Selesaikan {l}{2a_{1}+b=a_{1}+a_{3}}{a_{1}+a_{2}+a_{3}=42}{a_3^2=a_{1}a_{3}}

$$\left\{ \begin{array} { l } { 2 a _ { 1 } + b = a _ { 1 } + a _ { 3 } } \\ { a _ { 1 } + a _ { 2 } + a _ { 3 } = 42 } \\ { a _ { 3 } ^ { 2 } = a _ { 1 } a _ { 3 } } \end{array} \right.$$

Selesaikan untuk a_1, b, a_3, a_2

$$a_{1}=42-a_{2}\text{, }b=a_{2}-42\text{, }a_{3}=0$$

$$a_{1}=-\frac{a_{2}}{2}+21\text{, }b=0\text{, }a_{3}=-\frac{a_{2}}{2}+21$$