题目内容
4、设a1,a2,a3,a4,a5构成等比数列,若a2a5<0,则下列各式正确的是( )
分析:根据等比数列的通项公式结合已知条件可得q<0,逐一分析各个选项,找出正确答案即可.
解答:解:设公比为q,由等比数列的通项公式可得a2a5=a1q•a1q4=a12•q5<0,
∴q<0;
A、a1a3a4a5=a1•a1q2•a1q3•a1q4=a14q9<0,故错误;
B、a1a2a4a5=a1•a1q•a1q3•a1q4=a14q8>0,故错误;
C、a1a2a3a5=a1•a1q•a1q2•a1q4=a14q7<0,故错误;
D、a1a2a3a4=a1•a1q•a1q2•a1q3=a14q6>0,故正确;
故选D.
∴q<0;
A、a1a3a4a5=a1•a1q2•a1q3•a1q4=a14q9<0,故错误;
B、a1a2a4a5=a1•a1q•a1q3•a1q4=a14q8>0,故错误;
C、a1a2a3a5=a1•a1q•a1q2•a1q4=a14q7<0,故错误;
D、a1a2a3a4=a1•a1q•a1q2•a1q3=a14q6>0,故正确;
故选D.
点评:本题考查了等比数列的通项公式和有理数指数幂的运算性质,比较简单.

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