Prove the following


If $\mathrm{n}_{4}, \mathrm{n}_{5}$ and $\mathrm{nC}_{6}$ are in A.P., then $\mathrm{n}$ can be:

  1. 14

  2. 11

  3. 9

  4. 12

Correct Option: 1


$2 \cdot \mathrm{n}_{5}=\mathrm{n}_{4}+\mathrm{n}_{6}$

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.

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