已知各项不为0的等差数列{an},满足2a3-a12=0,a1=d,数列{bn}是等比数列,且b13=a2,b1=a1则b6b8A.72B.4C.8D.16
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A
解析分析:由2a3-a12=0,a1=d,可得2(a1+2d)-a12=0,由此求得 a1=d=6,an =6n.再由b13=a2,b1=a1,可得 b6b8 =b1?b13=a1?a2 ,运算求得结果.
解答:∵各项不为0的等差数列{an},满足2a3-a12=0,a1=d,∴2(a1+2d)-a12=0,即 2(3a1))-a12=0,∴a1=d=6,an =6n.又∵数列{bn}是等比数列,且b13=a2=12,b1=a1 =6,∴b6b8 =b1?b13=a1?a2=6×12=72,故选A.
点评:本题主要考查等差数列的定义和性质,等差数列的通项公式,以及等比数列的定义和性质的应用,属于中档题.