Let the first term a and the common ratio r of a geometric progression be positive integers with r > 1 . The sum of the first three terms of this geometric progression is 211 . If m arithmetic means are inserted between a and r such that the sum of these inserted means is 120 , then the value of m is

Step-by-step solution

The sum of the first three terms of the geometric progression is given by a + ar + ar^2 = 211 a(1 + r + r^2) = 211 Since r is a positive integer and r > 1 , we have r 2 , which implies 1 + r + r^2 7 . The number 211 is a prime number. Therefore, its only positive factors are 1 and 211 . This forces a = 1 and 1 + r + r^2 = 211 . Solving the quadratic equation: r^2 + r - 210 = 0 (r + 15)(r - 14) = 0 Since r > 0 , we get r = 14 . Now, m arithmetic means are inserted between a = 1 and r = 14 . The sum of m arithmetic m