问题
选择题
各项都是正数的等比数列{an}中,3a1,
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答案
∵3a1,
a3,2a2成等差数列1 2
∴3a1+2a2=a3,
∴3a1+2a1q=a1q2
∴q2-2q-3=0
∵q>0
∴q=3
∴
=a10+a12 a8+a10
=q2=9a8(q2+q4) a8(1+q2)
故选D
各项都是正数的等比数列{an}中,3a1,
|
∵3a1,
a3,2a2成等差数列1 2
∴3a1+2a2=a3,
∴3a1+2a1q=a1q2
∴q2-2q-3=0
∵q>0
∴q=3
∴
=a10+a12 a8+a10
=q2=9a8(q2+q4) a8(1+q2)
故选D