a1+a3=,当且仅当a2,q同为正时,a1+a3≥2a2成立;,所以;若a1=a3,则a1=a1q2,从而可知a1=a2或a1=-a2;若a3>a1,则a1q2>a1,而a4-a2=a1q(q2-1),其正负由q的符号确定,故可得结论.
【解析】
设等比数列的公比为q,则a1+a3=,当且仅当a2,q同为正时,a1+a3≥2a2成立,故A不正确;
,∴,故B正确;
若a1=a3,则a1=a1q2,∴q2=1,∴q=±1,∴a1=a2或a1=-a2,故C不正确;
若a3>a1,则a1q2>a1,∴a4-a2=a1q(q2-1),其正负由q的符号确定,故D不正确
故选B.