Question

determinar os termos da PA com seus termos nesta ordem (2x-7, x+2, 5x+1)

Answer 1

$\mathrm{PA=\{a_1,a_2,a_3\}=\{2x-7,x+2,5x+1\}}\\\\ \mathrm{r=a_2-a_1\ \to\ r=x+2-(2x-7)}\\ \mathrm{r=x+2-2x+7\ \to\ \boxed{\mathrm{r=9-x}}}\\\\ \mathrm{r=a_3-a_2\ \to\ 9-x=5x+1-(x+2)}\\ \mathrm{9-x=5x+1-x-2\ \to\ 9-x=4x-1}\\ \mathrm{9+1=4x+x\ \to\ 5x=10\ \to\ \boxed{\mathrm{x=2}}}\\\\ \mathrm{PA=\{2.2-7,2+2,5.2+1\}=\{4-7,4,10+1\}}\\\\ \boxed{\mathbf{PA=\{-3,4,11\}}}$

Answer 2

(x + 2) - (2x - 7) = (5x + 1) - (x+2)

x + 2 - 2x + 7 = 5x + 1 - x - 2
x - 2x - 5x + x = +1 - 2 - 2 - 7 
-5x = -10 *(-1)
5x = 10
x = 10 / 5
x = 2


2x-7 = 2*2 - 7 = 4 - 7 = -3

x+2 = 2+2 = 4

5x+1 = 5*2 + 1 = 10+1 = 11

Resposta: Esta é uma P.A. de razão 7, em que os termos são: -3, 4 e 11.