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r = \frac{1 + \frac{1}{2} \cdot \frac{2\sqrt{7}}{3}}{1 - \frac{3}{2} \cdot \frac{2\sqrt{7}}{3}} = \frac{1 + \frac{\sqrt{7}}{3}}{1 - \sqrt{7}}.
As before, this evaluates to $ -\frac{5 + 2\sqrt{7}}{9} $, but this is negative, which contradicts the premise unless the terms are not ordered. However, the ratio of third to first term is geometric in symmetric AP.
Alternatively, let the first term be $ A $, common difference $ d $, so terms: $ A, A+d, A+2d, A+3d $. Sum of squares of first and last:

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