A geometric progression has consecutive terms 3^(2x+1), 9^x, and 81. Find x, the common ratio, and the sum of the first 4 terms.

a) The first three consecutive terms of a geometric progression (GP) are $3^{2x+1}$, $9^x$, and 81.
In a GP, the ratio of consecutive terms is constant.
So, we can write:

$$\frac{9^x}{3^{2x+1}} = \frac{81}{9^x}$$

To solve for $x$, we can express all terms with a base of 3:
$9 = 3^2$ and $81 = 3^4$.
Substitute these into the equation:

$$\frac{(3^2)^x}{3^{2x+1}} = \frac{3^4}{(3^2)^x}$$ $$\frac{3^{2x}}{3^{2x+1}} = \frac{3^4}{3^{2x}}$$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$$3^{2x - (2x+1)} = 3^{4 - 2x}$$ $$3^{-1} = 3^{4 - 2x}$$

Since the bases are equal, the exponents must be equal:

$$-1 = 4 - 2x$$ $$2x = 4 + 1$$ $$2x = 5$$ $$x = \frac{5}{2}$$ $$x = 2.5$$

The value of $x$ is $\boxed{2.5}$.

b) To find the common ratio ($r$) of the series, we can use the value of $x$ found in part (a).
The terms are $3^{2x+1}$, $9^x$, 81.
Substitute $x = 2.5$:
First term ($T_1$): $3^{2(2.5)+1} = 3^{5+1} = 3^6 = 729$
Second term ($T_2$): $9^{2.5} = (3^2)^{2.5} = 3^5 = 243$
Third term ($T_3$): 81

The common ratio $r$ is $\frac{T_2}{T_1}$ or $\frac{T_3}{T_2}$.

$$r = \frac{243}{729} = \frac{1}{3}$$

Alternatively,

$$r = \frac{81}{243} = \frac{1}{3}$$

The common ratio of the series is $\boxed{\frac{1}{3}}$.

c) To calculate the sum of the first 4 terms ($S_4$) of the series, we use the formula for the sum of the first $n$ terms of a GP: $S_n = \frac{a(1-r^n)}{1-r}$, where $a$ is the first term and $r$ is the common ratio.
From parts (a) and (b):
First term, $a = 729$
Common ratio, $r = \frac{1}{3}$
Number of terms, $n = 4$

Substitute these values into the formula:

$$S_4 = \frac{729\left(1 - \left(\frac{1}{3}\right)^4\right)}{1 - \frac{1}{3}}$$ $$S_4 = \frac{729\left(1 - \frac{1}{81}\right)}{\frac{2}{3}}$$ $$S_4 = \frac{729\left(\frac{81-1}{81}\right)}{\frac{2}{3}}$$ $$S_4 = \frac{729\left(\frac{80}{81}\right)}{\frac{2}{3}}$$ $$S_4 = \frac{9 \times 80}{\frac{2}{3}}$$ $$S_4 = \frac{720}{\frac{2}{3}}$$ $$S_4 = 720 \times \frac{3}{2}$$ $$S_4 = 360 \times 3$$ $$S_4 = 1080$$

The sum of the first 4 terms of the series is $\boxed{1080}$.

d) We are given that the fifth and the seventh terms of the GP form the first two consecutive terms of an arithmetic sequence (AP).
First, find the fifth term ($T_5$) and the seventh term ($T_7$) of the GP. The general term of a GP is $T_n = ar^{n-1}$.
$a = 729$ and $r = \frac{1}{3}$.

Fifth term of GP ($T_5$):

$$T_5 = ar^{5-1} = ar^4 = 729 \times \left(\frac{1}{3}\right)^4$$ $$T_5 = 729 \times \frac{1}{81} = 9$$

Seventh term of GP ($T_7$):

$$T_7 = ar^{7-1} = ar^6 = 729 \times \left(\frac{1}{3}\right)^6$$ $$T_7 = 729 \times \frac{1}{729} = 1$$

These terms, $T_5=9$ and $T_7=1$, are the first two consecutive terms of an arithmetic sequence.
Let the first term of the AP be $A_1 = 9$.
Let the second term of the AP be $A_2 = 1$.

Now, find the common difference ($d$) of this arithmetic sequence:

$$d = A_2 - A_1 = 1 - 9 = -8$$

We need to calculate the sum of the first 20 terms ($S_{20}$) of this arithmetic sequence.
The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(2A_1 + (n-1)d)$.
Here, $n=20$, $A_1=9$, and $d=-8$.

Substitute these values into the formula:

$$S_{20} = \frac{20}{2}(2(9) + (20-1)(-8))$$ $$S_{20} = 10(18 + (19)(-8))$$ $$S_{20} = 10(18 - 152)$$ $$S_{20} = 10(-134)$$ $$S_{20} = -1340$$

The sum of the first 20 terms of the arithmetic sequence is $\boxed{-1340}$.