If the ${{3}^{rd}}$ , ${{4}^{th}}$ , ${{5}^{th}}$ and ${{6}^{th}}$ terms in the expansion of ${{\left( x+\text{a} \right)}^{n}}$ be respectively $a,b,c\text{ and }d,$ then prove that $\dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{5a}{3c}$.

Hint: For solving this question first, we will expand the given term using the binomial expansion formulae and then form some equations as per the given data and then prove the result.

Complete step-by-step answer:
Given:
If ${{3}^{rd}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $a$ , ${{4}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $b$ , ${{5}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $c$ and ${{6}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $d$ . And we have to prove that, $\dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{5a}{3c}$ .
Now, we will use the following binomial expansion result:
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}\cdot y+{}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{y}^{2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{y}^{3}}+{}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{y}^{4}}+{}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{y}^{5}}+....................+{}^{n}{{C}_{n}}{{y}^{n}}\]
We can apply the above result to expand ${{\left( x+\text{a} \right)}^{n}}$ . Then,
\[{{\left( x+\text{a} \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}\cdot \text{a}+{}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{\text{a}}^{2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{\text{a}}^{3}}+{}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{\text{a}}^{4}}+{}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{\text{a}}^{5}}+....................+{}^{n}{{C}_{n}}{{\text{a}}^{n}}\]
Where, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Now, let ${{T}_{r}}$ represent the ${{r}^{th}}$ term in the above expression. It is given to us that ${{3}^{rd}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $a$ , ${{4}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $b$ , ${{5}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $c$ and ${{6}^{th}}$ term in the expansion of ${{\left( x+y \right)}^{n}}$ is $d$ . Then,
${{T}_{3}}=a={}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{\text{a}}^{2}}=\dfrac{n\cdot \left( n-1 \right)}{2}\cdot {{x}^{n-2}}\cdot {{a}^{2}}$
${{T}_{4}}=b={}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{\text{a}}^{3}}=\dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)}{6}\cdot {{x}^{n-3}}\cdot {{\text{a}}^{3}}$
${{T}_{5}}=c={}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{\text{a}}^{4}}=\dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)\cdot \left( n-3 \right)}{24}\cdot {{x}^{n-4}}\cdot {{\text{a}}^{4}}$
${{T}_{6}}=d={}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{\text{a}}^{5}}=\dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)\cdot \left( n-3 \right)\cdot \left( n-4 \right)}{120}\cdot {{x}^{n-5}}\cdot {{\text{a}}^{5}}$
Now, we will calculate the expression of $\dfrac{b}{a},\dfrac{c}{b}$ and $\dfrac{d}{c}$ .
$\begin{align}
  & \dfrac{{{T}_{4}}}{{{T}_{3}}}=\dfrac{b}{a}=\dfrac{\left( \dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)}{6}\cdot {{x}^{n-3}}\cdot {{\text{a}}^{3}} \right)}{\left( \dfrac{n\cdot \left( n-1 \right)}{2}\cdot {{x}^{n-2}}\cdot {{a}^{2}} \right)} \\
 & \Rightarrow \dfrac{b}{a}=\left( \dfrac{n-2}{3} \right)\cdot \left( \dfrac{\text{a}}{x} \right)....................\left( 1 \right) \\
\end{align}$
$\begin{align}
  & \dfrac{{{T}_{5}}}{{{T}_{4}}}=\dfrac{c}{b}=\dfrac{\left( \dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)\cdot \left( n-3 \right)}{24}\cdot {{x}^{n-4}}\cdot {{\text{a}}^{4}} \right)}{\left( \dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)}{6}\cdot {{x}^{n-3}}\cdot {{\text{a}}^{3}} \right)} \\
 & \Rightarrow \dfrac{c}{b}=\left( \dfrac{n-3}{4} \right)\cdot \left( \dfrac{\text{a}}{x} \right)................\left( 2 \right) \\
\end{align}$
$\begin{align}
  & \dfrac{{{T}_{6}}}{{{T}_{5}}}=\dfrac{d}{c}=\dfrac{\left( \dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)\cdot \left( n-3 \right)\cdot \left( n-4 \right)}{120}\cdot {{x}^{n-5}}\cdot {{\text{a}}^{5}} \right)}{\left( \dfrac{n\cdot \left( n-1 \right)\cdot \left( n-2 \right)\cdot \left( n-3 \right)}{24}\cdot {{x}^{n-4}}\cdot {{\text{a}}^{4}} \right)} \\
 & \Rightarrow \dfrac{d}{c}=\left( \dfrac{n-4}{5} \right)\cdot \left( \dfrac{a}{x} \right)................\left( 3 \right) \\
\end{align}$
Now, we will try to evaluate $\dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}$ . Then,
$\dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{ab\left( \dfrac{b}{a}-\dfrac{c}{b} \right)}{bc\left( \dfrac{c}{b}-\dfrac{d}{c} \right)}$
Now, substitute the value of $\dfrac{b}{a},\dfrac{c}{b}$ and $\dfrac{d}{c}$ from (1), (2) and (3) in the above expression. Then,

$\begin{align}
  & \dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{ab\left( \dfrac{b}{a}-\dfrac{c}{b} \right)}{bc\left( \dfrac{c}{b}-\dfrac{d}{c} \right)}=\dfrac{ab\left( \left( \dfrac{n-2}{3} \right)\cdot \left( \dfrac{\text{a}}{x} \right)-\left( \dfrac{n-3}{4} \right)\cdot \left( \dfrac{\text{a}}{x} \right) \right)}{bc\left( \left( \dfrac{n-3}{4} \right)\cdot \left( \dfrac{\text{a}}{x} \right)-\left( \dfrac{n-4}{5} \right)\cdot \left( \dfrac{a}{x} \right) \right)} \\
 & \Rightarrow \dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{ab\left( \dfrac{n-2}{3}-\dfrac{n-3}{4} \right)}{bc\left( \dfrac{n-3}{4}-\dfrac{n-4}{5} \right)}=\dfrac{a\left( \dfrac{4n-8-3n+9}{12} \right)}{c\left( \dfrac{5n-15-4n+16}{20} \right)} \\
 & \Rightarrow \dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{a\left( \dfrac{n+1}{12} \right)}{c\left( \dfrac{n+1}{20} \right)}=\dfrac{20a}{12c} \\
 & \Rightarrow \dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{5a}{3c} \\
\end{align}$
Thus, from the above result, we can say that $\dfrac{{{b}^{2}}-ac}{{{c}^{2}}-bd}=\dfrac{5a}{3c}$ .
Hence Proved.

Note: Here, the student must proceed stepwise to prove the result and don’t skip any step and in such questions before doing the calculation first analyse the result which we have to prove in such questions after getting the idea about which term we can evaluate to prove the result without any mistake.