Question

Determine x real de modo que 2x-3, 2x e 2x+1 estejam em PG

Answer 1

$\Large\boxed{\begin{array}{l}\sf\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}\implies a_2^2=a_1\cdot a_3\\\\\sf (2x-3,2x,2x+1)\\\sf (2x)^2=(2x-3)\cdot(2x+1)\\\sf 4x^2=4x^2+2x-6x-3\\\sf 4x^2-4x^2+6x-2x=-3\\\sf 4x=-3\\\sf x=-\dfrac{3}{4}\blue{\checkmark}\end{array}}$