题目内容
16.在等比数列{an}中,有a3a11=4a7,数列{bn}是等差数列,且b7=a7,则b5+b9=8.分析 由a3a11=4a7,解出a7的值,由 b5+b9=2b7 =2a7 求得结果.
解答 解:等比数列{an}中,由a3a11=4a7,可知a72=4a7,∴a7=4,
∵数列{bn}是等差数列,∴b5+b9=2b7 =2a7 =8,
故答案为:8.
点评 本题考查等差数列、等比数列的性质,求出a7的值是解题的关键,是基础题.
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