A geometric series has common ratio $r$, and an arithmetic series has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. The first three terms of the geometric series are equal to the first, fourth and sixth terms respectively of the arithmetic series.

1. Express the first three terms of both the arithmetic and geometric series.

Let the first three terms of the geometric series be $G_1$, $G_2$, and $G_3$.
Since the first term of the arithmetic series is $a$, $G_1 = a$.
Then, $G_2 = ar$ and $G_3 = ar^2$.

Let the terms of the arithmetic series be $T_1$, $T_4$, and $T_6$.
$T_1 = a$
$T_4 = a + 3d$
$T_6 = a + 5d$

2. Equate the corresponding terms to form simultaneous equations.

Since the first three terms of the geometric series are equal to the first, fourth and sixth terms of the arithmetic series respectively:
$ar = a + 3d \quad \text{--- (1)}$
$ar^2 = a + 5d \quad \text{--- (2)}$

3. Express $d$ in terms of $a$ and $r$ using the first equation.

From equation (1):
$3d = ar - a$
$d = \frac{a(r - 1)}{3}$

4. Substitute the expression for $d$ into the second equation and rearrange to obtain the required quadratic equation.

Substitute $d$ into equation (2):
$ar^2 = a + 5\left( \frac{a(r - 1)}{3} \right)$
Multiply both sides by 3 to clear the fraction:
$3ar^2 = 3a + 5a(r - 1)$
$3ar^2 = 3a + 5ar - 5a$
$3ar^2 = 5ar - 2a$
Since $a$ is non-zero ($a \neq 0$), we can divide the entire equation by $a$:
$3r^2 = 5r - 2$
$3r^2 - 5r + 2 = 0 \quad \text{(Shown)}$

1. Solve the quadratic equation from part (i) to find the possible values of $r$.

$3r^2 - 5r + 2 = 0$
$(3r - 2)(r - 1) = 0$
$r = \frac{2}{3} \text{ or } r = 1$

2. Reject the invalid value of $r$ and deduce that the geometric series converges.

If $r = 1$, the first three terms of the geometric series are all $a$.
Then the 4th term of the arithmetic series is $a + 3d = a \implies d = 0$.
Since the question states $d$ is non-zero, $r = 1$ is rejected.
Thus, $r = \frac{2}{3}$.
Since $|r| = \left|\frac{2}{3}\right| < 1$, the geometric series is convergent.

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

$S_\infty = \frac{a}{1 - r}$
$S_\infty = \frac{a}{1 - \frac{2}{3}}$
$S_\infty = 3a$

1. Find the common difference $d$ in terms of $a$ using the common ratio $r$ obtained from the earlier parts.

From part (ii), the geometric series converges, so the common ratio is $r = \frac{2}{3}$.
The fourth term of the arithmetic series is equal to the second term of the geometric series:
$a + 3d = ar$
$3d = a\left(\frac{2}{3}\right) - a$
$3d = -\frac{1}{3}a \implies d = -\frac{1}{9}a$

2. Express the sum of the first $n$ terms, $S$, in terms of $n$ and $a$.

The sum of the first $n$ terms of an arithmetic series is given by:
$S = \frac{n}{2}[2a + (n-1)d]$
Substitute $d = -\frac{1}{9}a$:
$S = \frac{n}{2}\left[2a - \frac{a(n-1)}{9}\right]$

3. Formulate the inequality $S > 4a$ and simplify it to a quadratic inequality.

Given that $S > 4a$, we have:
$\frac{n}{2}\left[2a - \frac{a(n-1)}{9}\right] > 4a$
Since $a > 0$, we can divide both sides by $a$ without reversing the inequality sign:
$\frac{n}{2}\left[2 - \frac{n-1}{9}\right] > 4$
$n\left[\frac{18 - (n-1)}{9}\right] > 8$
$n(19 - n) > 72$
$19n - n^2 > 72$
$n^2 - 19n + 72 < 0$

4. Solve the quadratic inequality to find the range of possible values for $n$.

First, find the roots of the corresponding quadratic equation $n^2 - 19n + 72 = 0$:
$n = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(1)(72)}}{2(1)}$
$n = \frac{19 \pm \sqrt{361 - 288}}{2} = \frac{19 \pm \sqrt{73}}{2}$
Calculating the approximate values, we get $n \approx 5.23$ and $n \approx 13.77$.
Since the quadratic curve opens upwards and we require it to be less than zero, $n$ must lie between the roots:
$5.23 < n < 13.77$

5. Determine the set of possible integer values for $n$.

Since $n$ represents the number of terms, it must be a positive integer.
The integers strictly between $5.23$ and $13.77$ are $6, 7, 8, 9, 10, 11, 12$, and $13$.
The set of possible values of $n$ is $\{6, 7, 8, 9, 10, 11, 12, 13\}$.