$\begin{array}{1 1}21 \\ 20 \\ 11 \\ 19 \end{array} $

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- Sum of $n$ terms of an A.P. $=S_n=\large\frac{n}{2}$$(2a+(n-1)d)$

Let the A.P. be $a,\:a+d,\:a+2d,..........t_n$

Given that sum of first $4$ terms of the A.P.$=56$ and $a=11$

We know that sum of $n$ terms of an A.P. $=S_n=\large\frac{n}{2}$$(2a+(n-1)d)$

$\Rightarrow\:\large\frac{4}{2}$$(2\times 11+(4-1)d)=56$

$\Rightarrow\:22+3d=28$

$\Rightarrow\:3d=6$ or $d=2$

Step 2

Also given that sum of last $4$ terms $=112$

Now

If $ t_1,t_2,.........t_n$ is an A.P. with first term $a$, common difference $d$ and

last term $t_n$, then $t_n,t_{n-1}...........t_2,t_1$ is an A.P. with first term $t_n$ and common difference $-d$

$\therefore\:t_n+t_{n-1}+t_{n-2}+t_{n-3}=112$

$\Rightarrow\:\large\frac{4}{2}$$\big[2.t_n+(4-1)(-2)\big]=112$

$\Rightarrow\:2t_n-6=56$

$\Rightarrow\:t_n=62/2=31$

But we know that $t_n=a+(n-1)d$

$\Rightarrow\:11+(n-1).2=31$

$\Rightarrow\:n-1=20/2=10$ $\Rightarrow\:n=11$

$i.e., $ No. of terms $=11$.

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