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

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.

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

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

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

$\Rightarrow n - r = \frac{a}{b}$

Similarly $n - r + 1 = \frac{b}{c}$

$\begin{array}{ll} ∴ \frac{b}{c} - \frac{a}{b} = 1 \\ \Rightarrow b^{2} - a c = b c \Rightarrow b^{2} - (a + b) c = 0 . \end{array}$

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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