If  nPr−1a= nPrb= nPr+1c , then which of the following holds good

# If $\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r-1}}{a}=\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r}}{b}=\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r+1}}{c}$, then which of the following holds good

1. A
${c}^{2}=a\left(b+c\right)$
2. B
${a}^{2}=c\left(a+b\right)$
3. C
${b}^{2}=a\left(b+c\right)$
4. D
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

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### Solution:

Given, $\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r-1}}{a}=\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r}}{b}=\frac{{\text{\hspace{0.17em}}}^{n}{P}_{r+1}}{c}$

$⇒\frac{n!}{a\left(n-r+1\right)!}=\frac{n!}{b\left(n-r\right)!}=\frac{n!}{c\left(n-r-1\right)!}$

$⇒\frac{1}{a\left(n-r+1\right)\left(n-r\right)\left(n-r-1\right)!}=\frac{1}{b\left(n-r\right)\left(n-r-1\right)!}=\frac{1}{c\left(n-r-1\right)!}$

$⇒\frac{1}{a\left(n-r+1\right)\left(n-r\right)}=\frac{1}{b\left(n-r\right)}=\frac{1}{c}$

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