Question

Determine: a razão e os termos a2, a3 e a4 de uma PA crescente cujos termos a5 e a1 são as raízes da equação (x+3)^2+ (x-4)^2= x(x+10) +89​

Answer 1

1º vamos desenvolver a equação :

$\displaystyle (\text x+3)^2+(\text x-4)^2=\text x(\text x+10)+89 \\\\ \text x^2+6\text x+9+\text x^2-8\text x+16 = \text x^2+10\text x+89 \\\\ \text x^2-12\text x-64 = 0 \\\\ \text x = \frac{-(-12)\pm\sqrt{(-12)^2-4.1.(-64)}}{2} \\\\\\ \text x = \frac{12\pm\sqrt{144+256}}{2} \to \text x = \frac{12\pm20}{2} \\\\\\ \text x = \frac{12-20}{2} \to \boxed{\text {x'} = -4}\\\\ \text x = \frac{12+20}{2} \to \boxed{\text {x''}= 16}$

A questão diz que $\text a_5\ \text e \ \text a_1$ são raízes da equação, e também diz que é uma PA  crescente, ou seja, $\text a_1 < \text a_5$ , portanto :

$\text a_1 = -4 \\\\ \text a_5 = 16$

Achando a razão :

$\text a_5 = \text a_1 + 4.\text r \\\\ 16 = -4 +4.\text r \\\\ 4 = -1+\text r \\\\ \huge\boxed{\text r = 5 \ }\checkmark$

Achando os termos :

$\text a_2 = \text a_1+\text r \to \text a_2 = -4+5 \to \huge\boxed{\text a_2 = 1}\checkmark$

$\text a_3 = \text a_1+2\text r \to \text a_3 = -4+2.5 \to \huge\boxed{\text a_3 = 6}\checkmark$

$\text a_4 = \text a_1+3\text r \to \text a_4 = -4+3.5 \to \huge\boxed{\text a_4 =11\ }\checkmark$