Question

在等比数列｛an}中，a1+a2+a3+a4+a5=8,且1/a1+1/a2+1/a3+1/a4+1/a5=2,则a3=?

谢谢！

Answer 1

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

祝您学习愉快

Answer 2

a1+a2+a3+a4+a5=8
a1(1-q^5)/(1-q)=8 1式
1/A1+1/A2+1/A3+1/A4+1/a5=2
1/a1(1-1/q^5)/(1-1/q)=2
1/a1(1-q^5)/(q^5-q^4)=4 2式
1式除以2式
a1^2(q^5-q^4/(q-1))=4
a1^2*q^4=4
a1*q^2=a3=2