Question:
If $\mathrm{n}_{4}, \mathrm{n}_{5}$ and $\mathrm{nC}_{6}$ are in A.P., then $\mathrm{n}$ can be:
Correct Option: 1
Solution:
$2 \cdot \mathrm{n}_{5}=\mathrm{n}_{4}+\mathrm{n}_{6}$
2. $\frac{\lfloor n}{|5| n-5}=\frac{\mid n}{|4| n-4}+\frac{\lfloor n}{|6| n-6}$
$\frac{2}{5} \cdot \frac{1}{n-5}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$
$\mathrm{n}=14$ satisfying equation.