Let $\{x_n\}$ be a sequence in a metric space such that $\{x_{2n}\}$, $\{x_{2n+1}\}$ and $\{x_{3n}\}$ all convergent. Show that $\{x_{n}\}$ converges

Question

I don't know how to begin this. My thoughts are that since there are three convergent subsequences, we could use some knowledge about their limits to construct our convergent sequence but I'm not sure how to prove $x_n$ also converges. Is there a property of metric spaces I should use, or something about Bolzano-Weierstrass theorem (for metric spaces)?

EDIT: As noted in a comment I am not allowed to assume they all converge to the same limit.

Answer 1

Assume that $x_{3n}\rightarrow L$ and $x_{2n}\rightarrow M$, $x_{2n+1}\rightarrow N$, then the sequence $\{x_{6n}\}$ is a convergent subsequence of both $\{x_{3n}\}$ and $\{x_{2n}\}$, so $L=M$. The sequence $\{x_{6n+3}\}$ is a convergent subsequence of both $\{x_{3n}\}$ and $\{x_{2n+1}\}$, so $L=N$, so $L=M=N$.