高中数学等差数列题目

a1+a1+2d+a1+4d=3a1+6d=105,a1+2d=35
a1+d+a1+3d+a1+5d=3a1+9d=99,a1+3d=33
d=-2,a1=39
am<=0,a1+(m-1)d<=0
39-2m+2<=0
m>=41/2=21.5
m=21时，a21=39+20*-2=-1,a20=1
所以，S20=(39+1)*20/2=400，n=20
2、
a1+3d=1
(a1+a1+4d)*5/2=10,a1+2d=2
d=-1,a1=4
a5=4-4=0
S5=S4=10,取得最大值的n=4,n=5
3、
S17=(25+25+16d)*17/2=(25+25+8d)*9/2
50*17+16*17d=50*9+72d
200d=-400,d=-2
am<=0
25+(m-1)*(-2)<=0
25-2m+2<=0,m>=27/2=13.5
a13=25-24=1
S13=(25+1)*13/2=169

1、因为{an}为等差数列
a1+a3+a5=105 则有 3*a3 = 105，即a3 = 35
a2+a4+a6=99 同理得a4 = 33
所以公差d = -2，得a1 = 39 an = 41 - 2n;
要使sn达到最大就要是an>0,an+1<0,由于a20 = 1，a21 = -1，所以s20最大，也即 n = 20.

2、由于s5 = 5a3，所以： a3 = 2；但 a4 = 1 ==> d = -1 ==>an = 5 -n
==> a5 = 0 (s4 = s5) ==> n = 4 或 n = 5为题所求

3、由 S17=S9 得a10 + a11 + ... + a17 = 0 ==> a13 + a14 = 0 ==>
a1 + 12d + a1 + 13d = 0 ==> 2a1 + 25d = 0 ==> d = -2 ==> an = 27-2n ==> a13 = 1, a14 = -1 ==> s13最大

1.n=20;
2.n=4或5；
3.d=-2,sn的最大值为S13=13*25+13*12/2*(-2)=169