If it is given that the terms \[^{n}{{C}_{4}}{{,}^{n}}{{C}_{5}}\,and{{\,}^{n}}{{C}_{6}}\] are in A.P. then the value of n can be?
(a) 14
(b) 11
(c) 9
(d) 12

Hint: We know that if it is given that there are three terms in A.P. then the sum of the extreme terms will be equal to two times the value of the middle term. It is also known as arithmetic mean. So we have terms \[^{n}{{C}_{4}}{{,}^{n}}{{C}_{5}}\,and{{\,}^{n}}{{C}_{6}}\] in A.P. so according to above definition of arithmetic mean we get,
${{2.}^{n}}{{C}_{5}}{{=}^{n}}{{C}_{4}}{{+}^{n}}{{C}_{6}}$ now solving this expression further we will get the value of n. Also use the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .

Complete step by step answer:
We are given that the terms \[^{n}{{C}_{4}}{{,}^{n}}{{C}_{5}}\,and{{\,}^{n}}{{C}_{6}}\] are in A.P. and,
We know that if three terms are given in the A.P. then according to the definition of arithmetic mean sum of extreme terms would be equal to two times the middle term.
Hence we get,
${{2.}^{n}}{{C}_{5}}{{=}^{n}}{{C}_{4}}{{+}^{n}}{{C}_{6}}$
Now we know that the value of $^{n}{{C}_{r}}$ is given by,
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Now using this in the obtained equation and expanding the equation, we get
${{2.}^{n}}{{C}_{5}}{{=}^{n}}{{C}_{4}}{{+}^{n}}{{C}_{6}}$
$2\times \dfrac{n!}{5!\left( n-5 \right)!}=\dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{n!}{6!\left( n-6 \right)!}$
Cancelling out $n!$ from both sides, we get
$2\times \dfrac{1}{5!\left( n-5 \right)!}=\dfrac{1}{4!\left( n-4 \right)!}+\dfrac{1}{6!\left( n-6 \right)!}$
Now taking $\dfrac{1}{5!\left( n-5 \right)!}$ to the RHS of the equation we get,
$\begin{align}
  & 2=\dfrac{5!\left( n-5 \right)!}{4!\left( n-4 \right)!}+\dfrac{5!\left( n-5 \right)!}{6!\left( n-6 \right)!} \\
 & 2=\dfrac{5.4!\left( n-5 \right)!}{4!\left( n-4 \right)\left( n-5 \right)!}+\dfrac{5!\left( n-5 \right)\left( n-6 \right)!}{6.5!\left( n-6 \right)!} \\
\end{align}$
Cancelling out similar terms from the numerator and the denominator, we get
$2=\dfrac{5}{\left( n-4 \right)}+\dfrac{\left( n-5 \right)}{6}$
Taking LCM we get,
\[\begin{align}
  & 2=\dfrac{30+\left( n-4 \right)\left( n-5 \right)}{6.\left( n-4 \right)} \\
 & 12.\left( n-4 \right)=30+\left( {{n}^{2}}-9n+20 \right) \\
 & {{n}^{2}}-21n+98=0 \\
\end{align}\]
$\begin{align}
  & \left( n-7 \right)\left( n-14 \right)=0 \\
 & n=7,14 \\
\end{align}$
So we get two values of n as 7 and 14,
Hence the correct answer is the option (a).

Note:
While solving this question many students make mistakes while expanding the $^{n}{{C}_{r}}$ term and they interchange n and r in the equation so you need to be careful while expanding this term. And whenever factorials are included in the question then try to expand the factorial only upto that term which can be cancelled out from either numerator or denominator it will help you simplify the equations easily.