an是公差为d的等差数列,sn=a1+a2+...+an,当且仅当n=17时,sn取得最大值,且a15=9,(1),求d的范围
(2)若d属于Z,丨a1丨+丨a2丨+...+丨an丨=697,求n的值
人气:300 ℃ 时间:2020-02-05 23:10:20
解答
1.
an=a1+(n-1)d
a15=a1+14d=9
a1=9-14d
sn=na1+n(n-1)d/2
=n(9-14d)+n(n-1)d/2
=(1/2)dn^2+(9-29d/2)n
=(1/2)d[n+(9-29d/2)/d]^2-(1/2)d[(9-29d/2)/d]^2
=(1/2)d[n+(9-29d/2)/d]^2-(1/2)(9-29d/2)^2/d
当d<0,n+(9-29d/2)/d≈0时,才有最大值,
n=-(9-29d/2)/d
16<-(9-29d/2)/d<18
-16d<9-29d/2<-18d
-32d-18<-29d<-36d-18
36d+18<29d<32d+18
-6<d<-18/7;
2.
-6<d<-18/7
-5≤d≤-3
(1)d1=-5,a1=79,an=79-5(n-1)=84-5n
……
(2)d2=-4,a1=65,an=65-4(n-1)=69-4n
……
(3)d3=-3,a1=51,an=51-3(n-1)=54-3n
……
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