If three successive terms of a G.P. with common ratio $\mathrm{r}(\mathrm{r}>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to _____________.

<p>To solve this problem, let&#39;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 $r &gt; 1$. These three terms represent the lengths of the sides of a triangle.</p> <p>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:</p> <p>$$ 1) \ a + ar &gt; ar^2 \\\\ $$ <br/><br/>$2) \ a + ar^2 &gt; ar$ <br/><br/>$3) \ ar + ar^2 &gt; a$</p> <p>Given that $r &gt; 1$, inequalities 2 and 3 will always hold because:</p> <p>$ar &lt; ar^2 \ \text{and} \ a &lt; ar,$</p> <p>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:</p> <p>$a + ar &gt; ar^2$</p> <p>Simplifying this, we get:</p> <p>$a(1 + r) &gt; a r^2$</p> <p>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:</p> <p>$1 + r &gt; r^2$</p> <p>We can then subtract $r$ from both sides:</p> <p>$1 &gt; r^2 - r$</p> <p>Simplifying the right side by factoring $r$:</p> <p>$1 &gt; r(r - 1)$</p> <p>Given that $r &gt; 1$, 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.</p> <p>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).</p> <p>Since $1 &lt; r &lt; \sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.</p> <p>For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r &gt; 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).</p> <p>Now we can substitute these values into the expression:</p> <p>$3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1$</p> <p>Therefore $3[r] + [-r]$ is equal to 1.</p>