To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:

$$ 1) \ a + ar > ar^2 \\\\ $$

$$2) \ a + ar^2 > ar $$

$$ 3) \ ar + ar^2 > a $$

Given that $r > 1$, inequalities 2 and 3 will always hold because:

$$ ar < ar^2 \ \text{and} \ a < ar, $$

indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:

$$ a + ar > ar^2 $$

Simplifying this, we get:

$$ a(1 + r) > a r^2 $$

Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:

$$ 1 + r > r^2 $$

We can then subtract $r$ from both sides:

$$ 1 > r^2 - r $$

Simplifying the right side by factoring $r$:

$$ 1 > r(r - 1) $$

Given that $r > 1$, the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \sqrt{2})$ because $r(r - 1)$ increases with increasing $r$, and it would be 1 when $r = \sqrt{2}$. It should be greater than 1, and less than $\sqrt{2}$ such that $r^2 - r$ stays below 1.

Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).

Since $1 < r < \sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.

For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r > 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).

Now we can substitute these values into the expression:

$$ 3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1 $$

Therefore $3[r] + [-r]$ is equal to 1.