Is an extension of a consistent theory that assumes its own finite consistency consistent? Fix a proof system for first-order arithmetic and let $\text{Con}(T,k)$ be an encoding of the statement that there is no proof of a contradiction in theory $T$ shorter than $k$ bits.
Now define $U_k = \text{PA} + \text{Con}(U_k, k)$ using Gödel's fixed point theorem.  So each $U_k$ is the theory of PA with an additional axiom claiming that $U_k$ is consistent up to $k$-bit proofs.
Let's use ZFC as a meta-theory, so we know that PA is consistent.  Are all the $U_k$ also consistent?  
 A: It depends on the particular fixed-point that you use. 
First, on the negative side, I claim that for some fixed points and large enough $k$, the answer is no, they can be inconsistent. 
For example, let $\psi$ be any statement that can be refuted by PA in many fewer than $k$ bits, such as the statement $1\neq 1$. In this case, PA proves that $\psi$ is false, and it also proves that $\text{PA}+\psi$ proves a contradiction, and for large enough $k$ this proof will use fewer than $k$ bits. So PA proves that $\psi$ and $\text{Con}(\text{PA}+\psi,k)$ are equivalent, both being false. So it is one of the fixed points asserted to exist by the fixed-point lemma. But it is not consistent. 
Meanwhile, on the positive side, let us assume that PA is consistent. Let $\psi$ be the sentence $1=1$, which is provable in PA and consistent with PA. For any fixed $k$, it follows that there will be no proof of a contradiction from $\text{PA}+\psi$ in fewer than $k$ bits (since we assumed PA is consistent), and furthermore PA will prove this for each particular $k$, since it can note the fact specifically about each of the finitely many proofs. So PA proves that $\psi$ is true and also that $\text{Con}(\text{PA}+\psi,k)$ is true, and so it is a fixed point. And for this fixed point, we have consistency. 
Thinking about it a little more, I've come to realize that what is going on is the following:
Theorem. If PA proves or refutes $\psi$, then for any sufficiently large $k$, PA proves that $\psi$ is equivalent to $\text{Con}(\text{PA}+\psi,k)$, and so all such statements are fixed points. 
Proof. If PA refutes $\psi$, then that proof has some length, and so for $k$ large enough, PA also proves that PA+$\psi$ is inconsistent, and thus PA proves $\psi\leftrightarrow \text{Con}(\text{PA}+\psi,k)$, since both sides are proven false. 
If PA proves $\psi$, and without loss is consistent, then for any $k$ there will be no proof of a contradiction from PA+$\psi$ using fewer than $k$ bits, and PA will prove that by verifying each of those proofs. So in this case, PA proves $\psi\leftrightarrow \text{Con}(\text{PA}+\psi,k)$.
So in either case, $\psi$ is a fixed point.
QED
Update. Although one might be disappointed by the fixed-points provided by the previous theorem, I've now realized that all fixed points are exactly like that. 
Theorem. The following are equivalent for any sentence $\psi$:


*

*$\text{PA}\vdash\psi\leftrightarrow\text{Con}(\text{PA}+\psi,k)$ for some $k$.

*$\text{PA}\vdash\psi\leftrightarrow\text{Con}(\text{PA}+\psi,k)$ for all sufficiently large $k$. 

*$\psi$ is provable or refutable in PA.


Proof. I've already proved that if $\psi$ is provable or refutable, then $2$ holds, and this implies $1$. Conversely, suppose that $1$ holds. Since there are only finitely many possible proofs using $k$ at most $k$ bits, it follows for any particular $k$ that PA either proves or refutes $\text{Con}(\text{PA}+\psi,k)$, simply by inspecting the finitely many proofs. So PA proves or refutes $\psi$, since $\psi$ is equivalent to that statement. So all three statements are equivalent.
QED
In particular, there are no fixed points that are themselves independent of PA. 
