Question

If $\frac{{}^2{}^{0}P{}_r{}_{-}{}_1}{a}=\frac{{}^2{}^{0}P{}_r}{3}=\frac{{}^2{}^{0}P{}_r{}_{+}{}_1}{4}$, then $a=$

• A. $\frac{9}{7}$
• B. $\frac{7}{9}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{9}{7}$

From the known result,

$\frac{n_{P_r}}{n_{P_{r-1}}}=n-r+1$

It can be found that if $\frac{n_{P_{r-1}}}{a}=\frac{n_{P_r}}{b}=\frac{n_{P_{r+1}}}{c}$ then,

$b^2=a(b+c)$

Given : $b=3,c=4$

$\Rightarrow 9=a\times 7$

$\Rightarrow a=\frac{9}{7}$