If $\displaystyle \frac{^{n}P _{r-1}}{a}=\frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$,then which of the following holds goodÂ
$c^{2}=a(b+c)$
$a^{2}=c(a+b)$
$b^{2}=a(b+c)$
$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$\displaystyle \frac{^{n}p _{r-1}}{a}=\frac{^{n}P _{r}}{b}$$\displaystyle \Rightarrow n-r=\frac{b}{a}-1$and $\displaystyle \frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$$\displaystyle \Rightarrow n-r=\frac{c}{b}$On dividing (i) and (ii) we get$b^{2}=a(b+c)$