Question

The coefficient of 9th, 10th and 11th term in the expansion of ${\left( {1 + x} \right)^n}$ are in A.P then n =

Answer 1

Hint: The numbers of a series are in A.P if the difference between two consecutive numbers are the same for all throughout the series. If a, b and c are in A.P then
\[ \Rightarrow 2b = a + c\]
Secondly, the coefficient of rth term in the expansion ${\left( {1 + x} \right)^n}$ is equal to the $^n{C_{r - 1}}$.
By relating these two concepts we can calculate the value of n.

Complete step-by-step answer:
In this question we have given that 9th, 10th and 11th term coefficient in the expansion ${\left( {1 + x} \right)^n}$ are in A.P.
The coefficient of rth term in the expansion ${\left( {1 + x} \right)^n}$ will be equal to $^n{C_{r - 1}}$.
So, we can calculate the coefficient of the 9th, 10th and 11th term by putting the value of r in $^n{C_{r - 1}}$.
The coefficient of 9th term ${ = ^n}{C_8}$,
The coefficient of 10th term ${ = ^n}{C_9}$ and
The coefficient of 11th term ${ = ^n}{C_{10}}$.
Here, we will use the formula of combination i.e.
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{n - r!.r!}}$
Therefore,
${ \Rightarrow ^n}{C_8} = \dfrac{{n!}}{{n - 8!.8!}}$ …………….(1)
${ \Rightarrow ^n}{C_9} = \dfrac{{n!}}{{n - 9!.9!}}$ ……………(2)
${ \Rightarrow ^n}{C_{10}} = \dfrac{{n!}}{{n - 10!.10!}}$ …………..(3)

As 9th, 10th and 11th term coefficient in the expansion ${\left( {1 + x} \right)^n}$ are in A.P i.e. $^n{C_9}$, \[^n{C_9}\] and\[^n{C_{10}}\]. Therefore,
$ \Rightarrow {2^n}{C_9}{ = ^n}{C_8}{ + ^n}{C_{10}}$ …………..(4)
From (1), (2), (3) and (4)
$ \Rightarrow 2(\dfrac{{n!}}{{n - 9!.9!}}) = \dfrac{{n!}}{{n - 8!.8!}} + \dfrac{{n!}}{{n - 10!.10!}}$
By taking $n!$ common from each side and by cancelling them we get,
$ \Rightarrow 2(\dfrac{1}{{n - 9!.9!}}) = \dfrac{1}{{n - 8!.8!}} + \dfrac{1}{{n - 10!.10!}}$
By opening of the required factorial, we get,
$ \Rightarrow 2(\dfrac{1}{{(n - 9).n - 10!.9.8!}}) = \dfrac{1}{{(n - 8)(n - 9).n - 10!.8!}} + \dfrac{1}{{n - 10!.10.9.8!}}$
By taking common factorial from each term we get,
$ \Rightarrow 2 \times \dfrac{1}{{n - 10!.8!}} \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{n - 10!.8!}}(\dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{10.9}})$
By cancelling common factorial from each side, we get,
$ \Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{10.9}}$
\[
   \Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{1}{{(n - 8)(n - 9)}} + \dfrac{1}{{90}} \\
   \Rightarrow 2 \times \dfrac{1}{{(n - 9).9}} = \dfrac{{90 + (n - 8)(n - 9)}}{{90(n - 8)(n - 9)}} \\
\]
By cancelling \[(n - 9)\] and 9 with 90 from both sides we get,
\[ \Rightarrow 2 = \dfrac{{90 + (n - 8)(n - 9)}}{{10(n - 8)}}\]
By taking denominator of each side to other side to eliminate it, we get,
\[ \Rightarrow 2 \times 10 \times (n - 8) = 90 + (n - 8)(n - 9)\]
By opening the bracket, we get,
\[
   \Rightarrow 20n - 160 = 90 + {n^2} - 8n - 9n + 72 \\
   \Rightarrow 20n - 160 = {n^2} - 17n + 162 \\
\]
Taking all terms on one side and 0 on other we get,
\[ \Rightarrow {n^2} - 17n - 20n + 162 + 160 = 0\]
\[ \Rightarrow {n^2} - 37n + 322 = 0\]
We can find the value of n by factorization method.
$ \Rightarrow {n^2} - 23n - 14n + 322 = 0$
$ \Rightarrow (n - 23)(n - 14) = 0$
$ \Rightarrow n = 14,23$
Therefore, the value of n can be 14 or 23.

Note: To calculate the coefficient of the given expansion some make mistakes i.e. they take $^n{C_r}$ instead of $^n{C_{r - 1}}$. If you take that then your answer will get wrong. Also, while seeing expansion what will some students do? They just consider n as a term and write their expansion. In this way they got struck.
n and r are different things. Here n means power of expansion and r means the term. Take care of this thing. They have said their coefficients are in A.P not the complete term.