已知等差数列{an}满足A3+A6=－1/3，A1A3=－4/3且A1＞A8.求数列｛An｝的通项公式.

Question

已知等差数列{an}满足A3+A6=－1/3，A1A3=－4/3且A1＞A8.求数列｛An｝的通项公式.

Answer 1

因为a1a8,所以d0;
a1+a3=-1/3,a1=-1/3-a3,又 a1a3,所以a3-2/3;
a1*a3=-4/3,(-1/3-a3)*a3=-4/3,a3=1(舍去)a3=-4/3;a1=1 d=a3-a1/2=-7/6
an=a1+(n-1)d=1-7/6(n-1);