It is given that $x$, $y$, $z$ are the first three terms of a geometric progression. When the three terms are arranged in the order of $z$, $x$, $y$, they form three consecutive terms of an arithmetic progression.

1. Formulate an equation from the given arithmetic progression $z, x, y$.

Since $z$, $x$, $y$ are consecutive terms of an arithmetic progression, the common difference is constant:
$x - z = y - x$
$2x = y + z$

2. Formulate an equation from the given geometric progression $x, y, z$.

Since $x$, $y$, $z$ are consecutive terms of a geometric progression, the common ratio is constant:
$\frac{y}{x} = \frac{z}{y}$
$y^2 = xz$

3. Substitute the expression for $x$ from the AP equation into the GP equation.

From the AP equation, $x = \frac{y + z}{2}$. Substituting this into the GP equation gives:
$y^2 = \left(\frac{y + z}{2}\right)z$
$2y^2 = yz + z^2$

4. Divide by $y^2$ to obtain the required equation.

Dividing the entire equation by $y^2$ (assuming $y \neq 0$):
$2 = \frac{yz}{y^2} + \frac{z^2}{y^2}$
$2 = \frac{z}{y} + \left(\frac{z}{y}\right)^2$
Rearranging the terms yields:
$\left(\frac{z}{y}\right)^2 + \left(\frac{z}{y}\right) - 2 = 0 \quad \text{(Shown)}$

1. Identify the common ratio from the given equation and solve for its possible values.

Let $r$ be the common ratio of the geometric progression.
Since $x, y, z$ are the first three terms of the geometric progression, we have $r = \frac{y}{x} = \frac{z}{y}$.

Substituting $r = \frac{z}{y}$ into the equation from (i):
$r^2 + r - 2 = 0$
$(r+2)(r-1) = 0$
$r = -2 \quad \text{or} \quad r = 1$

2. Check the values of the common ratio against the condition for the convergence of a geometric series.

For the sum to infinity of a geometric progression to exist, the common ratio $r$ must satisfy the condition $|r| < 1$ (i.e., $-1 < r < 1$).

Since $|-2| = 2 \ge 1$ and $|1| = 1 \ge 1$, neither possible value of $r$ satisfies this condition.
Therefore, the sum to infinity does not exist.