An arithmetic series has positive first term $a$ and common difference $d$. The first, fifth and second terms of the arithmetic series are the first three consecutive terms of a geometric series respectively. It is also given that the terms of the arithmetic series are decreasing. It is given that the sum to infinity of the geometric series is $\dfrac{32}{7}$.

1. Express the first three terms of the geometric series using the $n$th term formula for an arithmetic series.

Let the first three terms of the geometric series be $g_1$, $g_2$, and $g_3$.
$g_1 = a$
$g_2 = a + 4d$
$g_3 = a + d$

2. Equate the common ratios of the geometric series to form an equation in terms of $a$ and $d$.

For a geometric series, the ratio between consecutive terms is constant:
$\frac{g_2}{g_1} = \frac{g_3}{g_2}$
$\frac{a + 4d}{a} = \frac{a + d}{a + 4d}$
$(a + 4d)^2 = a(a + d)$
$a^2 + 8ad + 16d^2 = a^2 + ad$
$16d^2 + 7ad = 0$

3. Solve the quadratic equation for $d$ and reject the invalid root based on the given conditions.

$d(16d + 7a) = 0$
$d = 0 \quad \text{or} \quad 16d + 7a = 0$
Since the terms of the arithmetic series are decreasing, the common difference $d < 0$.
Given that the first term $a$ is positive, $d = 0$ is rejected.
$16d = -7a$
$d = -\frac{7}{16}a$

1. Calculate the common ratio $r$ of the geometric series using the expression for the second term divided by the first term.

The common ratio $r$ is the ratio of the second term of the geometric series to the first term.
$r = \frac{T_5}{T_1} = \frac{a + 4d}{a}$
Substitute the result $d = -\frac{7}{16}a$ from part (a):
$r = \frac{a + 4\left(-\frac{7}{16}a\right)}{a}$
$r = 1 - \frac{7}{4} = -\frac{3}{4}$

2. State the condition for convergence to justify that the series converges.

For an infinite geometric series to converge, its common ratio must satisfy $|r| < 1$.
$|r| = \left|-\frac{3}{4}\right| = \frac{3}{4}$
Since $\frac{3}{4} < 1$, the geometric series converges.

1. Find the common ratio $r$ of the geometric series.

The first and second terms of the geometric series correspond to the first and fifth terms of the arithmetic series.
$g_1 = a$
$g_2 = a + 4d$
Using $d = -\frac{7}{16}a$ from part (a):
$r = \frac{a + 4d}{a} = \frac{a + 4(-\frac{7}{16}a)}{a} = \frac{a - \frac{7}{4}a}{a} = -\frac{3}{4}$

2. Use the sum to infinity of the geometric series to find the first term $a$.

$S_\infty = \frac{a}{1 - r} = \frac{32}{7}$
$\frac{a}{1 - (-\frac{3}{4})} = \frac{32}{7}$
$\frac{a}{\frac{7}{4}} = \frac{32}{7}$
$a = \frac{32}{7} \times \frac{7}{4} = 8$

3. Find the common difference $d$ and the sum of the first 20 terms of the arithmetic series.

$d = -\frac{7}{16}(8) = -\frac{7}{2}$
For the arithmetic series:
$S_{20} = \frac{20}{2}[2a + (20 - 1)d]$
$S_{20} = 10\left[2(8) + 19\left(-\frac{7}{2}\right)\right]$
$S_{20} = 10(16 - 66.5) = -505$

1. Identify the common ratio $r$ and first term $a$ of the geometric series from earlier parts.

From part (a), the first two terms of the geometric series are $T_1 = a$ and $T_2 = a+4d$.
With $d = -\frac{7}{16}a$, the common ratio $r$ is:
$r = \frac{a+4(-\frac{7}{16}a)}{a} = 1 - \frac{7}{4} = -0.75$
Given $S_\infty = \frac{32}{7}$, we find $a$:
$\frac{a}{1 - (-0.75)} = \frac{32}{7} \implies a = \frac{32}{7} \times 1.75 = 8$

2. Formulate an inequality expressing that the sum after the $n$th term is greater than $2\%$ of the sum of the first $n$ terms.

The sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r} = \frac{32}{7}(1 - (-0.75)^n)$.
The sum of all terms after the $n$th term is $S_\infty - S_n = \frac{32}{7} - \frac{32}{7}(1 - (-0.75)^n) = \frac{32}{7}(-0.75)^n$.
The given condition $S_\infty - S_n > 0.02 S_n$ translates to:
$\frac{32}{7}(-0.75)^n > 0.02 \left( \frac{32}{7}(1 - (-0.75)^n) \right)$

3. Simplify the inequality to isolate the exponential term.

Dividing both sides by $\frac{32}{7}$ (since it is positive):
$(-0.75)^n > 0.02(1 - (-0.75)^n)$
$(-0.75)^n > 0.02 - 0.02(-0.75)^n$
$1.02(-0.75)^n > 0.02$
$(-0.75)^n > \frac{0.02}{1.02} = \frac{1}{51}$

4. Solve for the possible values of $n$, noting the constraint on the parity of $n$.

Since $\frac{1}{51} > 0$, $(-0.75)^n$ must be strictly positive. This requires $n$ to be an even integer.
For even $n$, we have $(-0.75)^n = 0.75^n$:
$0.75^n > \frac{1}{51}$
Taking the natural logarithm on both sides:
$n \ln 0.75 > \ln \left(\frac{1}{51}\right)$
Since $\ln 0.75 < 0$, dividing by it flips the inequality sign:
$n < \frac{\ln(1/51)}{\ln 0.75} \approx 13.67$
Since $n$ is a positive even integer ($n \ge 1$), the possible values are:
$n = 2, 4, 6, 8, 10, 12$