急！在线等！高中等差数列题

设首项为a1,公差d，通项公式an=a1+(n-1)d;题中S4=a1+a2+a3+a4=4a1+1/2x4x(4-1)d=4a1+6d=12,得a=3-3d/2。再由a1.a2.a3.a4=a1.(a1+d)(a1+2d)(a1+3d)=(3-3d/2)(3-d/2)(3+d/2)(3+3d/2)=(9-9/4.d^2)(9-1/4.d^2)=180，设x=(d/2)^2,则上式化为：(1-x)(9-x)=20解得x=11,x=-1(舍掉)从而解得d=2倍正负根号11，a1=3-3倍正负根好11。所以通项公式为an=3+2n倍根号11-5倍根号11或an=3+5倍根号11-2n根号11