Convergence in distribution and Slutsky theorem

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When $\{X_1,X_2,X_3,\cdots,X_n\}$ is a random sample from the population distribution $(\mu,\sigma^2)$, according to the CLT, $\bar{X} \to \mathcal{N}(\mu,\sigma^2/n)$ in distribution sense.

Then what is the approximate distribution of $2n \bar{X} / (2n+1)$?

Does it follow the same distribution as $\bar{X}$?

Is it possible to use Slutsky theorem because $2n/(2n+1)$ approximate to 1 in probability?