等比数列{an},a1+a2+a3+a4+a5=8,1/(a1)+1/(a2)+1/(a3)+1/(a4)+1/(a5)=2,求a_数学解答

等比数列{an},a1+a2+a3+a4+a5=8,1/(a1)+1/(a2)+1/(a3)+1/(a4)+1/(a5)=2,求a3.(a后的数字都为下标,是序号)

人气：485 ℃　时间：2020-03-24 11:04:04

解答

a1+a2+a3+a4+a5=8.A
a1+a2+a4+a5=8-a3.B
a3^2=a1a5=a2a4.C
1/(a1)+1/(a2)+1/(a3)+1/(a4)+1/(a5)=2
a3^3(a1+a2+a4+a5)/a1a2a3a4a5=2.D
将B,C代入D得
a3^3(8-a3)/a3^5=2
8-a3=2a3^2
2a3^2+a3-8=0
a3=[-1±√65]/4