But there is no pattern at all!

The sequence $\{\color{#3D99F6}{2}, \color{#D61F06}{5}, \color{#20A900}{6}, \color{#3D99F6}{7}, \color{#D61F06}{11}, \color{#20A900}{13}, \color{#3D99F6}{12}, \color{#D61F06}{17}, \color{#20A900}{20}, \color{#3D99F6}{17}, \color{#D61F06}{23}, \color{#20A900}{27},...\}$ is composed of three arithmetic progressions as follows: \[\begin{array} {} \color{blue}{2}, & & & \color{blue}{7}, & & & \color{blue}{12}, & & & \color{blue}{17}, & & & \color{blue}{\cdots} \\ & \color{red}{5}, & & & \color{red}{11}, & & & \color{red}{17}, & & & \color{red}{23}, & & & \color{red}{\cdots} \\ & & \color{green}{6}, & & & \color{green}{13}, & & & \color{green}{20}, & & & \color{green}{27}, & & & \color{green}{\cdots} \end{array} \]

The first terms, common differences and numbers of terms for the first 50 terms of the sequence of the three AP are:

$\begin{cases} a_1 = 2, \ d_1 = 5, \ n_1 = 17 \\ a_2 = 5, \ d_1 = 6, \ n_2 = 17 \\ a_3 = 6, \ d_1 = 7, \ n_3 = \left \lfloor \frac{50}{3} \right \rfloor = 16 \end{cases}$

The sum of the first 50 terms of the sequence is:

$\begin{aligned} S_1(17)+S_2(17)+S_3(16) & = 17\left(\frac{2(2)+5(17-1)}{2} \right) + 17\left(\frac{2(5)+6(17-1)}{2} \right) + 16\left(\frac{2(5)+7(16-1)}{2} \right) \\ & = 714 + 901 + 936 \\ & = \boxed{2551} \end{aligned}$

We can break up this sequence into three parts:- $2,7,12,7,\cdots\\ 5,11,17,23,\cdots\\ 6,13,20,27,\cdots$ .

Now, these individual parts form an arithmetic progression. Since $50$ is not divisible by $3$ , the first too sequences would have one more term than the last one. Now, the number of terms in the last sequence is $\left \lfloor \dfrac{50}{3} \right \rfloor = \left \lfloor 16.666...... \right \rfloor = 16$ . So, the first two sequences would have $16 + 1 = 17$ terms.

Sum of first 17 terms of first sequence = $\dfrac{17}{2} × \left(2×2 + 16×5\right)\\ = \dfrac{17}{2} × 84\\ = 17 × 42\\ = 714$ .

Sum of first 17 terms of second sequence = $\dfrac{17}{2} × \left(2×5 + 16×6\right)\\ = \dfrac{17}{2} × 106\\ = 17 × 53\\ = 901$ .

Sum of first 16 terms of third sequence = $\dfrac{16}{2} × \left(2×6 + 15 × 7\right)\\ = \dfrac{16}{2} × 117\\ = 8 × 117\\ = 936$ .

So, the sum of the first $50$ terms of the given AP is $714 + 901 + 936\\ = \color{#69047E}{\boxed{2551}}$