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

Let, $a, ar, a{r}^{2} \to G . P .$

It is given that sum of any two sides is greater than the third side.

$⇒ a + ar > a{r}^{2}, a + a{r}^{2} > ar, ar + a{r}^{2} > a$

Case-I: ${r}^{2} - r - 1 < 0$

$⇒ r = \dfrac{1 \pm \sqrt{1 - 4 \times 1 \times \left(- 1\right)}}{2}$

$⇒ r = \dfrac{1 \pm \sqrt{5}}{2}$

$⇒ r \in \left(\dfrac{1 - \sqrt{5}}{2}, \dfrac{1 + \sqrt{5}}{2}\right) . . . \left(i\right)$

Case-II: ${r}^{2} - r + 1 > 0$ is always true.

Case-III: ${r}^{2} + r - 1 > 0$

$⇒ r \in \left(- \infty, \dfrac{- 1 - \sqrt{5}}{2}\right) \cup \left(\dfrac{- 1 + \sqrt{5}}{2}, \infty\right) . . . \left(ii\right)$

Taking intersection of \left(i\right)\text{and}\left(ii\right)$</p><p>$⇒ r \in \left(\dfrac{- 1 + \sqrt{5}}{2}, \dfrac{1 + \sqrt{5}}{2}\right)$</p><p>As$r > 1$</p><p>$⇒ r \in \left(1, \dfrac{1 + \sqrt{5}}{2}\right)$</p><p>$⇒ \left[r\right] = 1\text{and}\left[- r\right] = - 2$</p><p>$⇒ 3\left[r\right] + \left[- r\right] = 3 - 2$</p><p>$⇒ 3\left[r\right] + \left[- r\right] = 1$