My 2 cents (Please, accept my apologies, native English speakers - I will be grateful for stylistic corrections).
Firstly let me say, that when pointing minimum (best case) and maximum number (worst case) of comparisons as $1$ and $n$ and using "therefore" average is $\frac{n+1}{2}$ leaves impression, that you calculate arithmetical mean, which is wrong action.
Based on above, firstly, I introduce conception of average case: suppose we are considering some algorithm and define $T$ its running time or time complexity function as number of some particular operations, depending on input with size $n$. Suppose $T$ can obtain only finite amount of values $x_1,x_2,\cdots,x_k$ and for each value we know also its expectation i.e. we have numbers $p_1,p_2,\cdots,p_k$, where each $p_i$ characterizing expectation for $T$ to take value $x_i$. Not going deeply in soul of term "expectation", on first step, we can simply understand them as numbers with properties $\forall i,0 \leqslant p_i \leqslant 1$ and $\sum\limits_{i=1}^{k}p_i=1$ i.e shares, parts of unit. Now average case we define as:
$$AT=\sum\limits_{i=1}^{k}p_ix_i$$
In most simple case we can consider equal expectations i.e. $\forall i, p_i=\frac{1}{k}$ and call it uniform expectation. Not uniform expectations, sometimes, are, also, very useful and interesting.
Now coming back to linear search algorithm, let's assume, that we have array with length $n$ and we are searching for some particular value. Usually we consider $T$ as number of comparisons happened during searching. Obviously we have $n$ possible cases to find searched value respectively on places $1,2,\cdots,n$ or not find it at all, so we have $n+1$ possible cases altogether. Obviously $T$ get value $i$ if find happens on place $i$ and together with case to find happened on $n$-th $T$ get value $n$ also in case when search failed. So we have for $T$
$$\begin{cases}
\ \ 1\quad 2 \quad\cdots \quad n \quad n & \text{ values}\\
\ \frac{1}{n+1}\ \frac{1}{n+1} \cdots \ \frac{1}{n+1} \ \frac{1}{n+1} & \text{ expectations}
\end{cases}$$
Now average case is
$$AT=\sum\limits_{i=1}^{n}\frac{i}{n+1}+\frac{n}{n+1}=\frac{1}{n+1}\frac{n(n+1)}{2}+\frac{n}{n+1}=\frac{n}{2}+\frac{n}{n+1}$$
Now come time to remember definition of big-$O$ $$O(f)=\{g: \exists C>0, \exists N \in \mathbb{N}, \forall n>N, g(n) \leqslant Cf(n)\}$$
and we can conclude
$$AT=\frac{n}{2}+O(1)=O\left(\frac{n}{2}\right)=O(n)$$
And now we come to your last question and let me show, that $O\left(\frac{n}{2}\right)=O(n)$ i.e. $O\left(\frac{n}{2}\right) \subset O(n)$ and $O(n) \subset O\left(\frac{n}{2}\right)$:
- Suppose $f \in O\left(\frac{n}{2}\right)$, which means $\exists C,N$ such that $f(n) \leqslant C\frac{n}{2}=\frac{C}{2}n=C_1 n$, so $f \in O(n)$.
- Suppose $f \in O(n)$. This means $\exists C,N$ such that $f(n) \leqslant C n=2C \frac{n}{2} = C_1 \frac{n}{2} $ , so $f \in O\left(\frac{n}{2}\right)$.
Gift 1. This gift I would recommend read after understanding above fully. Now let's consider not uniform case, so lets assume that we expect to find searched value with expectation $p$, i.e. find it in any place from $1$ to $n$, and, so, not to find it with expectation $1-p$. Now we have for for $T$
$$\begin{cases}
\ \ 1\quad 2 \quad\cdots \quad n \quad n & \text{ values} \\
\ \frac{p}{n}\quad \frac{p}{n}\quad \cdots \ \ \frac{p}{n} \quad \ 1-p & \text{ expectations}
\end{cases}$$
Now average will be:
$$AT=\sum\limits_{i=1}^{n}\frac{ip}{n}+n(1-p)=\frac{p}{n}\frac{n(n+1)}{2}+n(1-p)=n\left(1- \frac{p}{2}\right)+\frac{p}{2} = O(n)$$
Gift 2. There can be considered linear search algorithm, which counts 2 comparisons on each step, based on counts as a comparison with the desired value, so a comparison when the end of the cycle is checked. In this case is very interesting so called sentinel method of searching. The average of this algorithm is half of the considered classical search. An interesting exercise to test your awareness of the issue.
