Discussion
If \(\frac{{}^{n}P_{r-1}}{a} = \frac{{}^{n}P_{r}}{b} = \frac{{}^{n}P_{r+1}}{c}\) , then
Answer: C
The relation is \(b^{2} = a(b + c)\).
Use \({}^{n}P_{r} = {}^{n}P_{r-1} \times (n - r + 1)\) and \({}^{n}P_{r+1} = {}^{n}P_{r} \times (n - r)\). From the equal ratios,
\[ \begin{aligned} \frac{b}{a} &= \frac{{}^{n}P_{r}}{{}^{n}P_{r-1}} = n - r + 1 \\ \frac{c}{b} &= \frac{{}^{n}P_{r+1}}{{}^{n}P_{r}} = n - r \end{aligned} \]
Subtracting, \(\frac{b}{a} - \frac{c}{b} = 1\). Multiply through by \(ab\): \(b^{2} - ac = ab\), so \(b^{2} = ab + ac = a(b + c)\).
A numerical check: with \(n = 5\), \(r = 2\) the permutations are 5, 20 and 60, and \(20^{2} = 400 = 5(20 + 60)\), while for example \(c^{2} = 3600\) is not \(b(a + c) = 1300\). The other options fail this check.
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