Question

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

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

Answer 1

The correct option is C $b^2=a(b+c)$

$\frac{{}^n{p}_{r-1}}{a}=\frac{{}^n{P}_r}{b}$
$\Rightarrow n-r=\frac{b}{a}-1$
and $\frac{{}^n{P}_r}{b}=\frac{{}^n{P}_{r+1}}{c}$
$\Rightarrow n-r=\frac{c}{b}$
On dividing (i) and (ii) we get
$b^2=a(b+c)$