Is the system is a geometric progression?

Question

I am trying to show that the following system is a geometric progression i.e. $a_2=a_1^2$, $a_3=a_1^3$ etc.

$a_{k}^{2}=a_{k-1}a_{k+1}$

$a_{k-1}^{2}=a_{k-2}a_{k}$

$\cdots$

$a_{2}^{2}=a_{1}a_{3}$

here $a_k>...>a_2>a_1$ is assumed.

Is there sufficient information to conclude that $a_i=a_{1}^{i}$ in general? I am seeing that I would probably need to assume $a_2=a_1^{2}$. Could the geometric progression be concluded otherwise?

Answer 1

$\frac{a_{k+1}}{a_k}=\frac{a_k}{a_{k-1}}=\cdots=\frac{a_3}{a_2}=\frac{a_2}{a_1}$
so $\{a_k\}$ forms a geometric sequence. let the common ratio be $r$.
then the sequence is $\{a_1,a_1 r,...\}$
if $(a_1,r)=(1,2)$ , then $a_i\neq a_1^i$.
if we assume $a_2=a_1^2 \iff r=a_1$
then the sequence is $\{a_1,a_1^2,a_1^3,...\}$

i have excluded the case $a_1=0$ , otherwise the sequence is constantly 0.