Since, a, $2 \mathrm{a}+2$ and $3 \mathrm{a}+3$ are in GP $\therefore(2 a+2)^{2}=a(3 a+3)$
$\Rightarrow 4 \mathrm{a}^{2}+4+8 \mathrm{a}=3 \mathrm{a}^{2}+3 \mathrm{a} \Rightarrow \mathrm{a}^{2}+5 \mathrm{a}+4=0$
$\Rightarrow a(a+4)+1(a+4)=0 \Rightarrow(a+4)(a+1)=0$
$\Rightarrow a+4=0$ or $a+1=0$
$\Rightarrow a=-4$ or $-1$
Let the fourth term be $\mathrm{x}$.
$\frac{a}{2 a+2}=\frac{3 a+3}{x}$
$\Rightarrow x=\frac{(3 a+3)(2 a+2)}{a}$
When $\mathrm{a}=-4$
$\mathrm{x}=-13.5$
and $\mathrm{a}=-1, \mathrm{x}=0$
So, the fourth term is $-13.5$