Question

If $\frac{nP_{r - 1}}{a} = \frac{nP_{r}}{b} = \frac{nP_{r + 1}}{c}$ then

• A. ab, b, ac are in A.P.
• B. ab, b, ac are in G.P.
• C. b² = a(b + c)
• D. None of these

Answer 1

Answer: (C)

b² = a(b + c)

Answer 2

$\frac{nP_{r - 1}}{a} = \frac{nP_{r}}{b} = \frac{nP_{r + 1}}{c}$

$\frac{1}{a}.\frac{n!}{(n - r + 1)!} = \frac{1}{b}.\frac{n!}{(n - r)!} = \frac{1}{c}\frac{n!}{(n - r - 1)!}$from first two

$\frac{b}{a}$ = n – r + 1

From last two $\frac{c}{b}$ = n – r

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

b² = ab + ac