Geometric Progression

Question

Three consecutive terms of a geometric progression are x-1, x²-1, and x+1. How do I find two possible values of x?

Answer 1

Set of proportion using the ratios of the terms:

(x² - 1) / (x - 1) = (x + 1) / (x² - 1)

Solve for x and answer

Answer 2

x-1, x^2 -1, x+1, ...are 3 terms of a geometic sequence

an=a1r^(n-1)= (x-1)r^(n-1)

a3 = (x-1)r^2 =x+1

a2 = (x-1)r=x^2-1, r=x+1

x-1=1/r

x^2-1=1

x=+/-sqr2,

r =1/(x-1)= 1/(+/sqr2 -1)= about 1/.414,-1/2.414

what are 2 possible values of x? x =sqr2,-sqr2

sequence is

about .414, 1, 2.414,...

or

about -2.414, 1,-.414,...