Question

If ^k + 5P_{k + 1} =\[\frac{11 (k - 1)}{2}\]. ^k + 3P_k , then the values of k are

विकल्प

• 7 and 11

• 6 and 7

• 2 and 11

• 2 and 6

Answer 1

 6 and 7

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

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