问题 解答题

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,(m,s,t∈[1,10],m,s,t互不相等)成等差数列时,求m+s+t的值.

答案

(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=-
34
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GP或a4,a7,a1成GP,则m+s+t=12,(11分)

同理:a2,a8,a5成GP或a5,a8,a2成GP,

则m+s+t=15,a3,a9,a6成GP或a6,a9,a3成GP,

则m+s+t=18,a4,a10,a7成GP或a7,a10,a4成GP,

则m+s+t=21,∴m+s+t的值为12,15,18,21. (15分)

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