题目内容
已知公差不为0的等差数列{an}的前n项和为Sn,且满足S5=3a5-2,又a1,a2,a5依次成等比数列,数列{bn}满足b1=-9,
(1)求数列{an},{bn}的通项公式;
(2)记数列an+bn的前n项和为Tn,若当且仅当n=3时,Tn取得最小值,求实数k的取值范围.
【答案】分析:(1)根据所给的等差数列的三项之间的关系,求出数列的首项和公差的关系,求出首项和公差,写出数列的通项,根据所给的数列的递推式,代入前面求出的数列的通项,整理仿写一个通项,连续两项做差,再利用累加得到要求的数列的通项.
(2)根据所求的两个数列的通项.构造新数列,连续两项做差,得到数列是一个递增数列,当n=3时,取得最小值,根据条件做出k的取值范围.
解答:解:(1)设等差数列{an}的公差为d,则S5=5a1+10d
∵S5=3a5-2=3(a1+4d)-2=3a1+12d-2
∴5a1+10d=3a1+12d-2
∴a1=d-1
∵a1,a2,a5依次成等比数列
∴a22=a1a5即(a1+d)2=a1(a1+4d)
化简得:d=2a1
∴a1=1,d=2
∴an=a1+(n-1)d=2n-1
∴
∴
当n≥2时,


∴
∴
当n=1时,b1=9满足上式
∴
(2)∵an=2n-1,
∴
∴数列an+bn是递增数列
∵当n=3时,Tn取得最小值
∴

解得
.
点评:本题考查数列的递推式和数列的求和,本题解题的关键是应用函数的思想来解决数列的问题,本题是一个综合题目.
(2)根据所求的两个数列的通项.构造新数列,连续两项做差,得到数列是一个递增数列,当n=3时,取得最小值,根据条件做出k的取值范围.
解答:解:(1)设等差数列{an}的公差为d,则S5=5a1+10d
∵S5=3a5-2=3(a1+4d)-2=3a1+12d-2
∴5a1+10d=3a1+12d-2
∴a1=d-1
∵a1,a2,a5依次成等比数列
∴a22=a1a5即(a1+d)2=a1(a1+4d)
化简得:d=2a1
∴a1=1,d=2
∴an=a1+(n-1)d=2n-1
∴

∴

当n≥2时,



∴

∴

当n=1时,b1=9满足上式
∴

(2)∵an=2n-1,

∴

∴数列an+bn是递增数列
∵当n=3时,Tn取得最小值
∴


解得

点评:本题考查数列的递推式和数列的求和,本题解题的关键是应用函数的思想来解决数列的问题,本题是一个综合题目.

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