Simply simple progression

Let the first term be $a$ , common difference be $d$ , $a_4$ be the fourth term, $a_{50}$ be ${50}^{\text{th}}$ term and sum of first 4 terms be $S_4$ .
$\begin{aligned} a_4 & = a + \left(4 - 1\right)×d\\ 22 & = a + 3d \longrightarrow \boxed{1}\\ \\ S_4 & = \dfrac{4}{2} × \left(2a + \left(4 - 1\right)×d\right)\\ 6d + 4 & = 2 × \left(a + a + 3d\right)\\ \dfrac{2\left(3d + 2\right)}{2} & = a + a + 3d\\ 3d + 2 & = a + a + 3d \longrightarrow \boxed{2}\\ \\ \text{Substituting} \boxed{1} \text{in} \boxed{2}:-\\ 3d + 2 & = a + 22\\ a - 3d & = -20 \longrightarrow \boxed{3}\\ \\ \text{Adding} \boxed{1} \text{and} \boxed{3}:-\\ 2a & = 2\\ a & = 1 \longrightarrow \boxed{4}\\ \\ \text{Substituting} \boxed{4} \text{in} \boxed{1}:-\\ 22 & = 1 + 3d\\ 3d & = 21\\ d & = 7\\ \\ a_{50} & = a + \left(50 - 1\right)d\\ & = 1 + 49×7\\ & = 1 + 343\\ & = \color{#69047E}{\boxed{344}} \end{aligned}$

Let the first term be $a$ and the common difference be $d$ . From the given conditions:

$T_4 = 22\\ a+3d = 22$

$S_4 - 6d = 4\\ \dfrac{4}{2}(2a+3d) - 6d = 4\\ 2(2a+3d) - 6d=4\\ 4a+6d-6d=4\\ 4a=4\\ a=1$

Substitute to find $d$ :

$1+3d=22\\ 3d=21\\ d=7$

Therefore,

$T_{50} = 1+(49)(7) = \boxed{344}$

We know that:

• $1-a_4=a_1+3d=22$

• $2- S_4=2(a_1+a_4)=6d + 4$

From $2$ :

$S_4=2(a_1+22)=6d+4$

$\implies S_4=a_1+22=3d+2$

$\implies\color{#3D99F6}{a_1=-20+3d}$

From $1$ :

$22=\color{#3D99F6}{-20+3d}+3d$

$\implies 42=6d \implies \color{#D61F06}{d=7}$

Now, from the formula for the $n$ th term of an arithmetic progression $a_n = a_1 + (n-1)d$ :

$a_{50}=a_1+49d$

$a_{50}=\color{#3D99F6}{-20+3d}+49(\color{#D61F06}{7})$

$a_{50}=-20+3(\color{#D61F06}{7})+343 = \boxed{344}$ .