1.首项设为a,公差设为d,3*a+3*d=12;a(a+d)(a+2d)=48;
a+d=4;a(a+2d)=a(4+d)=16-d^2=12.所以a=2,d=2或a=6;d=-2;(舍)
得a=2
2.a1+a4+a7=45=3a1+9d;
a2+a5+a8=39=3a1+12d;求出d=-2
a3+a6+a9=3a1+15d=a1+12d+3d=39-6=33
3.a3+a8+a13=12=3a1+21d,a1+7d=4
a3a8a13=28=(a1+2d)(a1+7d)(a1+12d),7=(a1+2d)(a1+12d)=(a1+7d-5d)(a1+7d+5d),
求出a1=-1/4,d=3/5,an=a1+(n-1)d
或a1=41/5,d=-3/5.an=a1+(n-1)d