Question

题目：已知等差数列{an}的公差d=1,前n项和为Sn.

（1）若2,a1,a5成等比数列,求{an}的通项公式；（2）若S7＞a1a12,求a1的取值范围.

Answer 1

（1）因为an公差为1，所以a5=a1+（5-1）*1=a1+4。
又因为2,a1,a5成等比数列；
所以2*a5=a1*a1
2（a1+4）=a1
a1*a1-2a1-8=0
(a1+2)(a1-4)=0
a1=4或a1=-2
an=a1+(n-1)d
an=n+3或an=n-3
（2）an=a1+(n-1)d
a7=a1+6;a12=a1+11
a1+a7）7/2>2a1(a1+11)
(2a1+6)7>2a1(a1+11)
7a1+21>a1^2+11
a1^2-7a1-10<0
(a1-2)(a1-5)<0
2<a1<5

Answer 2

解：a1^2=2a5
a1^2-2(a1+4d)=0
a1^2-2a1-8=0
(a1-4)(a1+2)=0
a1=4或a1=-2
an=4+n-1=n+3或-2+n-1=n-3

Answer 3

(1)an=n+3