Minimum number of states in the DFA
Let $L_n$ consist of all words over $\{a,b\}$ such that the $n$th symbol from the right is $a$. The minimum number of states in a DFA accepting $L_n$ is $2^n$.

First, let us construct a DFA with so many states. The states of the DFA correspond to strings of length $n$ over $\{a,b\}$. The starting state is $b^n$. When in state $\sigma_1\ldots \sigma_n$ and reading $\sigma$, we move to state $\sigma_2 \ldots \sigma_n \sigma$. A state is accepting if the first symbol is $a$.

Second, we use the Myhill–Nerode theorem to show that $2^n$ states are necessary. We claim that all words of length $n$ are pairwise inequivalent. Indeed, let $x \neq y$ be words of length $n$. Then $x_i \neq y_i$ for some $i$, say $x_i = a$ and $y_i = b$. This implies that $x b^{i-1} \in L_n$ whereas $y b^{i-1} \notin L_n$, showing that $x,y$ are inequivalent modulo $L_n$.