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

(b) 6 and 7

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

$$\begin{gathered} \Rightarrow \frac{\left(k+5\right)!}{\left(k+5-k-1\right)!}=\frac{11\left(k-1\right)}{2}\times \frac{\left(k+3\right)!}{\left(k+3-k\right)!} \\ \Rightarrow \frac{\left(k+5\right)!}{4!}=\frac{11\left(k-1\right)}{2}\times \frac{\left(k+3\right)!}{3!} \\ \Rightarrow \frac{\left(k+5\right)!}{\left(k+3\right)!}=\frac{11\left(k-1\right)}{2}\times \frac{4!}{3!} \\ \Rightarrow \left(k+5\right)\left(k+4\right)=22\left(k-1\right) \\ \Rightarrow k^2+9k+20=22k-22 \\ \Rightarrow k^2-13k+42=0 \\ \Rightarrow k=6,7 \end{gathered}$$