Question
Answers

If the coefficients of ${{5}^{th}},{{6}^{th}},{{7}^{th}}$ terms in the expansion of ${{\left( 1+x \right)}^{n}},n\in N$, are in A.P, then n is equal to
A. 5
B. 6
C. 7
D. 7 or 14

Answer Verified Verified
Hint: To solve this question, we should know the expression for the general term of the expansion ${{\left( 1+x \right)}^{n}},n\in N$. By writing the general term ${{T}_{r+1}}$, we get
${{T}_{r+1}}={}^{n}{{C}_{r}}{{\left( 1 \right)}^{n-r}}{{\left( x \right)}^{r}}$
We can compare that the coefficient of ${{(r+1)}^{th}}$ term is given by ${}^{n}{{C}_{r}}$. Using this concept, we can write the coefficients of ${{5}^{th}},{{6}^{th}},{{7}^{th}}$ terms. In the question, we can infer that the coefficients of ${{5}^{th}},{{6}^{th}},{{7}^{th}}$ terms are in A.P.
If a, b, c are in A.P, then we can write the relation between them as
$\left( b-a \right)=\left( c-b \right)$.
By using this relation and simplifying the equations, we get the value of n.

Complete step-by-step solution:
We know that the general term of the expansion ${{\left( 1+x \right)}^{n}},n\in N$ which is ${{T}_{r+1}}$ can be written as
${{T}_{r+1}}={}^{n}{{C}_{r}}{{\left( 1 \right)}^{n-r}}{{\left( x \right)}^{r}}$.
We can compare that the coefficient of ${{(r+1)}^{th}}$ term is given by ${}^{n}{{C}_{r}}$.
${{C}_{r+1}}={}^{n}{{C}_{r}}$
The given terms in the question are coefficients of ${{5}^{th}},{{6}^{th}},{{7}^{th}}$ terms.
We can write the coefficient of ${{5}^{th}}$ term as
${{C}_{5}}={}^{n}{{C}_{4}}$
Similarly, we can write the coefficients of ${{6}^{th}} and {{7}^{th}}$ terms as
$\begin{align}
  & {{C}_{6}}={}^{n}{{C}_{5}} \\
 & {{C}_{7}}={}^{n}{{C}_{6}} \\
\end{align}$
In the question, it is given that the coefficients of ${{5}^{th}},{{6}^{th}},{{7}^{th}}$ terms are in A.P.
If a, b, c are in A.P, then we can write the relation between them as
$\left( b-a \right)=\left( c-b \right)$.
Substituting the values of $a={{C}_{5}},b={{C}_{6}},c={{C}_{7}}$in the above equation, we get
$\left( {{C}_{6}}-{{C}_{5}} \right)=\left( {{C}_{7}}-{{C}_{6}} \right)$
By rearranging, we get
$2\times {{C}_{6}}={{C}_{5}}+{{C}_{7}}$
Substituting the values of ${{C}_{5}}= {}^{n}{{C}_{4}}, {{C}_{6}} ={}^{n}{{C}_{5}}, {{C}_{7}}= {}^{n}{{C}_{6}}$ in the above equation, we get
$2\times {}^{n}{{C}_{5}}={}^{n}{{C}_{4}}+{}^{n}{{C}_{6}}$
We know the relation that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using this in above equation, we get
$2\times \dfrac{n!}{5!\left( n-5 \right)!}=\dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{n!}{6!\left( n-6 \right)!}$
Cancelling n! in L.H.S and R.H.S , we get
$\begin{align}
  & 2\times \dfrac{1}{5!\left( n-5 \right)!}=\dfrac{1}{4!\left( n-4 \right)!}+\dfrac{1}{6!\left( n-6 \right)!} \\
 & \dfrac{2}{5!\left( n-5 \right)!}=\dfrac{1}{4!\times \left( n-4 \right)\times \left( n-5 \right)!}+\dfrac{1}{6!\left( n-6 \right)!} \\
\end{align}$
Multiplying by $5!\left( n-5 \right)!$ on both sides, we get
$\dfrac{2\times 5!\times \left( n-5 \right)!}{5!\left( n-5 \right)!}=\dfrac{5!\left( n-5 \right)!}{4!\times \left( n-4 \right)\times \left( n-5 \right)!}+\dfrac{5!\left( n-5 \right)!}{6!\left( n-6 \right)!}$
Simplifying the terms in the L.H.S and R.H.S, we get
$2=\dfrac{5}{\left( n-4 \right)}+\dfrac{\left( n-5 \right)}{6}$
Taking L.C.M and cross multiplying, we get
$\begin{align}
  & 2=\dfrac{5}{\left( n-4 \right)}+\dfrac{\left( n-5 \right)}{6} \\
 & 2=\dfrac{\left( 5\times 6 \right)+\left( n-4 \right)\times \left( n-5 \right)}{\left( n-4 \right)\times 6} \\
 & 2\times \left( n-4 \right)\times 6=30+{{n}^{2}}-4n-5n+20 \\
 & 12n-48={{n}^{2}}-9n+50 \\
 & {{n}^{2}}-21n+98=0 \\
\end{align}$
Factorisation gives,
$\begin{align}
  & {{n}^{2}}-14n-7n+98=0 \\
 & n\left( n-14 \right)-7\left( n-14 \right)=0 \\
 & \left( n-14 \right)\left( n-7 \right)=0 \\
\end{align}$
So, we know that when $ab=0$then $a=0\text{ or }b=0$
Here, a = n – 14, b = n – 7 , using this, we get
$\begin{align}
  & \left( n-14 \right)\left( n-7 \right)=0 \\
 & n-14=0\text{ or }n-7=0 \\
 & n=14\text{ or }7 \\
\end{align}$
$\therefore $ n = 7 or 14. The answer is option D.

Note: There is a chance of mistakes by students when they try to check options. The number n = 7 satisfies $2\times {}^{n}{{C}_{5}}={}^{n}{{C}_{4}}+{}^{n}{{C}_{6}}$ and there is an option with n = 7. But, by the calculation, we found out that n can also take the value of 14. So, in this type of question, we have to calculate and option verification might lead to a wrong answer.