MCQ
If $^n{C_4},{\,^n}{C_5},$ and ${\,^n}{C_6},$ are in $A.P.,$ then $n$ can be
- A$9$
- ✓$14$
- C$11$
- D$12$
$2.\frac{{\left| n \right.}}{{\left| {5\left| {n - 5} \right.} \right.}} = \frac{{\left| n \right.}}{{\left| {4\left| {n - 4} \right.} \right.}} + \frac{{\left| n \right.}}{{\left| {6\left| {n - 6} \right.} \right.}}$
$\frac{2}{5}.\frac{1}{{n - 5}} = \frac{1}{{\left( {n - 4} \right)\left( {n - 5} \right)}} + \frac{1}{{30}}$
$n=14$ satisfying equation.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.