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/a4+1/a5=211/48，求a3

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……

Answer 2

a1=a3/q2 a2=a3/q a4=a3q a5=a3q2
a1+a2+a3+a4+a5=211/27===>a3(1+q+q2+q3+q4)/q2==211/27-----1
1/a1+1/a2+1/a3+1/a4+1/a5=211/48===>1/a3 (1+q+q2+q3+q4)/q2==211/48----2
1/2===>(a3)2=48/27====>a3=4/3