设Sn是等比数列{an}的前n项和,S3,S9,S6成等差数列.(Ⅰ)求数列{an}的公比q;(Ⅱ)求证:a3,a9,a6成等差数列;(Ⅲ)当am,as,at(m,s,t∈[1,10],m,s,t互不相等)成等差数列时,求m+s+t的值.

问题描述:

设Sn是等比数列{an}的前n项和,S3,S9,S6成等差数列.
(Ⅰ)求数列{an}的公比q;
(Ⅱ)求证:a3,a9,a6成等差数列;
(Ⅲ)当am,as,at(m,s,t∈[1,10],m,s,t互不相等)成等差数列时,求m+s+t的值.

(Ⅰ)当q=1时,S3=3a1,S9=9a1,S6=6a1
∵2S9≠S3+S6,∴S3,S9,S6不成等差数列,与已知矛盾,
∴q≠1.(2分)
由2S9=S3+S6得:2•

a1(1−q9)
1−q
a1(1−q3)
1−q
+
a1(1−q6)
1−q
,(4分)
即2(1-q9)=(1-q3)+(1-q6)⇒2q6-q3-1=0,
q3=−
1
2
⇒q=−
3
1
2
,q3=1⇒q=1(舍去),∴q=−
3 4
2
(6分)
(Ⅱ)∵2a9-a3-a6=2a1q8-a1q2-a1q5=a1q2(2q6-1-q3)=0,
∴2a9=a3+a6,∴a3,a9,a6成等差数列.(9分)
(Ⅲ)S3,S9,S6成等差数列⇔2q6-q3-1=0⇔2q6=q3+1⇔2a1q6=a1q3+a1⇔2a7=a4+a1
∴a1,a7,a4成等差数列或a4,a7,a1成等差数列,则m+s+t=12,(11分)
同理:a2,a8,a5成等差数列或a5,a8,a2成等差数列,则m+s+t=15,
a3,a9,a6成等差数列或a6,a9,a3成等差数列,则m+s+t=18,
a4,a10,a7成等差数列或a7,a10,a4成等差数列,则m+s+t=21,
∴m+s+t的值为12,15,18,21.(15分)
答案解析:(Ⅰ)由题意设出等比数列的前n项和公式,由S3,S9,S6成等差数列建立方程求q即可;
(Ⅱ)由(Ⅰ)给出a3,a9,a6的表达式,验证是否构成等差数列即可;
(Ⅲ)am,as,at(m,s,t∈[1,10],m,s,t互不相等)成等差数列时,由等差数列的性质构建方程,讨论既得.
考试点:等比关系的确定;等差数列的性质;等比数列的性质.

知识点:本题考查等差与等比数列的综合运用,用到了分类讨论的思想,综合性较强.本题解题时容易因为讨论不全出错.