Arithmetic

From general terms of arithmetic sequence, $a_n=a_1+(n-1)d$

since,

$\begin{aligned}\color{#D61F06}{a_2}=&\ a_1+(2-1)d=&\ \color{#D61F06}{a_1+d}\\\color{#20A900}{a_3}=&\ a_1+(3-1)d=&\ \color{#20A900}{a_1+2d}\\\color{#3D99F6}{a_4}=&\ a_1+(4-1)d=&\ \color{#3D99F6}{a_1+3d}\end{aligned}$

That is,

$\begin{aligned}a_1+(\color{#20A900}{a_1+2d})=&\ 2\color{#69047E}{a_1}+2\color{#624F41}d=&\ 6&\Rightarrow(1)\\(\color{#D61F06}{a_1+d})+(\color{#3D99F6}{a_1+3d})=&\ 2\color{#69047E}{a_1}+4\color{#624F41}d=&\ 20&\Rightarrow(2)\\\text{Equation(2) - Equation(1);}\quad\quad\ 2\color{#624F41}d=&\ 14\Rightarrow \color{#624F41}{d=7}\end{aligned}$

instead $\color{#624F41}d$ with $\color{#624F41}7$ in equation (1);

$\begin{aligned}2\color{#69047E}{a_1}+2(\color{#624F41}7)=&\ 6\\2\color{#69047E}{a_1}+14=&\ 6\\2\color{#69047E}{a_1}=&\ -8\\\color{#69047E}{a_1}\color{#69047E}=&\ \color{#69047E}{-4}\end{aligned}$

Thus, $a_{15}=\color{#69047E}{a_1}+14\color{#624F41}d=\color{#69047E}{-4}+14(\color{#624F41}7)=-4+98=\boxed{94}$