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$If three successive terms of a G.P with common ratio r (r > 1), form the sides$ $of a triangle ABC and [r] denotes greater integer function then | [r] + [- r] | is$

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Solution

$a + a r > a r^{2} (\sin c e s u m o f t w o s i d e s o f a t r i a n g l e i s g r e a t e r t h a n t h e t h i r d s i d e)$
$\begin{array}{l} \to r^{2} - r - 1 < 0 \\ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \\ \Rightarrow 1 < r < \frac{1 + \sqrt{5}}{2} < 2 (since r > 1) \\ \Rightarrow [r] = 1 \\ Also - \frac{(1 + \sqrt{5})}{2} < - r < - 1 \to [- r] = - 2 \\ ∴ | [r] + [- r] | = = 1 \end{array}$

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Geometric progression

Remember concepts with our Masterclasses.

If three successive terms of a G.P with common ratio r(r>1),form the sides of a triangle ABC and [r] denotes greater integer function then |[r]+[−r]| is

a+ar>ar2 since sum of two sides of a triangle is greater than the third side→r2−r−1<01−52<r<1+52⇒1<r<1+52<2 since r>1⇒[r]=1 Also −(1+5)2<−r<−1→[−r]=−2∴|[r]+[−r]|==1

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$If three successive terms of a G.P with common ratio r (r > 1), form the sides$ $of a triangle ABC and [r] denotes greater integer function then | [r] + [- r] | is$