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# The second, third and fourth terms in the binomial expression ${({\text{x}} + {\text{a}})^{\text{n}}}$ are $240,720,1080$ respectively. Find the value of ${\text{x, a and n}}$.

Last updated date: 12th Aug 2024
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Hint:
By substituting all the terms in the binomial expression formula we will be getting three equations. By dividing the equations, we will be getting the next two equations. By equating the two equations we will be getting the one value. By substituting the one values in the equations we will be getting all the values.

Useful formula:
The formulae used in this question are,
The formula for binomial expression is,
${({\text{x}} + {\text{a}})^{\text{n}}} \Rightarrow {{\text{T}}_{{\text{r + 1}}}} = {{\text{n}}_{{{\text{C}}_{\text{r}}}}}{({\text{x}})^{{\text{n}} - {\text{r}}}}{({\text{a}})^{\text{r}}}$
Where,
${\text{a}}$ be the first of the expression
${\text{r}}$ be the term number
${\text{n}}$ be the exponent on the binomial
The formula for ${\text{n}}!$ is,
${\text{n}}! = {\text{n}} \times ({\text{n}} - 1)!$
Where,
${\text{n}}$ be the required number
The formula for number of combinations is,
${{\text{n}}_{{{\text{C}}_{\text{r}}}}} = \dfrac{{{\text{n!}}}}{{{\text{r!(n}} - {\text{r)!}}}}$
Where,
${\text{n}}$ be the number of objects in the expression
${\text{r}}$ be the number of choosing objects

Complete step by step solution:
The data given in the question are,
The second term in the above expression is $240$
The third term in the above expression is $720$
The fourth term in the above expression is $1080$

