1. ∵ 2a=b+c, ∴ 2sinA=sinB+sinC,又sin²A =sinB+sinC,∴ sin²A =2sinA,sinA(sinA-2)=0, sinA=0, A=90°, ∴ △ABC是Rt△.2. 由-24+9d>0且-24+8d≤0,得公差d∈(8/3,3]3. ∵ 2b²=a²+c²,欲证1/(b+c),1/(c+a),1/(a+b)也成等差数列,只需证明2/(c+a)=1/(b+c)+1/(a+b)=[(a+c)+2b]/[(b+c)(a+b)],即要证明(a+c)²+2b(a+c)=2b²+2(ab+bc+ca),即需证明a²+c²=2b²,此式已知成立. ∴ 1/(b+c),1/(c+a),1/(a+b)也成等差数列4. 由已知,得An=A(n-1)/[A(n-1)+1], ∴ (1/An)-[1/A(n-1)]=1,数列{1/An}是首项=1/(A1)=1/f(2)=3/2,公差=1的等差数列,∴ 1/An=3/2+(n-1)×1=(2n+1)/2,An=2/(2n+1), ∴ A10=2/21