If S is the sum to infinity of a decreasing geometric progression with common ratio x such that |x| <1 ; x 0 . The ratio of fourth term to the second term is 1 16 and the ratio of the third term to the square of the second term is 1 9 . Then the value of S is

Step-by-step solution

Let the first term of the geometric progression be a and the common ratio be x . The terms are a, ax, ax^2, ax^3, . Given the ratio of the fourth term to the second term is 1 16 , we have ax^3 ax = x^2 = 1 16 . Since the progression is decreasing and |x| Given the ratio of the third term to the square of the second term is 1 9 , we have ax^2 (ax)^2 = ax^2 a^2x^2 = 1 a = 1 9 . Thus, a = 9 . The sum to infinity S of a geometric progression is given by S = a 1 - x . Substituting the values a = 9 and x = 1 4 , we get S