Discussion

If k, 6 and k+5 are the first three terms of a geometric series, then the possible values of the common ratio are

A.\(\frac{1}{2}, 2\)
B.2,-2
C.\(\frac{1}{3}, 3\)
D.\(\frac{3}{2}, \frac{-2}{3}\)
E.\(\frac{2}{3}, \frac{-3}{2}\)

Answer: D

The possible ratios are \(\frac{3}{2}\) and \(-\frac{2}{3}\).

For three terms of a G.P., the middle term squared equals the product of the outer two.

\[ \begin{aligned} 36 &= k(k + 5) \\ k^{2} + 5k - 36 &= 0 \\ (k + 9)(k - 4) &= 0 \end{aligned} \]

If k = 4, the terms are 4, 6, 9 and r = \(\frac{6}{4} = \frac{3}{2}\). If k = -9, the terms are -9, 6, -4 and r = \(\frac{6}{-9} = -\frac{2}{3}\).

Option E inverts both ratios, and the others do not satisfy \(6^{2} = k(k + 5)\) for any k.

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