设数列{an}为等差数列,且a4^2+2a4a7+a6a8=2004,则a5a6=?

设数列{an}为等差数列,且a4^2+2a4a7+a6a8=2004,则a5a6=?

设a4=x,公差为d
则x^2+2*x*(x+3d)+(x+2d)*(x+4d)=2004
化简得4x^2+12xd+8d^2=2004
a5a6=x^2+3xd+2d^2
会了吧,就是501