Mathematics · Sequence and Series

JEE Main 2024 — 1 February, Shift 2 — Question 21

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

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

a, ar, ar2→\mathrm{ar}^{2} \rightarrow G.P.

Sum of any two sides >> third side

a+ar>ar2,a+ar2>ar,ar+ar2>aa+a r>a r^{2}, a+a r^{2}>a r, a r+a r^{2}>a

r2−r−1<0\mathrm{r}^{2}-\mathrm{r}-1<0

r∈(1−52,1+52)\mathrm{r} \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)

r2r+1>0\mathrm{r}^{2}\mathrm{r}+1>0

always true r2+r−1>0r∈(−∞,−152)∪(−1+52,∞)\begin{aligned} & \mathrm{r}^{2}+\mathrm{r}-1>0 &\mathrm{r} \in\left(-\infty, \frac{-1\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right) \end{aligned}

Taking intersection of (1), (2)

r∈(−1+52,1+52)r \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)

As r>1\mathrm{r}>1

r∈(1,1+52)\mathrm{r} \in\left(1, \frac{1+\sqrt{5}}{2}\right)

[r]=1[−r]=−2[\mathrm{r}]=1[-\mathrm{r}]=-2

3[r]+[−r]=13[\mathrm{r}]+[-\mathrm{r}]=1

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression