Question

If P(n,r-1)/a = P(n,r)/b = P(n,r+1)/c, prove that b^2 = a(b + c)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\red{\rm :\longmapsto\:\dfrac{P(n, r - 1)}{a} = \dfrac{P(n, r )}{b} = \dfrac{P(n, r + 1)}{c} }$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: P(n, r) = \frac{n!}{(n - r)!} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{n!}{a(n - r + 1)!} = \dfrac{n!}{b(n - r)!} = \dfrac{n!}{c(n - r - 1)!}$

can be rewritten as on cancelation n!,

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)!} = \dfrac{1}{b(n - r)!} = \dfrac{1}{c(n - r - 1)!}$

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)(n - r - 1)!} = \dfrac{1}{b(n - r)(n - r - 1)!} = \dfrac{1}{c(n - r - 1)!}$

can be further rewritten as on cancelation (n - r - 1)!,

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)} = \dfrac{1}{b(n - r)} = \dfrac{1}{c}$

So, Taking first and second member, we have

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)} = \dfrac{1}{b(n - r)}$

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)} = \dfrac{1}{b}$

$\bf :\longmapsto\:\dfrac{b}{a} = n - r + 1 - - - (1)$

Now, Taking second and third member, we get

$\rm :\longmapsto\:\dfrac{1}{b(n - r)} = \dfrac{1}{c}$

$\bf :\longmapsto\:\dfrac{c}{b} = n - r - - - (2)$

On Substituting equation (2) in equation (1), we get

$\bf :\longmapsto\:\dfrac{b}{a} = \dfrac{c}{b} + 1$

$\bf :\longmapsto\:\dfrac{b}{a} = \dfrac{c + b}{b}$

$\red{\rm \implies\:\boxed{ \tt{ \: {b}^{2} = a(b + c) \: }}}$

• Hence, Proved

Answer 2

_______________________

$\large\underline{\sf{Solution-}}$

Given expression is

$\red{\rm :\longmapsto\:\dfrac{P(n, r - 1)}{a} = \dfrac{P(n, r )}{b} = \dfrac{P(n, r + 1)}{c} }$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: P(n, r) = \frac{n!}{(n - r)!} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{n!}{a(n - r + 1)!} = \dfrac{n!}{b(n - r)!}$ $\rm :\longmapsto\:= \dfrac{n!}{c(n - r - 1)!}$

can be rewritten as on cancelation n!,

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)!} = \dfrac{1}{b(n - r)!}$ $\rm :\longmapsto\:= \dfrac{1}{c(n - r - 1)!}$

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)(n - r - 1)!}$$\rm :\longmapsto\: = \dfrac{1}{b(n - r)(n - r - 1)!}= \dfrac{1}{c(n - r - 1)!}$

can be further rewritten as on cancelation (n - r - 1)!,

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)} = \dfrac{1}{b(n - r)} = \dfrac{1}{c}$

So, Taking first and second member, we have

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)(n - r)} = \dfrac{1}{b(n - r)}$

$\rm :\longmapsto\:\dfrac{1}{a(n - r + 1)} = \dfrac{1}{b}$

$\bf :\longmapsto\:\dfrac{b}{a} = n - r + 1 - - - (1)$

Now, Taking second and third member, we get

$\rm :\longmapsto\:\dfrac{1}{b(n - r)} = \dfrac{1}{c}$

$\bf :\longmapsto\:\dfrac{c}{b} = n - r - - - (2)$

On Substituting equation (2) in equation (1), we get

$\bf :\longmapsto\:\dfrac{b}{a} = \dfrac{c}{b} + 1$

$\bf :\longmapsto\:\dfrac{b}{a} = \dfrac{c + b}{b}$

$\red{\rm \implies\:\boxed{ \tt{ \: {b}^{2} = a(b + c) \: }}}$

Hence, Proved

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