Resposta :

PeH
[tex]\bullet \ \text{P.A.} = (x + 1, x - 3, 2x) \\\\ --------------- \\\\ (x - 3) - (x + 1) = 2x - (x - 3) \\ x - 3 - x - 1 = 2x - x + 3 \\ -4 = x + 3 \\ x = -4 - 3 \\\\ \boxed{x = -7} \\\\ ou \\\\ x - 3 = \frac{(x + 1) + 2x}{2} \\\\ 2 \cdot (x - 3) = 3x + 1 \\ 2x - 6 = 3x + 1 \\\\ \boxed{x = -7} \\\\ --------------- \\\\ \text{Assim:} \\\\ \text{P.A.} = (x + 1, x - 3, 2x) \rightarrow (-7 + 1, -7 - 3, 2 \cdot -7) \rightarrow \boxed{(-6, -10, -14)} \\\\ \boxed{\bullet \ r = -4 \rightarrow \text{P.A. decrescente}}[/tex]
x -  3 - (x + 1 ) = 2x - (x-3)

x - 3 - x - 1 = 2x - x + 3

- 4 - 3 = x

   x = - 7


x+1 = - 7 + 1 = - 6
x - 3= - 7 - 3 = - 10
2x = 2(-7) = - 14 


(-6, -10, - 14,.........)

Outras perguntas