Question

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

Answer 1

At first we’ll find the 9th, 10th, and 11th terms in the expansion${(1 + x)^n}$using the${(r + 1)^{th}}$term formula in the expansion of ${(a + b)^n}$i.e. ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
After that, we’ll find their coefficient, as they are mentioned in A.P., then using the property of common difference of A.P. we get an equation in n, solving that equation we’ll get the value for n.

Complete step-by-step answer:
Given data: ${T_9}$ , ${T_{10}}$and ${T_{11}}$of ${(1 + x)^n}$are in A.P.
It is well known that in the expansion of ${(a + b)^n}$the ${(r + 1)^{th}}$term is given by
${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Therefore, from the above formula, we can say that in the expansion of${(1 + x)^n}$,
${T_{r + 1}} = {}^n{C_r}{(1)^{n - r}}{x^r}$
On Substituting $r = 8$, we get the ninth term,
i.e., ${T_9} = {}^n{C_8}{(1)^{n - 8}}{x^8}$
On simplification we get,
$\Rightarrow {T_9} = {}^n{C_8}{x^8}$
On Substituting $r = 9$, in ${T_{r + 1}} = {}^n{C_r}{(1)^{n - r}}{x^r}$, we get,
\Rightarrow $$${T_{10}} = {}^n{C_9}{(1)^{n - 9}}{x^9}$ On simplification, we get, $ \Rightarrow {T_{10}} = {}^n{C_9}{x^9}$ On Substitutingr = 10, in ${T_{r + 1}} = {}^n{C_r}{(1)^{n - r}}{x^r}$, we get, \Rightarrow $$${T_{11}} = {}^n{C_{10}}{(1)^{n - 10}}{x^{10}}$On simplification, we get,$ \Rightarrow {T_{11}} = {}^n{C_{10}}{x^{10}}$From the above values, we can say that the coefficients of${T_9}$,${T_{10}}$and${T_{11}}are $${}^n{C_8}$$ , {}^n{C_9}$and${}^n{C_{10}}$respectively Since it is given that coefficient of${T_9}$,${T_{10}}$and${T_{11}}$are in A.P., therefore the common difference will be constant i.e.${T_{10}} - {T_9} = {T_{11}} - {T_{10}}$$ \Rightarrow 2{T_{10}} = {T_{11}} + {T_9}$Now, substituting the value of${T_9}$,${T_{10}}$and${T_{11}}$$ \Rightarrow 2{}^n{C_9} = {}^n{C_{10}} + {}^n{C_8}$We know that,${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$$ \Rightarrow 2\dfrac{{n!}}{{9!\left( {n - 9} \right)!}} = \dfrac{{n!}}{{10!\left( {n - 10} \right)!}} + \dfrac{{n!}}{{8!\left( {n - 8} \right)!}}$
Dividing both sides by n! , we get,

$\Rightarrow \dfrac{2}{{9!\left( {n - 9} \right)!}} = \dfrac{1}{{10!\left( {n - 10} \right)!}} + \dfrac{1}{{8!\left( {n - 8} \right)!}}$
Using $n! = n(n - 1)!$, we get,
$\Rightarrow \dfrac{2}{{9 \times 8!\left( {n - 9} \right)\left( {n - 10} \right)!}} = \dfrac{1}{{10 \times 9 \times 8!\left( {n - 10} \right)!}} + \dfrac{1}{{8!\left( {n - 8} \right)(n - 9)(n - 10)!}}$
Multiplying both sides by $8!\left( {n - 10} \right)!$, we get,
$\Rightarrow \dfrac{2}{{9\left( {n - 9} \right)}} = \dfrac{1}{{10 \times 9}} + \dfrac{1}{{\left( {n - 8} \right)\left( {n - 9} \right)}}$
Now, simplifying the right-hand side by taking LCM, we get,
$\Rightarrow \dfrac{2}{{9\left( {n - 9} \right)}} = \dfrac{{\left( {n - 8} \right)\left( {n - 9} \right) + 90}}{{10 \times 9\left( {n - 8} \right)\left( {n - 9} \right)}}$
Multiplying both sides by $9\left( {n - 9} \right)$, we get,
$\Rightarrow 2 = \dfrac{{{n^2} - 17n + 72 + 90}}{{10(n - 8)}}$
On cross-multiplication, we get,
$\Rightarrow 20(n - 8) = {n^2} - 17n + 162$
On further simplification, we get,
$\Rightarrow 20n - 160 = {n^2} - 17n + 162$
$\Rightarrow 0 = {n^2} - 37n + 322$
Now splitting the coefficient of n such that are the factors of the product of the other two coefficient, we get,
$\Rightarrow 0 = {n^2} - (23 + 14)n + 322$
$\Rightarrow 0 = {n^2} - 23n - 14n + 322$
Taking common factors from the first two and last two terms, we get,
$\Rightarrow 0 = n(n - 23) - 14(n - 23)$
Taking (n-23) common from both the terms, we get,
$\Rightarrow (n - 23)(n - 14) = 0$
Therefore $n = 23$ or $n = 14$.

Note: In the formula of the ${(r + 1)^{th}}$term in the expansion of ${(a + b)^n}$ is given by
${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$ , but most of the students forget to ‘r+1’ and instead take it as ‘r’ and make the equation as ${T_r} = {}^n{C_r}{a^{n - r}}{b^r}$which is wrong and will approach us to the wrong answer.
So always remember that we use the formula for ${(r + 1)^{th}}$ term in the expansion of ${(a + b)^n}$i.e. ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$.