If $A$ is a commutative unital ring and $E$ is a finite rank projective $A$-module there is a surjective $A$-linear map $\phi: A^n \rightarrow E$, with kernel $F:=ker(\phi)$ and $F\oplus E \cong A^n$ an isomorphism of $A$-modules. We get a short exact sequence of projective finite rank $A$-modules
$$ 0 \rightarrow F \rightarrow A^n \rightarrow E \rightarrow 0 .$$
There is a canonical projection map (an $A$-linear map)
$$ \psi: A^n \rightarrow F$$
with $ker(\psi)=E$. Using a "splicing-technique" we get an infinite exact sequence of free $A$-modules
$$ \cdots \rightarrow A^n \rightarrow A^n \rightarrow A^n \rightarrow E \rightarrow 0.$$
Hence $E$ has an infinite free resolution by projective (and free) $A$-modules of finite rank.
If $k$ is a field and $X \subseteq \mathbb{P}^n_k$ is a regular projective scheme there is for any coherent sheaf $F$ an infinite resolution by finite direct sums of invertible sheaves:
Hartshorne Corr. II.5.18 says: If $X \subseteq \mathbb{P}^n_k$ is projective, it follows any coherent sheaf $F$ is a quotient
$$ f_1:\oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$
It follows $X:=Proj(k[x_1,..,x_n]/I)$ with $B:=k[x_i]/I$ noetherian. Hence if $F$ is a coherent module it follow any submodule $E \subseteq F$ is coherent. It follows $\oplus \mathcal{O}(n(1)_i)$ is coherent hence $ker(f_1)$ is a coherent module. Hence there is a surjection
$$ f_2: \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow ker(f_1)$$
and continuing this process you get a long exact sequence of $\mathcal{O}_X$-modules
$$S1.\text{ }\cdots \rightarrow \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)}) \rightarrow \oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$
Example: If $X$ is projective space and $F$ a finite rank locally free sheaf it follows the sequence S1 terminates - it is finite. A similar property holds for the grassmannian. As a consequence it follows the Grothendieck group of projective space (and the grassmannian) is generated by classes of invertible sheaves.
General question: If $A \subseteq \mathbb{P}^n_K$ is an abelian variety with $K$ a number field - give an explicit example of a finite rank non-trivial vector bundle $F$ on $A$ with a resolution as in S1 that is infinite.
By "non-trivial" I mean: $F$ is not a direct sum of invertible sheaves.
Note: If $k:=\overline{K}$ is the algebraic closure of $K$ and if $A\subseteq \mathbb{P}^n_k$ is an abelian variety, any homogeneous finite rank vector bundle $E$ is on the form
$$ E\cong \oplus_{L\in Pic^0(A)} L\otimes U_L$$
where $U_L$ is an iterated extension of trivial bundles. Hence in the grothendieck group $K_0(A)$ we get
$$ [L\otimes U_L]=[L]*[U_L]$$
and $[U_L]=rk(U_L)[\mathcal{O}_A]$ hence
$$[L\otimes U_L]=[L]rk(U_L)[\mathcal{O}_A]=rk(U_L)[L]$$
and
$$[E]= \oplus_{L\in Pic^0(A)} rk(U_L)[L].$$
Hence for a homogeneous finite rank vector-bundle $E$ it follows the sequence S1 is finite. Hence in this case it follows the class $[E]\in K_0(A)$ lives in the sub group generated by classes of invertible sheaves. I'm looking for explicit examples where this does not hold.
Such coherent sheaves exist more generally by the following post: