Geometric Progression?

Let a and r be the first term and the common ratio respecitvely.
Then the first four terms are a, ar, ar² and ar³.

ar³ = 1 …… [1]
ar - ar² = 2 …… [2]

[2]/[1] :
(ar - ar²)/ar³ = 2
(1 - r)/r² = 2
1 - r = 2r²
2r² + r - 1 = 0
(2r - 1)(r + 1) = 0
r = 1/2 or r = -1 (rejected for r > 0)

Common ratio = 1/2

I'll call the 2nd, 3rd, and 4th terms by the names x, y, and z.
You know that z/y = y/x and x - y = 2 and z = 1. And hence,
1/y = y/(y+2) =>
y + 2 = y^2 =>
0 = y^2 - y - 2 =>
(y-2)(y+1) = 0 => y = 2 or -1.
In the first case, the x, y, and z are 4, 2, and 1, and the common ratio is 1/2.
In the 2nd case, the x, y, and z would be 1, -1, and 1, but this case is excluded because the problem says to assume that the common ratio is positive.

Answer: 1/2.

Let "a" be the initial term and "r" be the common ratio.

First term: a
Second term: ar
Third term: ar^2
Fourth term: ar^3.

However, we also know that ar^3 = 1, and ar = ar^2 + 2.
We use this to solve for a and r.

Since a = 1/r^3 from the first equation, substituting this into the second equation yields
(1/r^3) * r = (1/r^3) * r^2 + 2
==> 1/r^2 = 1/r + 2.

Multiply both sides by r^2:
1 = r + 2r^2
==> 2r^2 + r - 1 = 0
==> (2r - 1)(r + 1) = 0
==> r = 1/2, since we are given that r > 0.

I hope this helps!

#1 = a
#2 = ar = 2 + ar^2
#3 = ar^2
#4 = ar^3 = 1
#n = ar^(n-1)

Hence dividing
ar^3/ar = 1 / ( 2+ ar^2)
r^2 = 1/(2 + ar^2)
2r^2 - ar^4 = 1
Complete the Square for 'r^2'
ar^4 - 2r^2 = 1
r^4 - (2/a)r^2 = 1/a
(r^2 - 1/a)^2 - (1/a)^2 = 1/a
(r^2 - 1/a)^2 = 1/a + (1/a)^2
r^2 - 1/a = +/-sqrt(1/a + 1/a)^2)
r^2 = 1/a +/- sqrt[1/a + (1/a)^2]
r = +/- sqrt{ 1/a +/- sqrt[1/a + (1/a)^2]}