Wait, is this even possible?

Here is a formal proof for your solution.

Let there be a sequence $a_1, a_2, a_3, \cdots,a_n, \cdots$ which is in both AP and GP.
Then $a_{n - 1}$ , $a_n$ and $a_{n + 1}$ are consecutive terms of this progression.
Since they are in AP, $2a_n = a_{n - 1} + a_{n + 1} \longrightarrow \boxed{1}$ .

Now, these terms are also in GP. Let $r$ be the common ratio between its terms.
$a_{n - 1} = ar^{n - 2}\\ a_n = ar^{n - 1}\\ a_{n + 1} = ar^n$

Putting the above values in $\boxed{1}$ we get:-
$2ar^{n - 1} = ar^{n - 2} + ar^n\\ ar^{n - 2}\left(2r\right) = ar^{n - 2}\left(1 + r^2\right)\\ 2r = r^2 + 1\\ r^2 - 2r + 1 = 0\\ {\left(r - 1\right)}^2 = 0\\ r - 1 = 0\\ r = 1$

So, $a_1 = a_1\\ a_2 = a_1×r = a_1 × 1 = a_1\\ a_3 = a_1 × r^2 = a_1 × 1^2 = a_1\\ \vdots$

So, this progression is a constant progression where everu term is equal to the first term. It is in AP with common difference $0$ as well as in GP with common ratio $1$ . So, $\color{#69047E}{\boxed{\text{Yes}}}$ there exists a sequence which is in both AP and GP.

True (+1). It is called constant arithmetic progression. I want to put in a solution, but I cant :(.