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 or equal to r, then 3 r + − r is equal to:

Step-by-step solution

Let, a , a r , a r 2 → G . P . It is given that sum of any two sides is greater than the third side. ⇒ a + a r > a r 2 , a + a r 2 > a r , a r + a r 2 > a Case-I: r 2 − r − 1 < 0 ⇒ r = 1 ± 1 - 4 × 1 × - 1 2 ⇒ r = 1 ± 5 2 ⇒ r ∈ 1 − 5 2 , 1 + 5 2 . . . i Case-II: r 2 − r + 1 > 0 is always true. Case-III: r 2 + r − 1 > 0 ⇒ r ∈ − ∞ , − 1 − 5 2 ∪ − 1 + 5 2 , ∞ . . . i i Taking intersection of i and i i ⇒ r ∈ − 1 + 5 2 , 1 + 5 2 As r > 1 ⇒ r ∈ 1 , 1 + 5 2 ⇒ r = 1 and − r = − 2 ⇒ 3 r + − r = 3 - 2 ⇒ 3 r + − r = 1