iiser problem

A.an acute angled triangle

how can you say that????

how to solve it????

$Let $\mathbf{A=ln(a)-ln(2b),A+D=ln(2b)-ln(3c)}$ and $\mathbf{A+2D=ln(3c)-ln(a)}$\\\\ Now $\mathbf{3(A+D)=ln(a)-ln(2b)+ln(2b)-ln(3c)+ln(3c)-ln(a)=0}$\\\\ So $\mathbf{3.(A+D)=0\Leftrightarrow A+D=0\Leftrightarrow ln(2b)-ln(3c)=0\Leftrightarrow ln\left(\frac{2b}{3c}\right)=0\Leftrightarrow \frac{2b}{3c}=1}$\\\\ So $\boxed{\mathbf{c=\frac{2}{3}b}}$ and $\mathbf{b^2=ac}$(bcz $\mathbf{a,b,c}$ are in $\mathbf{G.P}$)\\\\\\ So $\boxed{\mathbf{b=\frac{2}{3}a}}$ and $\boxed{\mathbf{c=\frac{4}{9}a}}$\\\\\\ So $\mathbf{a>b>c}$\\\\\\ and $\mathbf{b^2+c^2=\left(\frac{2}{3}\right)^2.a^2+\left(\frac{2}{3}\right)^4.a^2=\frac{52}{81}a^2<a^2}$\\\\ So $\mathbf{b^2+c^2-a^2<0}$\\\\ Now Using Cosine formula $\mathbf{cos\;A=\frac{b^2+c^2-a^2}{2.b.c}<0}$\\\\ So $\mathbf{A}$ is an \underline{\underline{Obtuse angle}}.