Approximate distribution for the sample mean? 
A random variable $X$ is said to follow a discrete uniform distribution if its probability function is given by
  $$p_X(x) = \left\{ \begin{array}{ll}\frac{1}{\theta}, & x = 1, 2, \ldots, \theta\\
0, & \text{otherwise}, \end{array}\right.$$
  where $\theta$ is a positive integer. This distribution is denoted $\text{dunif[1, $\theta$]}$ in the following parts of the question.
  
Given that $\sum\limits_{j = 1}^n j = \frac{1}{2}n(n + 1)$ and $\sum\limits_{j = 1}^n j^2 = \frac{1}{6} n(n + 1)(2n + 1)$, find the mean and variance of this distribution.
Determine the cumulative distribution function $F_X(x)$, ensuring that it is defined properly for all $x$.
The random variables $X$ and $Y$ are independent, with $X \sim \text{dunif}[1, \theta]$ and $Y \sim \text{dunif}[1, \theta]$. Carefully apply the convolution formula to determine the probability function of $W = X + Y$.
A $20$-sided fair die is rolled as many times as necessary. Write down the probability expression that the third face $10$ turns up on the $n$th roll.
For a large $n$, what is an approximate distribution for the sample mean $\overline{X}$? Ensure that all the distribution details are given.
  

I've got the previous parts.
For the last part, my attempt is that it's about CLT (is this correct?) but then I am stuck.
Is the approximate distribution of sample mean Normal here?
And what are the details of such distribution? 
Thank you for your help. (This is the complete question)
 A: What you are being asked is as follows. Imagine tossing the die $n$ times. Let $X_i$ be the number obtained on the $i$-th toss, and let 
$$\bar{X}=\frac{X_1+X_2+\cdots+X_n}{n}$$
For large $n$, what is the approximate distribution of $\bar{X}$?
First we find the mean $\mu$ and the variance $\sigma^2$ of $X_i$.
The random variable $X_i$ has discrete uniform distribution, taking on values $1$ to $20$ each with probability $\frac{1}{20}$. It follows that 
$$\mu=\frac{1+2+\cdots+20}{20}.$$
By a formula you were given earlier in the problem, the sum $1+2+\cdots +20$ is $\frac{(20)(21)}{2}$.  It follows that $\mu=\frac{21}{2}$.
For $\sigma^2$, use the fact this is $E(X_i^2)-(E(X_i))^2$. We need to find $E(X_i^2)$. This is 
$$\frac{1^2+2^2+\cdots +20^2}{20}.$$
To find the sum of the squares, use the formula given to you for the sum of the first $n$ squares. The sum of the squares is $\frac{(20)(21)(41)}{6}$. 
Now you have all the information needed to find $\sigma^2$. Divide the sum of squares by $20$, and subtract $(10.5)^2$.
Finally, the mean of $\bar{X}$ is $\mu$, and the variance of $\bar{X}$ is $\frac{\sigma^2}{n}$. For largish $n$, and it does not have to be terribly large, $\bar{X}$ is approximately normal with mean, variance given in this paragraph. And yes, it is by the Central Limit Theorem that we conclude that $\bar{X}$ has approximately normal distribution. 
A: A normal distribution is determined by its mean and variance - these are the "details" you are being asked about.
So, you need to compute the expectation and variance of the sample mean and take the limit.
Your $X$ is discrete uniform; it has finite variance, so, by CLT, $\bar X$ converges to a Gaussian normal.
By linearity of expectation,
$$E(\bar X)=E(\frac{\sum_{i=1}^nX}{n})=\frac{\sum_iE(X)}{n}=E(X)\frac{n}{n}=E(X)$$
and one can compute the expectation
$$E(X)=\sum_{i=1}^\theta i\frac{1}{\theta}=\frac{1}{\theta}\frac{\theta(\theta-1)}{2}=\frac{\theta-1}{2}$$
Similarly for the variance - but I will leave it as an exercise to the reader (too late now, sorry).
