Let the $G.P.$ be $T_{1}, T_{2}, T_{3}, T_{4} \ldots . T_{2 n}$

Number of terms $=2 n$

According to the given condition,

$T_{1}+T_{2}+T_{3}+\ldots .+T_{2 n}=5\left[T_{1}+T_{3}+\ldots .+T_{2 n-1}\right]$

$\Rightarrow T_{1}+T_{2}+T_{3}+\ldots .+T_{2 n}-5\left[T_{1}+T_{3}+\ldots . .+T_{2 n-1}\right]=0$

$\Rightarrow T_{2}+T_{4}+\ldots .+T_{2 n}=4\left[T_{1}+T_{3}+\ldots . .+T_{2 n-1}\right]$

Let the $G.P.$ be $a, a r, a r^{2}, a r^{3} \dots$

$\therefore \frac{\operatorname{ar}\left(r^{n}-1\right)}{r-1}=\frac{4 \times a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow a r=4 a$

$\Rightarrow r=4$

Thus, the common ratio of the $G.P.$ is $4$