First slide

Permutations

Question

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

Moderate

A

0
B

1
C

-1
D

-2

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App