Question

Em um PG formada por (x -1), (x + 3), (6x), nessa ordem, determine sua razão.

Answer 1

Explicação passo-a-passo:

Em uma $\sf PG(a_1,~a_2,~a_3)$, temos:

$\sf (a_2)^2=a_1\cdot a_3$

Assim:

$\sf (x+3)^2=(x-1)\cdot6x$

$\sf x^2+6x+9=6x^2-6x$

$\sf 6x^2-x^2-6x-6x-9=0$

$\sf 5x^2-12x-9=0$

$\sf \Delta=(-12)^2-4\cdot5\cdot(-9)$

$\sf \Delta=144+180$

$\sf \Delta=324$

$\sf x=\dfrac{-(-12)\pm\sqrt{324}}{2\cdot5}=\dfrac{12\pm18}{10}$

$\sf x'=\dfrac{12+18}{10}~\Rightarrow~x'=\dfrac{30}{10}~\Rightarrow~\red{x'=3}$

$\sf x"=\dfrac{12-18}{10}~\Rightarrow~x"=\dfrac{-6}{10}~\Rightarrow~\red{x"=-\dfrac{3}{5}}$

=> Para $\sf x=3$:

• $\sf a_1=x-1$

$\sf a_1=3-1$

$\sf \red{a_1=2}$

• $\sf a_2=x+3$

$\sf a_2=3+3$

$\sf \red{a_2=6}$

• $\sf a_3=6x$

$\sf a_3=6\cdot3$

$\sf \red{a_3=18}$

=> $\sf \red{PG(2,6,18)}$

Assim:

$\sf q=\dfrac{a_2}{a_1}$

$\sf q=\dfrac{6}{2}$

$\sf \red{q=3}$

=> Para $\sf x=\dfrac{-3}{5}$:

• $\sf a_1=x-1$

$\sf a_1=\dfrac{-3}{5}-1$

$\sf a_1=\dfrac{-3-5}{5}$

$\sf \red{a_1=-\dfrac{8}{5}}$

• $\sf a_2=x+3$

$\sf a_2=\dfrac{-3}{5}+3$

$\sf a_2=\dfrac{-3+15}{5}$

$\sf \red{a_2=\dfrac{12}{5}}$

• $\sf a_3=6x$

$\sf a_3=6\cdotdfrac{-3}{5}$

$\sf \red{a_3=-\dfrac{18}{5}}$

=> $\sf PG\Big(-\dfrac{8}{5},\dfrac{12}{5},-\dfrac{18}{5}\Big)$

Assim:

$\sf q=\dfrac{a_2}{a_1}$

$\sf q=\dfrac{\frac{12}{5}}{-\frac{8}{5}}$

$\sf q=\dfrac{12}{5}\cdot\Big(-\dfrac{5}{8}\Big)$

$\sf q=-\dfrac{60}{40}$

$\sf \red{q=-\dfrac{3}{2}}$

Logo, $\sf q=3~ou~q=-\dfrac{3}{2}$