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[单选题] 在各项都为正数的等比数列{an}中,首项a1=3,前3项和为21,则a3+a4+a5=( )
A
33
B
72
C
84
D
189
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试题答案:
C
答案解释:
因为S3=21,即a1+a2+a3=21,所以a1(1+q+q2)=21,又a1=3,所以1+q+q2=7,q2+q-6=0,q=2或q=-3(舍去),所以a3+a4+a5=q2(a1+a2+a3)=4×21=84.故选C
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