If three successive terms of a G.P.

Question

If three successive terms of a G.P. with common ratio $r(r>1)$ 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 \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

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 $a{r}^{2}$, 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:

$1)a+ar>a{r}^{2}\, \, \, \,$

$2)a+a{r}^{2}>ar$

$3)ar+a{r}^{2}>a$

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

$arindicating that both $a+a{r}\^{2}$ and $ar+a{r}\^{2}$ 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:

$a+ar>a{r}\^{2}$

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 $1For $[-r]$, 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.

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