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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/23 15:02:17
![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](/uploads/image/z/1842208-16-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94s13%EF%BC%9Es6%EF%BC%9Es14%2Ca2%EF%BC%9D24%E2%91%A0%E6%B1%82%E5%85%AC%E5%B7%AEd%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E2%91%A1%E9%97%AE%E6%95%B0%E5%88%97%EF%BD%9Bsn%EF%BD%9D%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E6%9C%80%E5%A4%A7%E9%A1%B9)
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
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
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
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
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-...
全部展开
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;
收起