Problem: If a1β,a2β,a3β,β¦ is a sequence of positive numbers such that an+2β= anβan+1β for all positive integers n, then the sequence a1β,a2β,a3β,β¦ is a geometric progression
Answer Choices:
A. for all positive values of a1β and a2β
B. if and only if a1β=a2β
C. if and only if a1β=1
D. if and only if a2β=1
E. if and only if a1β=a2β=1
Solution:
The second through the fifth terms of {anβ} are a2β,a1βa2β,a1βa2β2,a1β2a2β3. If these terms are in geometric progression, then the ratios of successive terms must be equal: a1β=a2β=a1βa2β. Since a1β and a2β are positive, it is necessary that a1β=a2β=1. Conversely, if a1β=a2β=1, then {anβ} is the geometric progression 1, 1, 1, β¦.