JEE Main 2024 — Sequences and Series Question with Solution

Let, $a,ar,a{r}^2\to G.P.$

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

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

Case-I: $r^2-r-1<0$

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

$\Rightarrow r=\frac{1\pm \sqrt{5}}{2}$

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

Case-II: $r^2-r+1>0$ is always true.

Case-III: $r^2+r-1>0$

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

Taking intersection of $\left(i\right) \text{and} \left(ii\right)$

$\Rightarrow r\in \left(\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right)$

As $r>1$

$\Rightarrow r\in \left(1,\frac{1+\sqrt{5}}{2}\right)$

$\Rightarrow \left[r\right]=1 \text{and} \left[-r\right]=-2$

$\Rightarrow 3\left[r\right]+\left[-r\right]=3-2$

$\Rightarrow 3\left[r\right]+\left[-r\right]=1$