The sum of first four terms of an A.P.
$\begin{aligned}
&=11+(11+d)+(11+2 d)+(11+3 d) \\
&=\frac{4}{2}[11 \times 2+(4-1) d] \\
&=2[22+3 d]=44+6 d=56 \quad \ldots(i)\\
6 d &=56-44=12 \\
\therefore \quad d &=2
\end{aligned}$
Let the last four terms are $T_{n-3}, T_{n-2}, T_{n-1}, T_n$ then the sum of four term of an A.P.
$=T_{n-3}+T_{n-2}+T_{n-1}+T_n$
Now, $\quad T_{n-3}=a+(n-4) d=11+(n-4) d$
Sum of last four term
$\begin{aligned}
&=\frac{4}{2}[2(11+(n-4) d)+(4-1) d] \\
&=2[22+2 n d-8 d+3 d]
\end{aligned}$
$112=2[22+2 n d-5 d]$
$\Rightarrow \quad 22+2 n d-5 d=56 \quad \ldots(ii)$
Put $d=2$ in (ii)
$\quad 22+4 n-10=56 \Rightarrow 4 n=56-12$
$\Rightarrow \quad 4 n=44 \Rightarrow n=11$