A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places,

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,

$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n}=5\left[\mathrm{~T}_{1}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n-1}\right]$

$\Rightarrow \mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n}-5\left[\mathrm{~T}_{1}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n-1}\right]=0$

$\Rightarrow \mathrm{T}_{2}+\mathrm{T}_{4}+\ldots+\mathrm{T}_{2 n}=4\left[\mathrm{~T}_{1}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n-1}\right]$

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

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

$\Rightarrow a r=4 a$

$\Rightarrow r=4$

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