Arithmetic progressions

Let $a$ be the first term, $d$ be the common difference and $n$ terms be there in this AP.
Sum of first four terms = $\dfrac{4}{2} (2a + (4-1)d) = 2(2×11 + 3d) = 56\\ 22 + 3d = 28\\ 3d = 6\\ d = 2$

Sum of last 4 terms = $\dfrac{4}{2} (a + (n-4)d + a + (n-1)d) = 2(2a +(2n - 5)d) = 112\\ 2×11 + (2n - 5)×2 = 56\\ 22 + (2n - 5)×2 = 56\\ (2n-5)×2 = 34\\ 2n - 5 = 17\\ 2n = 22\\ n = \color{#69047E}{\boxed{11}}$

Let the number of terms be $n$ , and the common difference be $d$ . Given that $a=11$ and $S_4=56$ :

$\dfrac{4}{2}(2(11)+(4-1)(d))=56\\ 2(22+3d)=56\\ 22+3d=28\\ 3d=6 \implies d=2$

Now, let the last 4 terms be $T_{n-3}, T_{n-2}, T_{n-1}$ and $T_n$ respectively. We know that:

$T_n=11+(n-1)(2) = 9+2n\\ T_{n-1}=11+(n-2)(2) = 7+2n\\ T_{n-2}=11+(n-3)(2) = 5+2n\\ T_{n-3}=11+(n-4)(2) = 3+2n$

These $4$ terms add up to $112$ , therefore:

$9+2n + 7+2n + 5+2n + 3+2n = 112\\ 8n+24=112\\ 8n=88 \implies n=\boxed{11}$

Because first four terms' average is 14 and first term is 11 , so 1.5 d = 3, d=2

And, last four temr's average is 28 so, last term is 28+3=31

So the number of terms in this progression is $1+ \frac {1} {2} × ( 31-11) = 11$