Question

Vamos construir um P.G em que a soma do 3° com o 5° termo é 5/4 e a soma do 7° com o 9° termo é 20.

Answer 1

$(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9)\ \to\ \bigg(\dfrac{x}{q^4},\dfrac{x}{q^3},\dfrac{x}{q^2},\dfrac{x}{q},x,xq,xq^2,xq^3,xq^4\bigg)$

$a_3+a_5=\dfrac{5}{4}\ \to\ \dfrac{x}{q^2}+x=\dfrac{5}{4}\ \to\ 4x+4xq^2=5q^2\ \to\ q^2=\dfrac{4x}{5-4x}$

$a_7+a_9=20\ \to\ xq^2+xq^4=20\ \to\ xq^2(q^2+1)=20\ \to\\\\ \to\ x\bigg(\dfrac{4x}{5-4x}\bigg)\bigg(\dfrac{4x}{5-4x}+1\bigg)=20\ \to\ \dfrac{4x^2}{5-4x}\bigg(\dfrac{5}{5-4x}\bigg)=20\ \to\\\\ \to\ 20x^2=20(5-4x)^2\ \to\ x^2=25-40x+16x^2\ \to\\\\ \to\ 15x^2-40x+25=0\ \to\ 3x^2-8x+5=0\\\\ x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(3)(5)}}{2(3)}=\dfrac{8\pm\sqrt{64-60}}{6}=\dfrac{8\pm2}{6}=\dfrac{4\pm1}{3}\\\\ \boxed{x=\dfrac{5}{3}}\ \ \textrm{ou}\ \ \boxed{x=1}$

$q^2=\dfrac{4x}{5-4x}\ \to\ q=\pm\sqrt{\dfrac{4x}{5-4x}}\\\\ \mathrm{Se}\ x=\dfrac{5}{3}\ \to\ q=\pm\sqrt{\dfrac{\frac{20}{3}}{\frac{15}{3}-\frac{20}{3}}}=\pm\sqrt{\dfrac{20}{3}\bigg(\dfrac{-3}{5}\bigg)}=\pm\sqrt{-4}=\pm2i,\ q\notin \mathbb{R}\\\\ \mathrm{Se}\ x=1\ \to\ q=\pm\sqrt{\dfrac{4}{5-4}}=\pm\sqrt{4}=\pm2\ \to\ \boxed{q=\pm2}$

$\textrm{P.G.:}\ \bigg(\dfrac{x}{q^4},\dfrac{x}{q^3},\dfrac{x}{q^2},\dfrac{x}{q},x,xq,xq^2,xq^3,xq^4\bigg)\\\\ \mathrm{Se}\ x=1\ \mathrm{e}\ q=2\ \to\ \textrm{P.G.:}\ \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2},1,2,4,8,16\bigg)\\\\ \mathrm{Se}\ x=1\ \mathrm{e}\ q=-2\ \to\ \textrm{P.G.:}\ \bigg(\dfrac{1}{16},-\dfrac{1}{8},\dfrac{1}{4},-\dfrac{1}{2},1,-2,4,-8,16\bigg)$