Question

Obtenha uma P.A de três termos tais
Que sua soma seja 21 e seu produto seja 91

Answer 1

$\mathrm{PA=\{a_1,a_2,a_3\}=\{a_1,a_1+r,a_1+2r\}}\\ \mathrm{S_n=21\ \ \|\ \ P_n=91}\\\\ \mathrm{a_1+a_2+a_3=21\ \to\ a_1+a_1+r+a_1+2r=21}\\ \mathrm{3a_1+3r=21\ \to\ 3.(a_1+r)=21\ \to\ a_1+r=7}\\\\\\ \mathrm{a_1.a_2.a_3=91\ \to\ a_1.(a_1+r).(a_1+r+r)=91}\\ \mathrm{a_1.7.(7+r)=91\ \to\ a_1.(7+r)=13\ \to\ a_1=\dfrac{13}{(7+r)}}$

$\mathrm{a_1+r=7\ \to\ \dfrac{13}{(7+r)}+r=7\ \to\ \dfrac{13+r.(7+r)}{(7+r)}=7}\\ \mathrm{13+r.(7+r)=7.(7+r)\ \to\ 13+7r+r^2=49+7r}\\ \mathrm{13+r^2=49\ \to\ r^2=36\ \to\ r=\sqrt{36}\ \to\ \boxed{\mathbf{r=6}}}\\\\ \mathrm{a_1+r=7\ \to\ a_1+6=7\ \to\ a_1=7-6\ \to\ \boxed{\mathbf{a_1=1}}}\\\\ \mathrm{PA=\{a_1,a_1+r,a_1+2r\}=\{1,1+6,1+12\}}\\\\ \mathrm{\boxed{\mathbf{PA=\{1,7,13\}}}}$