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=? 答案：2.

Answer 1

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

Answer 2

1/a1+1/a2+1/a3+1/a4+1/a5
=(a2a3a4a5+a1a3a4a5+a1a2a4a5+a1a2a3a5+a1a2a3a4)/(a1a2a3a4a5)
=(a1)^4q^6(q^4+q^3+q^2+q+1)/(a1)^4q^10
=(q^4+q^3+q^2+q+1)/a1q^4
=a1(q^4+q^3+q^2+q+1)/(a1q^4*a1)
=a1(q^4+q^3+q^2+q+1)/(a3)^2
=(a1+a2+a3+a4+a5)/(a3)^2
=8/(a3)^2

8/(a3)^2=2
(a3)^2=4
a3=±2

Answer 3

a1+a2+a3+a4+a5=a1+a1*q+a1*q^2+a1*q^3+a1*q^4=a1(1+q+q^2+q^3+q^4)
=a1*(1-q^5)/(1-q)即(q^5-1)/(q-1)=8/a1
1/a1+1/a2+1/a3+1/a4+1/a5=1/a1+1/(a1*q)+1/(a1*q^2)+1/(a1*q^3)+1/(a1*q^4)
=1/a1*(1+1/q+1/q^2+1/q^3+1/q^4)
=1/a1*[1-(1/q)^5]/(1-1/q)
=1/a1*(q^5-1)/[q^4*(q-1)]=1/(a1*q^4)*8/a1=8/(a1^2*q^4)
1/a1+1/a2+1/a3+1/a4+1/a5=2 所以8/(a1^2*q^4)=2得a1=2/q^2
a3=a1*q^2=2/q^2*q^2=2