(2011•顺义区二模)已知等比数列{an}中,各项都是正数,且a1,12a3,2a2成等差数列,则a6+a7a8+a9等于(  ) A.1+2 B.1−2 C.3+22 D.3−22

问题描述:

(2011•顺义区二模)已知等比数列{an}中,各项都是正数,且a1

1
2
a3,2a2成等差数列,则
a6+a7
a8+a9
等于(  )
A. 1+
2

B. 1−
2

C. 3+2
2

D. 3−2
2

设等比数列{an}的公比为q,
∵各项都是正数,且a1

1
2
a3,2a2成等差数列,
∴a3=a1+2a2,即 a1q2=a1+2a1q,解得 q=1+
2
,或q=1-
2
(舍去).
a6+a7
a8+a9
=
a6+a6q
a6q2+a6q3
=
1+q
q2(1+q)
=
1
q2
=3-2
2

故选D.