Let $(X,d)$ be a metric space and $(x_n)_{n\in \mathbb{N}}$ a sequence in $X$ and $x\in X$. Then $$\lim \limits _{n\to \infty} x_n = x \iff \lim \limits _{n\to \infty} x_{2n}=\lim \limits _{n\to \infty} x_{2n+1}=\lim \limits _{n\to \infty} x_{3n}=x$$
The exercise doesn't especify the metric that is being used, is this even necessary?
Well, $\Rightarrow$ is trivial.
On $\Leftarrow$ I thought: I wan't to prove that given $\epsilon$ then $\exists N; n\geq N \rightarrow d(x_n,x)<\epsilon$. But, I know that there is some $N$ like that for even index or odd index, hence if I take the greatest $N$ of each case I'll have a natural that the statment holds for all $n$.
But it's weird because I didn't used the fact that $x_{3n}\rightarrow x$ (of course, if my ideia is right than this converges to $x$)
I'd be grateful if you help me! (and i'm sorry if its duplicated)
Your idea is correct and based on that you can write a formal proof without using the sub sequence $\{ x_{3n}\}$ because it is not really needed for the convergence of your sequence.