Find x such that x-4, x, 3x-8 are three consecutive terms in a geometric sequence.

Question

this is for precalulus honors and if you can show me how you get the answer would be great 

Answer 1

the i'th term of a geometric sequence can be represented as

 

a_i=a₀rⁱ

 

The ratio of one term to the previous term is

 

(a_i=a₀rⁱ)/(a_(i-1)=a₀r^(i-1))=r

 

Therefore

r=(3x-8)/x=x/(x-4)

 

x²=(x-4)(3x-8)=3x²-8x-12x+32

 

2x²-20x+32=0

 

x²-10x+16=0

 

(x-8)(x-2)=0

 

x=2 and x=8

 

Answer 2

x/(x-4) = (3x-8)/x

 

x^2=(x-4)(3x-8)

 

     =3x^2-20x+32

 

2x^2-20x+32=0

 

x^2-10x+16=0

 

(x-8)(x-2)=0

 

x=2  or. x=8