Given that ${\text{a}}_1,{\text{a}}_2,{\text{a}}_3$ is an A.P in that order, ${\text{a}}_1+{\text{a}}_2+{\text{a}}_3=15;{\text{b}}_1,{\text{ b}}_2,{\text{ b}}_3$ is a G.P. in that order, and ${\text{b}}_1{\text{ b}}_2{\text{ b}}_3=27.$ If ${\text{a}}_1+{\text{b}}_1,{\text{a}}_2+{\text{b}}_2,{\text{a}}_3+{\text{b}}_3$ are positive integers and form a G.P. in that order, then

Let ${\text{a}}_1=5-\text{d},{\text{a}}_2=5,{\text{a}}_3=5+\text{d},{\text{b}}_1=\frac{3}{\text{q}}, {\text{b}}_2=3,{\text{ b}}_3=3\text{q,}$ then the given conditions indicate that $5-d+\frac{3}{q}$ and $5+d+3q$ are all positive integers and $(5-d+\frac{3}{q})(5+d+3q)=64.$

It is easy to check that for getting maximum positive d, there are only four possibilities for $(5-d+\frac{3}{q},5+d+3q):(1,64),(2,32),(4,16)$ and (8, 8).
(i) By solving the system of equations corresponding to $(1,64),$ it follows that

$\begin{array}{rr} & 3\text{q}+\frac{3}{\text{q}}=55\Rightarrow {\text{q}}_{1,2}=\frac{55\pm 7\sqrt{61}}{6},{\text{ d}}_{1,2}=4+\frac{3}{{\text{q}}_{1,2}}\\ & \Rightarrow {\text{d}}_{\text{max}}=4+\frac{3.6}{55-7\sqrt{61}}=4+\frac{55+7\sqrt{61}}{2}\end{array}$

(ii) By solving the system of equations corresponding to $(2,32),$ it follows that

$\begin{array}{rr} & q+\frac{1}{q}=8\Rightarrow q_{1,2}=\frac{8\pm \sqrt{60}}{2}=4\pm \sqrt{15},d_{1,2}=3+\frac{3}{q_{1,2}}\\ & \Rightarrow d_{\text{max}}=3+\frac{3}{4-\sqrt{15}}=15+3\sqrt{15}.\end{array}$

(iii) By solving the system of equation corresponding to $(4,16),$ it follows that

$3\text{q}+\frac{3}{\text{q}}=10\Rightarrow \text{q}=3\text{ or }\frac{1}{3},{\text{ d}}_{1,2}=1+\frac{3}{{\text{q}}_{1,2}}=2$ or $10\Rightarrow {\text{d}}_{\text{max}}=10$  
(iv) By solving the system of equations corresponding to $(8,8),$ it follows that $\text{q}+\frac{1}{\text{q}}=2\Rightarrow \text{q}=1,\text{ d}=0.$
In summary, $d_{\text{max}}=4+\frac{55+7\sqrt{61}}{2},$ hence the maximum possible value of $a_3$ is $9+\frac{55+7\sqrt{61}}{2}$