In an art lesson, wires are cut and bent to create a series of shapes.

1. Identify the arithmetic progression and state the sum formula.

Let $n$ be the number of squares.
The perimeters of the squares form an arithmetic progression with first term $a = 5$ and common difference $d = 3$.
The total length of wire used for $n$ squares is $S_n = \frac{n}{2}[2a + (n-1)d]$.

2. Formulate the inequality for the maximum number of squares.

Since the total wire length available is 100 cm, $S_n \le 100$.
$\frac{n}{2}[2(5) + (n-1)(3)] \le 100$
$\frac{n}{2}(10 + 3n - 3) \le 100$
$n(3n + 7) \le 200$
$3n^2 + 7n - 200 \le 0$

3. Solve the quadratic inequality to find the maximum integer $n$.

Solving the equation $3n^2 + 7n - 200 = 0$ using the quadratic formula:
$n = \frac{-7 \pm \sqrt{7^2 - 4(3)(-200)}}{2(3)}$
$n = \frac{-7 \pm \sqrt{2449}}{6}$
$n \approx 7.081 \text{ or } n \approx -9.414$
For $3n^2 + 7n - 200 \le 0$, we have $-9.414 \le n \le 7.081$.
Since $n$ must be a positive integer, the maximum value of $n$ is $7$.

1. Identify the circumferences as a geometric progression and find the first term and common ratio.

The circumferences form a geometric progression.
First term, $a = 2\pi x$
Common ratio, $r = \frac{2}{3}$

2. Set up the equation for the total length of the wire using the sum of the first 12 terms.

$S_{12} = \frac{a(1 - r^{12})}{1 - r}$
$\frac{2\pi x \left( 1 - \left( \frac{2}{3} \right)^{12} \right)}{1 - \frac{2}{3}} = 100$

3. Simplify the equation to isolate $x$.

$\frac{2\pi x \left( 1 - \left( \frac{2}{3} \right)^{12} \right)}{\frac{1}{3}} = 100$
$6\pi x \left( 1 - \left( \frac{2}{3} \right)^{12} \right) = 100$

4. Solve for $x$ and give the answer to 3 significant figures.

$x = \frac{100}{6\pi \left( 1 - \left( \frac{2}{3} \right)^{12} \right)}$
$x = 5.3463... \approx 5.35$