If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio
A. 0, 8 B. -1, \(\frac{5}{3}\) C. 2, 3 D. 1, -1
Correct Answer: B
Explanation
\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\) \((k + 1)(3k + 1) = (2k - 1)(2k - 1)\) \(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\) \(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\) \(k^{2} - 8k = 0\) \(k(k - 8) = 0\) \(\therefore \text{k = 0 or 8}\) The terms of the sequence given k = 0: (1, -1, 1) \(\implies \text{The common ratio r = -1}\) The terms of the sequence given k = 8: (9, 15, 25) \(\implies \text{The common ratio r = } \frac{5}{3}\) The possible values of the common ratio are -1 and \(\frac{5}{3}\).