等差数列{an}中,如a1+a2+a3=6,a10+a11+a12=9,则a1+a2+…+a12=________.
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30
解析分析:a1+a2+a3=6,a10+a11+a12=9,两式相加结合等差数列的性质可得,3(a1+a2)=15,而a1+a2+…+a12=6(a1+a12)
解答:由a1+a2+a3=6,a10+a11+a12=9,可得a1+a2+a3+a10+a11+a12=15由等差数列的性质可得,3(a1+a2)=15∴a1+a2+…+a12=6(a1+a12)=30故