If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is

Step-by-step solution

Let us consider four positive consecutive terms of a G.P. as follow: a , a r , a r 2 , a r 3 a , r > 0 Product = a 4 r 6 = 1296 ⇒ a 2 r 3 = 36 ⇒ a = 6 r 3 / 2 Sum = a + a r + a r 2 + a r 3 = 126 ⇒ 1 r 3 / 2 + r r 3 / 2 + r 2 r 3 / 2 + r 3 r 3 / 2 = 126 6 ⇒ r - 3 / 2 + r 3 / 2 + r 1 / 2 + r - 1 / 2 = 21 Let r 1 / 2 + r - 1 / 2 = A ⇒ r - 3 / 2 + r 3 / 2 = ( r 1 2 + r - 1 2 ) 3 - 3 ( r 1 2 + r - 1 2 ) = A 3 - 3 A Therefore, A 3 - 3 A + A = 21 ⇒ A 3 - 2 A = 21 ⇒ A = 3 &#875