题目内容
(2010•江西模拟)已知正项数列{an}满足:①对任意n∈N*,都有an•an+2=an+12; ②lga1+lga2+…+lga9=27,则lga11+lga19-lga152的值为( )
分析:由an•an+2=an+12,知正项数列{an}是等比数列,所以lga11+lga19-lga152=lg(
)=lg1,由此能求出其结果.
a11•a19 |
a15 2 |
解答:解:∵an•an+2=an+12,
∴正项数列{an}是等比数列,
∴lga11+lga19-lga152
=lg(
)
=lg1
=0.
故选C.
∴正项数列{an}是等比数列,
∴lga11+lga19-lga152
=lg(
a11•a19 |
a15 2 |
=lg1
=0.
故选C.
点评:本题考查等比数列的性质,是基础题.解题时要认真审题,注意对数运算法则的灵活运用.
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