Question

设｛an｝是等比数列,且各项均为正,又a1+a2+aa53+a4+a5=211/27,1/a1+1/a2+1/a3+1/

Answer 1

a1+a2+a3+a4+a5

=a3/q²+a3/q+a3+a3q+a3q²

=a3(1/q²+1/q+1+q+q²)

=211/27 ①

1/a1+1/a2+1/a3+1/a4+1/a5

=q²/a3+q/a3+1/a3+q/a3+q²/a3

=1/a3*(1/q²+1/q+1+q+q²)

=211/48 ②

① ÷②：

a3²=48/27=16/9

==>a3=±4/3

又∵an>0

∴a3=4/3

楼上的错了……