An equilateral triangle of side length $x$ has perimeter $3x$ and area $\sqrt{3}x^2/4$. If the preceding term of the sequence is $T_{n-1}$, then its successor is
$$T_n = \frac{\sqrt{3}(T_{n-1}/3)^2}{4} = \frac{T_{n-1}^2 \sqrt{3}}{36}.$$
Thus the first three terms are
$$a, \frac{a^2\sqrt{3}}{36}, \frac{a^4\sqrt{3}}{15552}, \quad \text{or} \quad a, \frac{a^2}{2^{23/3}}, \frac{a^4}{2^{63/2}}.$$
Since the terms are in arithmetic progression, their common difference may be computed in two ways
$$a - \frac{a^2}{2^2 3^{3/2}} = \frac{a^2}{2^2 3^{3/2}} - \frac{a^4}{2^6 3^{9/2}},$$
leading to
$$\frac{a^4}{2^6 3^{9/2}} - 2 \frac{a^2}{2^2 3^{3/2}} + a = 0, \quad \text{i.e.} \quad a \left( \left( \frac{a}{2^2 3^{3/2}} \right)^3 - 2 \frac{a}{2^2 3^{3/2}} + 1 \right) = 0.$$
Letting $y = a/(2^2 3^{3/2})$, 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 $a = 2^2 3^{3/2}y$ we deduce that the positive values of the initial terms are
$$a = 2^2 3^{3/2} = 4\sqrt{27} = 12\sqrt{3} \quad \text{or} \quad a = 6\sqrt{3} (\sqrt{5} - 1).$$