The Central Limit Theorem and the Scaled Sample Mean My introductory probability book gives the following two theorems:
Theorem 1.

For large $n$, the distribution of $\bar{X}_n$ is approximately $N(\mu, \sigma^2/n)$.

Theorem 2.

The CLT says that the sample mean $\bar{X}_n$ is approximately Normal, but since the sum $W_n = X_1 + ... + X_n = n\bar{X}_n$ is just a scaled version of $\bar{X}_n$, the CLT also implies $W_n$ is approximately Normal. If the $X_j$ have mean $\mu$ and variance $\sigma^2$, $W_n$ has mean $n\mu$ and variance $n\sigma^2$. The CLT then states that for large $n$,
$W_n \sim N(n\mu, n\sigma^2)$.

It seems to me like the theorem 2 is inconsistent with theorem 1. If $W_n = n\bar{X}_n$, meaning it is just a scaled version of the sample mean, and the sample mean itself is approximately Normal, then, according to theorem 1, for large $n$, the distribution of $W_n = n\bar{X}_n$ should also approximately be $N(\mu, \sigma^2/n)$, no?
I would greatly appreciate it if people could please take the time to clarify this.

$$\Bbb{E}(W_n) = \Bbb{E}(n \bar{X}_n) = n \Bbb{E}(\bar{X}_n) \stackrel{\text{Thm 1}}{\approx} n \mu \\
\mathrm{var}(W_n) = \mathrm{var}(n \bar{X}_n) = n^2 \mathrm{var}(\bar{X}_n) \stackrel{\text{Thm 1}}{\approx} n^2 \cdot \frac{\sigma^2}{n} = n \sigma^2$$
