设等差数列{an}的公差为d
∵a3=1,a4+a10=18,
∴a1+2d=1,a1+3d+a1+9d=18
解得a1=-3,d=2
故答案为:-3
已知数列{an}是等差数列,a3=1,a4+a10=18,则首项a1=_.
已知数列{an}是等差数列,a3=1,a4+a10=18,则首项a1=______.
数学人气:620 ℃时间:2020-04-03 17:19:38
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