已知等差数列{an}前三项的和为-3,前三项的积为8.(1)若a2,a3,a1成等比数列,求数列{|an|}的前n项和.(2)若a2,a3,a1不成等比数列,求数列{1anan+1}的前n项和.-高三

题目简介

已知等差数列{an}前三项的和为-3,前三项的积为8.(1)若a2,a3,a1成等比数列,求数列{|an|}的前n项和.(2)若a2,a3,a1不成等比数列,求数列{1anan+1}的前n项和.-高三

题目详情

已知等差数列{an}前三项的和为-3,前三项的积为8.
(1)若a2,a3,a1成等比数列,求数列{|an|}的前n项和.
(2)若a2,a3,a1不成等比数列,求数列{
1
anan+1
}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1)设等差数列{an}的公差为d,
由题意得
3a1+3d=-3
a1(a1+d)(a1+2d)=8
解得
a1=2
d=-3
a1=-4
d=3

∴an=2-3(n-1)=-3n+5或an=-4+3(n-1)=3n-7.
当an=3n-7时,a2,a3,a1分别为-1,2,-4,成等比数列,满足条件.
设数列{|an|}的前n项和为Sn.
∴当n=1,2时,|an|=7-3n,Sn=
n(4+7-3n)
2
=-class="stub"3
2
n2+class="stub"11
2
n;
当n≥3时,|an|=3n-7,
Sn=-a1-a2+a3+a4+…+an
=5+
(n-2)(2+3n-7)
2

=class="stub"3
2
n2-class="stub"11
2
n+10

综上可得:|an|=|7-3n|=
-3n+7,n=1,2
3n-7,n≥3

Sn=
-class="stub"3
2
n2+class="stub"11
2
n,n=1,2
class="stub"3
2
n2-class="stub"11
2
n+10,n≥3

(2)当an=-3n+5时,a2,a3,a1分别为-1,-4,2,不成等比数列.
class="stub"1
anan+1
=class="stub"1
(3n-5)(3n-2)
=class="stub"1
3
(class="stub"1
3n-5
-class="stub"1
3n-2
)

∴Tn=class="stub"1
3
[(-class="stub"1
2
-1)+(1-class="stub"1
4
)+…+(class="stub"1
3n-5
-class="stub"1
3n-2
)]

=class="stub"1
3
[-class="stub"1
2
-class="stub"1
3n-2
]

=class="stub"n
-6n+4

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