IRL ABooklet

A square of side length $x$ has perimeter $4x$ and area $x^2$. If the preceding term of the sequence is $T_{n-1}$, then its successor is $T_n = (T_{n-1}/4)^2$. Thus the first three terms are
$$a, \quad \frac{a^2}{16} = \frac{a^2}{2^4} \quad \text{and} \quad \frac{a^4}{4096} = \frac{a^4}{2^{12}}.$$
Since the terms are in arithmetic progression, their common difference may be computed in two ways
$$a - \frac{a^2}{2^4} = \frac{a^2}{2^4} - \frac{a^4}{2^{12}}$$
leading to
$$\frac{a^4}{2^{12}} - 2\frac{a^2}{2^4} + a = 0, \quad \text{i.e.} \quad a\left(\left(\frac{a}{2^4}\right)^3 - 2\frac{a}{2^4} + 1\right) = 0.$$
Letting $y = a/2^4$, this equation can be rewritten as $y^3 - 2y + 1 = 0$, since $a > 0$. Because
$$y^3 - 2y + 1 = (y - 1)(y^2 + y - 1) = (y - 1)\left(y + \frac{1 - \sqrt{5}}{2}\right)\left(y + \frac{1 + \sqrt{5}}{2}\right),$$
the positive solutions correspond to $y = 1$ and $y = (\sqrt{5} - 1)/2$. Using $a = 2^4 y$ we deduce that the positive values of the initial terms are
$$a = 2^4 = 16 \quad \text{or} \quad a = 8(\sqrt{5} - 1).$$