Question

8- Se (2x, x+1, 3x) é uma P.A. Qual o valor de r?

Answer 1

Explicação passo-a-passo:

Numa $\sf PA(a_1,a_2,a_3)$, temos que:

$\sf 2\cdot a_2=a_1+a_3$

$\sf PA(2x,x+1,3x)$

$\sf 2\cdot(x+1)=2x+3x$

$\sf 2x+2=5x$

$\sf 5x=2x+2$

$\sf 5x-2x=2$

$\sf 3x=2$

$\sf x=\dfrac{2}{3}$

Assim:

$\sf a_1=2x~\longrightarrow~a_1=2\cdot\dfrac{2}{3}~\longrightarrow~a_1=\dfrac{4}{3}$

$\sf a_2=x+1~\longrightarrow~a_2=\dfrac{2}{3}+1~\longrightarrow~a_2=\dfrac{5}{3}$

$\sf a_3=3x~\longrightarrow~a_3=3\cdot\dfrac{2}{3}~\longrightarrow~a_3=\dfrac{6}{3}$

$\sf PA\left(\dfrac{4}{3},\dfrac{5}{3},\dfrac{6}{3}\right)$

A razão dessa PA é:

$\sf r=a_2-a_1$

$\sf r=\dfrac{5}{3}-\dfrac{4}{3}$

$\sf r=\dfrac{1}{3}$