1. **State the problem:** We are given the 4th and 5th terms of a geometric progression (G.P.) as $-13.5$ and $40.5$ respectively, and we need to find the first term $a$. 2. **Recall the formula for the $n$th term of a G.P.:** $$T_n = a r^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. 3. **Write equations for the 4th and 5th terms:** $$T_4 = a r^{3} = -13.5$$ $$T_5 = a r^{4} = 40.5$$ 4. **Divide the equation for $T_5$ by $T_4$ to find $r$:** $$\frac{T_5}{T_4} = \frac{a r^{4}}{a r^{3}} = r = \frac{40.5}{-13.5} = -3$$ 5. **Substitute $r = -3$ back into the equation for $T_4$ to find $a$:** $$a (-3)^3 = -13.5$$ $$a (-27) = -13.5$$ $$a = \frac{-13.5}{-27} = 0.5$$ 6. **Final answer:** The first term of the G.P. is $\boxed{0.5}$.