GP

Let the required numbers be $\left(\dfrac ar , a, ar\right)$ where $a$ is the second term of this G.P. and $r$ is the common ratio.

From the problem,

$\begin{aligned} & \dfrac ar \cdot a \cdot ar = 27 \\ \implies & a^3 = 27 \\ \implies & a = 3 \end{aligned}$

And,

$\begin{aligned} & \dfrac ar + a + ar = 9 \\ \implies & 3 \left( \dfrac 1r + 1 + r \right) = 9 \\ \implies & \dfrac{1+r+r^2}{r} = 3 \\ \implies & r^2 - 2r + 1 = 0 \\ \implies & r=1 \end{aligned}$

Thus, the first term of the G.P. is, $\dfrac ar = \dfrac 31 = \boxed{3}$

Note:

$a^3 - 27 = 0$ will have three values for $a$ of which the only real value is $a=3$ . Since, most problems in progressions deal with only the real values therefore only the real values have been considered in this problem.