If `\ ^(k+5)P_(k+1)=(11(k-1))/2dot\ ^(k+3)P_(k\ )P_k` then the values of `k` are `7\ a n d\ 11` b. `6\ a n d\ 7` c. `2\ a n d\ 11` d. `2\ a n d\ 6`

Explanation

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

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

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

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

$$\Rightarrow(k+5)(k+4)=22(k-1)$$

$$\Rightarrow k^{2}+9 k+20=22 k-22$$

$$\Rightarrow k^{2}-13 k+42=0$$
$$\Rightarrow k=6,7$$