If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio
\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\)
\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)
\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)
\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)
\(k^{2} - 8k = 0\)
\(k(k - 8) = 0\)
\(\therefore \text{k = 0 or 8}\)
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 \(\frac{5}{3}\).
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