Question

If ${}^{K+5}{P}_{K+1}=\frac{11(K-1)}{2}.{}^{K+3}{P}_K$, then the values of K are

• A. 7 and 11
• B. 6 and 7
• C. 2 and 11
• D. 2 and 6

Answer 1

Answer: (B)

6 and 7

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

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

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

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

$k^2+5k+4k+20=22k-22$

$k^2+9k-22k+42=0$

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

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

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

(k-7)(k-6) = 0

k = 7,6