设Sn是等比数列{an}的前n项和,S3,S9,S6成等差数列.
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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)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=−
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)
∵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)
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