Is it an equilateral triangle?

Let $b = ar, c = ar^2$ be in G.P. Also let:

$log(\frac{3b}{5c}) = log(\frac{5c}{a}) + d$ (i)

$log(\frac{a}{3b}) = log(\frac{5c}{a}) + 2d$ (ii)

be in A.P. Substitution of (i) into (ii) (for the common difference $d$ ) gives: $log(\frac{a}{3b}) = log(\frac{5c}{a}) + 2[log(\frac{3b}{5c}) - log(\frac{5c}{a})] \Rightarrow log(\frac{a}{3b}) = 2log(\frac{3b}{5c}) - log(\frac{5c}{a})$ . Ultimately we arrive at the equation $\frac{a}{3b} = \frac{9b^2}{25c^2} \cdot \frac{a}{5c} \Rightarrow 125c^3 = 27b^3 \Rightarrow c = \frac{3b}{5} \Rightarrow ar^2 = \frac{3}{5}ar \Rightarrow r = \frac{3}{5}.$

We now have $a, \frac{3}{5}a, \frac{9}{25}a$ as the candidate sides of a triangle; however, $\frac{3}{5}a + \frac{9}{25}a < a$ (which fails the Triangle Inequality). No such triangle exists under these conditions.