Question

已知{an}是等比数列1）a5^2=a3·a7是否成立?a5^2=a1·a9是否成立?2)an^2=a(n-2)·a(n

Answer 1

1 a5=a1q^4

a3=a1q^2,a7=a1q^6

所以a5^2=a1^2*q^8=a3*a7

a9=a1q^8

所以a5^2=a1*a9

2 an=a1q^(n-1),a(n-2)=a1q^(n-3),a(n+2)=a1q^(n+1)

所以an^2=a1^2*q^(2n-2)=a(n-2)·a(n+2)

a(n-k)=a1q^(n-k-1),a(n+k)=a1q^(n+k-1)

a(n-k)a(n+k)=a1^2*q(2n-2)=an^2