Let $E_{p,q}^r$ be a spectral sequence which converges to $A_n$. Let $E_{p,q}^r=0$ for $q\ge 2$. How to construct a long exact sequence
$$\cdots \to A_{p+1}\to E_{p+1,0}^2\xrightarrow{d_{p+1,0}^2} E_{p-1,1}^2\to A_p\to E_{p,0}^2\to \cdots \text{?}$$
I would say, for all $p\ge 0$ the maps $A_{p}\to E_{p,0}^2$ have to be the canonical projection, because: $E_{p,q}^r$ converges to $A_n$ means that there exists a sequence $$0=F_0A_n\subseteq F_1A_n\subseteq \cdots \subseteq F_{n-1}A_n\subseteq F_nA_n=A_n$$ such that $\bigcup\limits_{l=0,\ldots,n}F_lA_n=A_n$ and $E_{pq}^\infty\cong \frac{F_pA_{p+q}}{F_{p-1}A_{p+q}}$.
In our case is: $E_{p+1,0}^r\cong \frac{F_{p+1}A_{p+1}}{F_p A_{p+1}}=\frac{A_{p+1}}{F_p A_{p+1}}$ for all $r\ge 2$. Therefore we can take $A_{p+1}\to E_{p+1,0}^2$ the canonical projection I would say. But I'm not sure if its range is the kernel of $d_{p+1,0}^2$.
The next step is then to consider $E_{p-1,1}^2\to A_p$. Again, I have $E_{p-1,1}^2\cong \frac{F_{p-1}A_{p}}{F_{p-2}A_{p}}$ and we have the inclusion $F_{p-1} A_p \to A_p$. But here I'm stuck. How to continue to construct the long exact sequence? Best.