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If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then which of the following holds good

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c^{2} = a (b + c)

a^{2} = c (a + b)

b^{2} = a (b + c)

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

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detailed solution

Correct option is C

Given,

\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}

$\Rightarrow \frac{n!}{a (n - r + 1)!} = \frac{n!}{b (n - r)!} = \frac{n!}{c (n - r - 1)!}$

$\Rightarrow \frac{1}{a (n - r + 1) (n - r) (n - r - 1)!} = \frac{1}{b (n - r) (n - r - 1)!} = \frac{1}{c (n - r - 1)!}$

$\Rightarrow \frac{1}{a (n - r + 1) (n - r)} = \frac{1}{b (n - r)} = \frac{1}{c}$

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If nPr−1a= nPrb= nPr+1c , then which of the following holds good

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If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then which of the following holds good