Sequence defined by $s_{n+1}=\sqrt{s_{n}s_{n+2}}$ 
Let $(s_n)$ be the sequence defined by:
  $$  s_0,s_1\in \mathbb{R}^{+},\quad \forall n\in \mathbb{N};\quad s_{n+1}=\sqrt{s_{n}s_{n+2}}  $$
  
  
*
  
*$(s_n)$ is arithmetic sequence
  
*$(s_n)$ is geometric sequence
  
*$(s_n)$ is neither arithmetic nor geometric
  
*we can not conclude anything
  

note that from french source $\mathbb{R}^{+}=\{x\in R \mid x \geq 0\}$ which means $0 \in \mathbb{R}^{+}$ so here we talk about $0\in\mathbb{R}^{+}$
My thoughts:
first if $s_0=0, s_1=1$ then $1=0$ is undefined so let take   $s_0,s_1\in \mathbb{R}^{+}_{*}=\{x \mid x  \text{is a strictly positive real number} \}$
and let 's check every suggestions:


*

*Yes, indeed:
If $s_0=0$, else, $s_{n+1}=b$, $s_n=b-r$ and $ s_{n+2}=b+r$ implies 
$b=\sqrt{b^2-r^2}$, thus the conclusion : $r=0$.

*Yes, indeed:
for all $n, \frac{s_{n+1}}{s_n}=q>0$ and $q\ne 1$ (for example by induction on $n$).

*nope from solution of question 1 and 2 

*nope of course!


Am i right is there other nice proofs.  
Add some thoughts :
if $s_{0}=s_{1}$ then $(s_n)$ is a constant sequence and by that we know that it's an arithmetic sequence with ratio equal $0$ and it's a geometric sequence with an common ratio equal $1$
then answer $1$ and $2$ are both correct
-I would to prove also that others suggestions are worng so Correct that if i am wrong please :
here is my tired


*

*If $s_{0}=0$ then $s_{n}=s_{n+1}$ so the sequence $v_{n}$ is constant and thus :
$A$) True $B$) True $C$) False $D$) false

*If $s_{0}\neq 0$ then 
$$
\begin{align*}\forall n\in \mathbb{N};\quad s_{n+1}=\sqrt{s_{n}s_{n+2}}
 &\Longleftrightarrow s^{2}_{n+1}={s_{n}s_{n+2}} \\
& \Longleftrightarrow \frac{s_{n+1}}{s_n} = \frac{s_{n+2}}{s_{n+1}}\\
&\Longleftrightarrow S_{n} \text{ is geometric }
\end{align*}
$$
and thus : $A$) False  $B$) true  $C$) False $D$) False
 A: Note that
$$s_{n+1}^2 = s_ns_{n+2}$$
$$\frac{s_{n+1}}{s_n} = \frac{s_{n+2}}{s_{n+1}}$$
The ratio between any term and the preceding one remains the same hence it must be a geometric progression.

...unless the common ratio is equals to one, in which case all terms in the sequence are equal.
A: If you rewrite the recursion as $s_{n+1}=s_n^2/s_{n-1}$ and take logarithms, you get a recursion
$$t_{n+1}=\log s_{n+1}=2\log s_n-\log s_{n-1}=2t_n-t_{n-1}$$
The recursion in $t_n$ has the generic solution
$$t_n=An+B$$
hence
$$s_n=e^{An+B}=Cr^n\quad\text{with } C=e^B\text{ and }r=e^A$$
which is a geometric sequence.
A: Well, if   $s_0,s_1\in \mathbb{R}^{+},\quad \forall n\in \mathbb{N};\quad s_{n+1}=\sqrt{s_{n}s_{n+1}}$, you can simplify by $\sqrt{s_{n+1}}$, then square both sides, to find $s_{n+1}=s_n$, which seems rather strange. I think there's a typo, and that we should read $s_0,s_1\in \mathbb{R}^{+},\quad \forall n\in \mathbb{N^{*}};\quad s_{n+1}=\sqrt{s_{n}s_{n-1}}$
If we use that definition, then the sequence is neither arithmetic nor geometric in general, but if $s_0=s_1$, it trivially is both.
