If three successive terms of a G.P. with common ratio r(r>1) are the length of the sides of a triangle and [r] denotes the greatest integer less than…

Let, $a,ar,a{r}^{2}\rightarrow G.P.$a,ar,ar2→G.P.

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

$$\Rightarrow a+ar>a{r}^{2},a+a{r}^{2}>ar,ar+a{r}^{2}>a$$⇒a+ar>ar2,a+ar2>ar,ar+ar2>a

Case-I: ${r}^{2}-r-1<0$r2−r−1<0

$$\Rightarrow r=\frac{1\pm \sqrt{1-4\times 1\times (-1)}}{2}$$⇒r=21±1−4×1×(−1)​​

$$\Rightarrow r=\frac{1\pm \sqrt{5}}{2}$$⇒r=21±5​​

$$\Rightarrow r\in (\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})...(i)$$⇒r∈(21−5​​,21+5​​)...(i)

Case-II: ${r}^{2}-r+1>0$r2−r+1>0 is always true.

Case-III: ${r}^{2}+r-1>0$r2+r−1>0

$$\Rightarrow r\in (-\infty ,\frac{-1-\sqrt{5}}{2})\cup (\frac{-1+\sqrt{5}}{2},\infty )...(ii)$$⇒r∈(−∞,2−1−5​​)∪(2−1+5​​,∞)...(ii)

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

$$\Rightarrow r\in (\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$$⇒r∈(2−1+5​​,21+5​​)

As $r>1$r>1

$$\Rightarrow r\in (1,\frac{1+\sqrt{5}}{2})$$⇒r∈(1,21+5​​)

$$\Rightarrow [r]=1\text{and}[-r]=-2$$⇒[r]=1and[−r]=−2

$$\Rightarrow 3[r]+[-r]=3-2$$⇒3[r]+[−r]=3−2

$$\Rightarrow 3[r]+[-r]=1$$⇒3[r]+[−r]=1