If a, b, c are in GP and loga - 2b, 2b - 3c and 3c - loga are in AP...

Given, $a, b, c$ are in $GP$.

$∴ b^{2} = ac$

and $2\left(\log 2b - \log 3c\right) = loga - \log 2b + \log 3c - loga$

$⇒ b^{2} = ac$ and $2b = 3c$

$∵ a + b = \dfrac{5a}{3} > c, b + c = \dfrac{10a}{9} > a$

and $c + a = \dfrac{13a}{9} > b$

$∴ a, b, c$ are the sides of a triangle.

Also, $a$ is the greatest side

$∴ cosA = \dfrac{b^{2} + c^{2} - a^{2}}{2bc} = - \dfrac{29}{48} < 0$

$∴ ∆ ABC$ is an obtuse angled triangle.

Hence, option (d) is correct.