Proof in sequent calculus Is it actually possible to prove this in sequent calculus?

if $\Gamma \vdash \lnot A$ , then $\Gamma \vdash A ⊃ B$

where $\Gamma$ is a set of formulas and A and B are formulas.
 A: Details will depend on the system in play. But you should be able e.g. to argue from  (i) $F \vdash \neg A$ to (ii) $F, A \vdash \bot$ to (iii) $F, A \vdash B$ to (iv) $F  \vdash A \to B$.  
A: I will use the LK system [see Gaisi Takeuti, Proof Theory (2nd ed - 1987), page 9-on] :

$A \rightarrow A$ --- Axiom
$\Gamma, A \rightarrow A$ --- by n Weakening-left
$\Gamma , A \rightarrow A, B$ --- by Weakening-right
$\Gamma \rightarrow A, A \supset B$ --- by $\supset$-right
$\Gamma, \lnot A \rightarrow A \supset B$ --- by $\lnot$-left.

At this point, we introduce the sequent :

$\Gamma \rightarrow \lnot A$

and then, by Cut (with cut-formula : $\lnot A$ ) :


$\Gamma \rightarrow A \supset B$.


A: @MauroALLEGRANZA has given an answer that are valid in classical logic, but the result actually holds intuitionistically too.  This is just a simple, intuitionistically valid modification of his proof:
\begin{eqnarray}
A&\vdash& A \\
A,\neg A & \vdash  & \\
A, \neg A & \vdash & B \\
\neg A & \vdash & A\to B 
\end{eqnarray}
At this point, as in @Mauro's proof, you introduce $\Gamma\vdash \neg A$, and apply the cut rule. 

Note I originally commented that @Peter Smith's answer was also classical, but on reflection, I see that his reasoning is intuitionistically valid (and is basically the same as what I have above). He does not use the Law of the Excluded Middle.
A: $(A \rightarrow B)$ is a logical consequence of $\lnot A$, so if your sequent calculus is complete, it is certainly possible to find such a proof.
