A sequence whose first term is positive has the property that any given term is the area of an equilateral triangle whose perimeter is the preceding term. If the first three terms form an arithmetic progression, determine all possible values of the first term.
1. Let the first term of the sequence be \( x \).
2. The perimeter of the equilateral triangle corresponding to the first term is \( x \). Therefore, the side length of the triangle is \( \frac{x}{3} \).
3. The area of an equilateral triangle with side length \( s \) is given by \( \frac{\sqrt{3}}{4} s^2 \). Hence, the second term of the sequence is:
   \[
   \frac{\sqrt{3}}{4} \left( \frac{x}{3} \right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{x^2}{9} = \frac{\sqrt{3}}{36} x^2
   \]
4. Similarly, the third term of the sequence is the area of an equilateral triangle whose perimeter is the second term. The side length of this triangle is \( \frac{\frac{\sqrt{3}}{36} x^2}{3} = \frac{\sqrt{3}}{108} x^2 \). Therefore, the third term is:
   \[
   \frac{\sqrt{3}}{4} \left( \frac{\sqrt{3}}{108} x^2 \right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{3 x^4}{108^2} = \frac{3 \sqrt{3} x^4}{4 \cdot 108^2} = \frac{3 \sqrt{3} x^4}{4 \cdot 11664} = \frac{\sqrt{3} x^4}{15552}
   \]
5. Given that the first three terms form an arithmetic progression, we have:
   \[
   2 \left( \frac{\sqrt{3}}{36} x^2 \right) = x + \frac{\sqrt{3} x^4}{15552}
   \]
   Simplifying this equation:
   \[
   \frac{2 \sqrt{3}}{36} x^2 = x + \frac{\sqrt{3} x^4}{15552}
   \]
   \[
   \frac{\sqrt{3}}{18} x^2 = x + \frac{\sqrt{3} x^4}{15552}
   \]
   Multiplying through by 15552 to clear the fraction:
   \[
   864 \sqrt{3} x^2 = 15552 x + \sqrt{3} x^4
   \]
   \[
   \sqrt{3} x^4 - 864 \sqrt{3} x^2 + 15552 x = 0
   \]
   Dividing through by \( \sqrt{3} \):
   \[
   x^4 - 864 x^2 + 5184 x = 0
   \]
   Factoring out \( x \):
   \[
   x (x^3 - 864 x + 5184) = 0
   \]
   Since \( x > 0 \), we discard \( x = 0 \) and solve the cubic equation:
   \[
   x^3 - 864 x + 5184 = 0
   \]
6. Let \( t = \frac{\sqrt{3}}{18} x \). Then:
   \[
   t^3 - 8t + 8 = 0
   \]
   Checking for roots, we find \( t = 2 \) is a root. Factoring the cubic equation:
   \[
   (t - 2)(t^2 + 2t - 4) = 0
   \]
   Solving the quadratic equation \( t^2 + 2t - 4 = 0 \):
   \[
   t = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}
   \]
   Since \( t > 0 \), we have \( t = -1 + \sqrt{5} \).

7. Thus, the possible values of \( t \) are \( t = 2 \) and \( t = -1 + \sqrt{5} \). Converting back to \( x \):
   \[
   \frac{\sqrt{3}}{18} x = 2 \implies x = \frac{36}{\sqrt{3}} = 12\sqrt{3}
   \]
   \[
   \frac{\sqrt{3}}{18} x = -1 + \sqrt{5} \implies x = 18 \cdot \frac{-1 + \sqrt{5}}{\sqrt{3}} = 6\sqrt{15} - 6\sqrt{3}
   \]

The final answer is \( \boxed{ x = 12\sqrt{3} } \) or \( x = 6\sqrt{15} - 6\sqrt{3} \).