最佳答案:∵{an}是等差数列,∴2a(r+1)=a(r)+a(r+2),即a(r)- 2a(r+1)+a(r+2)=0故当x=-1时,a(r)x^2+2a(r+1)x+
最佳答案:a1+a2=a3a1a2=a4即a1+a1+d=a1+2da1(a1+d)=a1+3d解得a1=d=2an=2n
最佳答案:是x²-a3x+a4=0,还是x²-a3+a4=0?x^2-(a1+2d)x+(a1+3d)=0将根a1,a1+d分别代入a1^2-a1^2-2da1+a1+3