For three positive integers p , q , r , x p q 2 = y q r = z p 2 r and r = p q + 1 such that 3 , 3 log y x , 3 log z y , 7 log x z are in A.P. with common difference 1 2 . The r - p - q is equal to

Step-by-step solution

Let x p q 2 = y q r = z p 2 r = λ Then, p q 2 = log x λ q r = log y λ p 2 r = log z λ So, log y λ log x λ = q r p q 2 ⇒ log y x = r p q       . . . 1 Similarly, log x z = p q 2 p 2 r = q 2 p r       . . . . 2 log z y = p 2 r q r = p 2 q       . . . 3 So, 3 ,   3 log y x ,   3   log z y ,   7 log x z     are in A.P., hence 3 , 3 r p q , 3 p 2 q , 7 q 2 p r are in A.P. Therefore, 3 r p q - 3 = 1 2 r = 7 6 p