题目内容
设数列{an}是等差数列,且a42+2a4a7+a6a8=4,则a5a6= .
考点:等差数列的通项公式
专题:等差数列与等比数列
分析:设等差数列{an}的公差为d,由通项公式和已知化简可得a12+9a1d+20d2=1,再由通项公式可得a5a6=a12+9a1d+20d2=1
解答:
解:设等差数列{an}的公差为d,
∵a42+2a4a7+a6a8=4,
∴(a1+3d)2+2(a1+3d)(a1+6d)+(a1+5d)(a1+7d)=4,
∴4a12+36a1d+80d2=4,∴a12+9a1d+20d2=1,
∴a5a6=(a1+4d)(a1+5d)=a12+9a1d+20d2=1,
故答案为:1.
∵a42+2a4a7+a6a8=4,
∴(a1+3d)2+2(a1+3d)(a1+6d)+(a1+5d)(a1+7d)=4,
∴4a12+36a1d+80d2=4,∴a12+9a1d+20d2=1,
∴a5a6=(a1+4d)(a1+5d)=a12+9a1d+20d2=1,
故答案为:1.
点评:本题考查等差数列的通项公式,属基础题.
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