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If $\frac{1}{a} (^{n} P_{r + 1}) = \frac{1}{b} (^{n} P_{r}) = \frac{1}{c} (^{n} P_{r - 1})$ then $b^{2} - (a + b) c$ is equal to

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Correct option is A

nPr+1 nPr=ab⇒(n−r)!(n−r−1)!=ab⇒ n−r=abSimilarly n−r+1=bc∴ bc−ab=1⇒ b2−ac=bc⇒b2−(a+b)c=0.

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If 1a nPr+1=1b nPr=1c nPr−1 then b2−(a+b)c is equal to

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If $\frac{1}{a} (^{n} P_{r + 1}) = \frac{1}{b} (^{n} P_{r}) = \frac{1}{c} (^{n} P_{r - 1})$ then $b^{2} - (a + b) c$ is equal to