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

Let the geometric progression be $a, ar, ar^2, ar^3, \dots, ar^{2n-1}$, where the number of terms is $2n$.

The sum of all terms is $S = a + ar + ar^2 + \dots + ar^{2n-1} = a \dfrac{r^{2n} - 1}{r - 1}$.

The terms occupying odd places are $a, ar^2, ar^4, \dots, ar^{2n-2}$. This is a geometric progression with first term $a$, common ratio $r^2$, and $n$ terms.

The sum of terms at odd places is $S_{odd} = a + ar^2 + ar^4 + \dots + ar^{2n-2} = a \dfrac{(r^2)^n - 1}{r^2 - 1} = a \dfrac{r^{2n} - 1}{r^2 - 1}$.

Given that $S = 5 S_{odd}$, we have $a \dfrac{r^{2n} - 1}{r - 1} = 5 a \dfrac{r^{2n} - 1}{r^2 - 1}$.

Assuming $r \neq 1$ and $r^{2n} \neq 1$, we can cancel $a(r^{2n} - 1)$ from both sides:

$\dfrac{1}{r - 1} = \dfrac{5}{r^2 - 1}$.

Since $r^2 - 1 = (r - 1)(r + 1)$, the equation becomes $\dfrac{1}{r - 1} = \dfrac{5}{(r - 1)(r + 1)}$.

Multiplying both sides by $(r - 1)$, we get $1 = \dfrac{5}{r + 1}$.

$r + 1 = 5$, which implies $r = 4$.

Answer: 4