=0
由此可得a1=-7b则a8=0所以s7=s8
则要sn最大,n等于7或8

n=7或8

{an}为递减等差数列,(an+1-an)-(an-an-1)=k,k为常数,且k<0
an+1+an-1=2an+k
s5=s10,s10-s5=a10+a9+a8+a7+a6=0
由an+1+an-1=2an+k:a10+a8=2a9+k,所以2a9+k+a9+a7+a6=3(a9+a7)-2a7+a6+k=0
又a9+a7=2a8+k:3(2a8+k)-2a7...

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{an}为递减等差数列,(an+1-an)-(an-an-1)=k,k为常数,且k<0
an+1+an-1=2an+k
s5=s10,s10-s5=a10+a9+a8+a7+a6=0
由an+1+an-1=2an+k:a10+a8=2a9+k,所以2a9+k+a9+a7+a6=3(a9+a7)-2a7+a6+k=0
又a9+a7=2a8+k:3(2a8+k)-2a7+a6+k=6a8+3k-2a7+a6+k=6a8-2a7+a6+4k=0,
又a8+a6=2a7+k得a6-2a7=k-a8,所以6a8+k-a8+4k=0,a8=-k>0,a7>a8=-k
a9+a7=2a8+k=-k>0,a9=-k-a7<0,递减等差数列{an}从a9开始都是负数,要sn最大,n等于8

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