已知 $a$，$b$，$c$ 是一个等比数列的前三项。同时也已知 $a$，$c$，$b$ 是一个等差数列的前三项。

1. Derive the relationship from the geometric progression.

Since $a$, $b$, $c$ are the first three terms of a geometric progression, the ratio between consecutive terms is constant:
$\dfrac{b}{a} = \dfrac{c}{b}$
$b^2 = ac \quad \text{(Shown)}$

2. Derive the relationship from the arithmetic progression.

Since $a$, $c$, $b$ are the first three terms of an arithmetic progression, the difference between consecutive terms is constant:
$c - a = b - c$
$2c = a + b$
$c = \dfrac{a + b}{2} \quad \text{(Shown)}$

1. Substitute the expression for $c$ from the arithmetic progression into the geometric progression equation.

Substitute $c = \dfrac{a+b}{2}$ into $b^2 = ac$:
$b^2 = a\left(\frac{a+b}{2}\right)$

2. Rearrange the equation and divide by $a^2$ to obtain the required quadratic form in terms of $(b/a)$.

$2b^2 = a^2 + ab$
$2b^2 - ab - a^2 = 0$
Divide the entire equation by $a^2$ (since for a valid geometric progression, $a \neq 0$):
$\frac{2b^2}{a^2} - \frac{ab}{a^2} - \frac{a^2}{a^2} = 0$
$2\left(\frac{b}{a}\right)^2 - \left(\frac{b}{a}\right) - 1 = 0 \quad \text{(shown)}$

1. Find the possible values of the common ratio $r = \frac{b}{a}$.

Let $r = \frac{b}{a}$ be the common ratio of the geometric progression.
From (ii), we have the equation $2r^2 - r - 1 = 0$.
Factorising the quadratic equation gives:
$(2r + 1)(r - 1) = 0$
$r = -\frac{1}{2}$ or $r = 1$

2. Select the valid common ratio based on the condition for sum to infinity.

For an infinite geometric progression to converge and have a sum to infinity, its common ratio must satisfy $|r| < 1$.
Therefore, we reject $r = 1$ and accept $r = -\frac{1}{2}$.

3. Calculate the sum to infinity in terms of $a$.

The sum to infinity is given by $S = \frac{a}{1 - r}$.
Substitute $r = -\frac{1}{2}$ into the formula:
$S = \frac{a}{1 - \left(-\frac{1}{2}\right)}$
$S = \frac{a}{\frac{3}{2}}$
$S = \frac{2a}{3}$