Question

已知等差数列an的前n项和为sn，且s13＞s6＞s14，a2＝24①求公差d的取值范围②问数列｛sn｝是否存在最大项

Answer 1

A(n)=A1+(n-1)d.
S(n)=na1+n(n-1)d/2.
A2=A1+d=24
因为S13>S6>S14 即13A1+78d>6A1+15d>14A1+91d
将A1=24-d代入上述不等式，
13(24-d) +78d>6(24-d)+15d>14(24-d)+91d
解得-3<d<-48/17

因为d为负数, 即{An}递减.
令An =A1+（n-1）d≥0，
将A1=24-d代入得：24-d+（n-1）d≥0，
（n-2）d≥-24，n-2≤-24/d, n≤-24/d+2,
因为-3<d<-48/17，所以-17/48<1/d<-1/3,
10<-24/d+2<10.5
而n≤-24/d+2,所以n≤10时，An≥0，
即数列{An}前10项为正，以后各项为负，
所以Sn最大时，n=10.

Answer 2

1：an=a1+(n-1)d =>a2=a1+d=24 =>a1=24-d;
s13>s6 =>13a1+13*12/2d>6a1+6*5/2d =>d>-3
s6>s14 =>6a1+6*5/2d>14a1+14*13/2d =>d<-48/17
d∈（-3，-48/17）；
2: sn=na1+n*(n-1)/2*d=n(24-d)+d/2(n^2-n)=d/2*n^2+(24-d-d/2)n=
d/2((n^2+(48/d-3)n)=d/2(n+24/d-3/2)^2-(24/d-3/2)^2
抛物线的中点为n=3/2-24/d∈（-∞，9.5）∨(10，+∞) 开口向下
所以当n=9时 sn有最大值 sn=d/2*81+(24-3/2d)*9=27d+216;