Question

If ${}^{k+5}{P}_{k+1}=\frac{11(k-1)}{2}{,}^{k+3}{P}_k$ then the value of $k=....$.

Answer 1

${}^{(k+5)}{P}_{(k+1)}=\frac{11(k-1)}{2}{.}^{(k+3)}{P}_{\left(k\right)}$

$\frac{(k+5)!}{(k+5)-(k+1)!}=\frac{11(k-1)}{2}\times \frac{(k+3)!}{(k+3-k)!}$

$\frac{(k+5)(k+4)(k+3)!}{4!}=\frac{11(k-1)}{2}\frac{(k+3)!}{3!}$

$k^2+9k+20=22k-22$

$k^2-13k+42=0$

$k^2-6k-7k+42=0$

$\to k(k-6)-7(k-6)=0$

$\to k(k-6)(k-7)=0$

$\Rightarrow k=6$ or 7

$\therefore$ The value of k can be 6 or 7