To find the value of ${\text{x, a and n}}$in the given ${({\text{x}} + {\text{a}})^{\text{n}}}$ expression,
The second term in the above expression $=$$240$
$\Rightarrow {{\text{T}}_2} = 240$
The above can also be written as,
$\Rightarrow {{\text{T}}_{{\text{1 + 1}}}} = 240$
By seeing the above equation, we can say that ${\text{r}} = 1$, and substituting in the formula we get,
$\Rightarrow {{\text{n}}_{{{\text{C}}_1}}}{({\text{x}})^{{\text{n}} - 1}}{({\text{a}})^1} = 240...............(1)$
The third term in the above expression $=$ $720$
$\Rightarrow {{\text{T}}_3} = 720$
The above can also be written as,
$\Rightarrow {{\text{T}}_{{\text{2 + 1}}}} = 720$
By seeing the above equation, we can say that ${\text{r}} = 2$, and substituting in the formula we get,
$\Rightarrow {{\text{n}}_{{{\text{C}}_2}}}{({\text{x}})^{{\text{n}} - 2}}{({\text{a}})^2} = 720...............(2)$
The fourth term in the above expression $=$ $1080$
$\Rightarrow {{\text{T}}_4} = 1080$
The above can also be written as,
$\Rightarrow {{\text{T}}_{{\text{3 + 1}}}} = 1080$
By seeing the above equation, we can say that ${\text{r}} = 3$, and substituting in the formula we get,
$\Rightarrow {{\text{n}}_{{{\text{C}}_3}}}{({\text{x}})^{{\text{n}} - 3}}{({\text{a}})^3} = 1080...............(3)$
By dividing $(2)\,{\text{and (1)}}$,
$\Rightarrow \dfrac{{{{\text{n}}_{{{\text{C}}_2}}}{{({\text{x}})}^{{\text{n}} - 2}}{{({\text{a}})}^2}}}{{{{\text{n}}_{{\text{C}}_1}}{{({\text{x}})}^{{\text{n}} - 1}}{{({\text{a}})}^1}}} = \dfrac{{720}}{{240}}$
By using the combination formula, we can get,
$\Rightarrow \dfrac{{\dfrac{{{\text{n!}}}}{{2!({\text{n}} - {\text{2}})!}}\, \times {{({\text{x}})}^{{\text{n}} - 2 - ({\text{n - 1}})}} \times {\text{a}}}}{{\dfrac{{{\text{n!}}}}{{1!({\text{n}} - 1)!}}}} = 3$
By simplifying we get,
$\Rightarrow \dfrac{{{\text{n!}}}}{{2({\text{n}} - {\text{2}})!}}\, \times \dfrac{{{\text{1!(n}} - 1){\text{!}}}}{{{\text{n}}!}} \times {({\text{x}})^{ - {\text{1}}}} \times {\text{a}} = 3$
By using ${\text{n!}}$ formula we get,
$\Rightarrow \dfrac{{{\text{(n}} - 1)({\text{n}} - 2){\text{!}}}}{{{\text{2(n}} - 2)!}} \times \dfrac{{\text{a}}}{{\text{x}}} = 3$
By cancelling we get,
$\Rightarrow \dfrac{{{\text{(n}} - 1)}}{{\text{2}}} \times \dfrac{{\text{a}}}{{\text{x}}} = 3$
$\Rightarrow {\text{(n}} - 1) \times \dfrac{{\text{a}}}{{\text{x}}} = 6$
By simplifying the above we get the below equation,
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{{{\text{(n}} - 1)}}...............({\text{A}})$
By dividing $(3)\,{\text{and (2)}}$,
$\Rightarrow \dfrac{{{{\text{n}}_{{\text{C}}_3}}{{({\text{x}})}^{{\text{n}} - 3}}{{({\text{a}})}^3}}}{{{{\text{n}}_{{\text{C}}_2}}{{({\text{x}})}^{{\text{n}} - 2}}{{({\text{a}})}^2}}} = \dfrac{{1080}}{{720}}$
By using the combination formula, we can get,
$\Rightarrow \dfrac{{\dfrac{{{\text{n!}}}}{{3!({\text{n}} - 3)!}}\, \times {{({\text{x}})}^{{\text{n}} - 3 - ({\text{n}} - 2)}} \times {\text{a}}}}{{\dfrac{{{\text{n!}}}}{{2!({\text{n}} - 2)!}}}} = \dfrac{3}{2}$
By simplifying we get,
$\Rightarrow \dfrac{{{\text{n!}}}}{{3!({\text{n}} - 3)!}}\, \times \dfrac{{{\text{2!(n}} - 2){\text{!}}}}{{{\text{n}}!}} \times {({\text{x}})^{ - {\text{1}}}} \times {\text{a}} = \dfrac{3}{2}$
By using ${\text{n!}}$ formula we get,
$\Rightarrow \dfrac{{{\text{2!(n}} - 2)({\text{n}} - 3){\text{!}}}}{{{\text{3!(n}} - 3)!}} \times \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}$
By using ${\text{n!}}$ formula we get,
$\Rightarrow \dfrac{{{\text{2(n}} - 2)}}{{3 \times 2}} \times \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}$
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} \times \dfrac{{{\text{(n}} - 2)}}{3} = \dfrac{3}{2}$
By simplifying the above we get the below equation,
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{9}{{{\text{2(n}} - 2)}}...............({\text{B}})$
By equating $({\text{A}})\,{\text{and (B)}}$,
$\Rightarrow \dfrac{6}{{{\text{n}} - 1}} = \dfrac{9}{{{\text{2(n}} - 2)}}$
By cancelling we get,
$\Rightarrow \dfrac{2}{{{\text{n}} - 1}} = \dfrac{3}{{{\text{2(n}} - 2)}}$
$\Rightarrow \dfrac{4}{{{\text{n}} - 1}} = \dfrac{3}{{{\text{n}} - 2}}$
Doing cross multiplication, we get,
$\Rightarrow 4({\text{n}} - {\text{2}}) = 3({\text{n}} - {\text{1}})$
By solving we get,
$\Rightarrow 4{\text{n}} - 8 = 3{\text{n}} - 3$
The value of ${\text{n}}$ is,
$\Rightarrow {\text{n}} = 5$
Substitute the value of ${\text{n}}$ in $({\text{A}})$
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{{5 - 1}}$
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{4}$
After solving we get,
$\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}$
The value of ${\text{a}}$ is,
$\Rightarrow {\text{a}} = \dfrac{3}{2}{\text{x}}$
Substitute the value of ${\text{a}} = \dfrac{3}{2}{\text{x}}\,{\text{and n}} = 5$ in the equation $(1)$
$\Rightarrow {5_{{{\text{C}}_1}}}{({\text{x}})^{5 - 1}}(\dfrac{3}{2}{\text{x}}) = 240$
By solving we get,
$\Rightarrow 5{({\text{x}})^4}(\dfrac{3}{2}{\text{x}}) = 240$
$\Rightarrow {{\text{x}}^5} = \dfrac{{240 \times 2}}{{5 \times 3}}$
By cancelling the above we get,
$\Rightarrow {{\text{x}}^5} = 16 \times 2 = {2^5}$
The value of ${\text{x}}$ is,
$\Rightarrow {\text{x}} = 2$
$\therefore$ The value of ${\text{x,}}\,{\text{n and a}}$ is $2,\,5\,{\text{and 3}}$ .

Hence, the values of ${\text{x,}}\,{\text{n and a}}$ in ${({\text{x}} + {\text{a}})^{\text{n}}}$
binomial expression are $2,\,5\,{\text{and 3}}$.

Note:
We can divide the equations in any order, we will be getting the same values. While substituting all the values in the binomial expression term formula one by one we will be getting the required terms. This is a small trick for rechecking.