(a(n-1))/(an)=(2a(n-1)+1)/(1-2an)
=>a(n-1)*(1-2an)=an*(2a(n-1)+1)
=>a(n-1)-2ana(n-1)=2ana(n-1)+an
=>a(n-1)-an=4ana(n-1)
=>1/an-1/a(n-1)=4
设bn=1/an
=>bn-b(n-1)=4,且b1=1/a1=5
=>bn=5+4(n-1)=4n+1
=>an=1/bn=1/(4n+1)
=>a1a2=1/5*1/9=1/45=1/(4*11+1
=>n=11
=>第十一项