由题意可得：{an}为正项等差数列,a2=a1+d,a3=a1+2d又S3=3a1+3d=12,即a1+d=a2=4a(2)^2=2a(1)*[a(3)+1]=2a(1)*a(3)+2a(1),所以a(2)^2=2[a(2)-d]*[a(2)+d]+2[a(2)-d]所以2（4-d)*(4+d)+2(4-d)=16,解得d=3或-4（舍去）所...