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[tex]\text{the three consecutive terms in an arithmetic sequence}\\a;\ a+d;\ a+2d\\\\\text{system of equals}\\\\ \left\{\begin{array}{ccc}a+a+d+a+2d=27\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}3a+3d=27&|:3\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}a+d=9&\to d=9-a\\a(a+d)(a+2d)=585\end{array}\right\\[/tex]
[tex]\text{substitute to the second equation}\\\\a(a+9-a)(a+2(9-a))=585\\\\a(9)(a+18-2a)=585\\9a(18-a)=585\ \ \ |:9\\a(18-a)=65\\18a-a^2=65\ \ \ |\text{change the signs}\\a^2-18a=-65\\a^2-2a\cdot9=-65\ \ \ |+9^2\\\underbrace{a^2-2a\cdot9+9^2}_{(a-b)^2=a^2-2ab+b^2}=-65+9^2\\(a-9)^2=16\to a-9\pm\sqrt{16}\\a-9=-4\ \vee\ a-9=4\ \ \ |+9\\a=5\ \vee\ a=13\\\\d=9-5=4\ \vee\ d=9-13=-4[/tex]
[tex]Answer:\ a=5;\ a+d=9; a+2d=13\ vee\ a=13;\ a+d=9;\ a+2d=5[/tex]

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