A geometric sequence has first term a = 3/128 and sixth term T6 = 3/4. Find the common ratio and the sum of the first 20 terms.

The first term is $a = \frac{3}{128}$.
The sixth term is $T_6 = \frac{3}{4}$.
The formula for the $n$-th term of a geometric sequence is $T_n = a \cdot r^{n-1}$.

Divide both sides by $\frac{3}{128}$:

Take the fifth root of both sides:

The common ratio is $\boxed{2}$.

2.1.2 The sum of the first 20 terms, correct to the nearest integer.

First term $a = \frac{3}{128}$.
Common ratio $r = 2$ (from 2.1.1).
Number of terms $n = 20$.
The formula for the sum of the first $n$ terms of a geometric sequence is $S_n = \frac{a(r^n - 1)}{r - 1}$.

The sum of the first 20 terms is $\boxed{24,576}$.

2.2 Prove that $a = \frac{\sqrt{3}(\sqrt{5}-1)}{2}$.

Given that $a, b, \sqrt{3}$ are three consecutive terms of a geometric sequence, the common ratio is constant.

Given that $a, b, \sqrt{3}$ are the sides of a right-angled triangle, and $a < b < \sqrt{3}$, then $\sqrt{3}$ must be the hypotenuse.
By the Pythagorean theorem:

Rearrange into a quadratic equation:

Here, $A=1$, $B=\sqrt{3}$, $C=-3$.

Since $a$ represents a side length, it must be positive.

The other solution, $a = \frac{-\sqrt{3} - \sqrt{15}}{2}$, is negative and thus not valid for a side length.

Factor out $\sqrt{3}$ from the numerator:

This matches the required expression.

We have $a = \frac{\sqrt{3}(\sqrt{5}-1)}{2}$.
From $b^2 = a\sqrt{3}$:

Since $\sqrt{5} \approx 2.236$, $\sqrt{3} \approx 1.732$.
$a \approx \frac{1.732(2.236-1)}{2} = \frac{1.732(1.236)}{2} \approx 1.070$
$b^2 \approx \frac{3(2.236-1)}{2} = \frac{3(1.236)}{2} \approx 1.854 \implies b \approx \sqrt{1.854} \approx 1.361$
The condition $a < b < \sqrt{3}$ becomes $1.070 < 1.361 < 1.732$, which is true.

Thus, we have proven that $a = \boxed{\frac{\sqrt{3}(\sqrt{5}-1)}{2}}$.