Question

If ⁿC₄ , ⁿC₅ and ⁿC₆ are in A.P., then find n.

Answer 1

Since ⁿC₄ , ⁿC₅ and ⁿC₆ are in AP.

∴ 2. ⁿC₅ = ⁿC₄ + ⁿC₆ 

\[\Rightarrow 2 \times \frac{n!}{5!\left( n - 5 \right)!} = \frac{n!}{4!\left( n - 4 \right)!} + \frac{n!}{6!\left( n - 6 \right)!}\]
\[ \Rightarrow \frac{2}{5 \times 4!\left( n - 5 \right)\left( n - 6 \right)!} = \frac{1}{4!\left( n - 5 \right)\left( n - 4 \right)\left( n - 6 \right)!} + \frac{1}{6 \times 5 \times 4!\left( n - 6 \right)!}\]
\[ \Rightarrow \frac{2}{5\left( n - 5 \right)} = \frac{1}{\left( n - 5 \right)\left( n - 4 \right)} + \frac{1}{30}\]
\[ \Rightarrow \frac{2}{5\left( n - 5 \right)} - \frac{1}{\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{30}\]
\[ \Rightarrow \frac{2n - 8 - 5}{5\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{30}\]
\[ \Rightarrow \frac{2n - 13}{\left( n - 5 \right)\left( n - 4 \right)} = \frac{1}{6}\]
\[ \Rightarrow 12n - 78 = n^2 - 9n + 20\]
\[ \Rightarrow n^2 - 21n + 98 = 0\]
\[ \Rightarrow \left( n - 7 \right)\left( n - 14 \right) = 0\]
\[ \therefore n = 7 \text{and} 14\]