QJEEM24Maaf0a: JEE Main 2024 Mathematics - Class 11 JEEM24Maaf0a Solution: If the second, third and fourth terms in the expansion of are…

$\begin{aligned}& T_2={ }^n C_1 y^1 \cdot x^{n-1}=135 \\& T_3={ }^n C_2 y^2 \cdot x^{n-2}=30 \\& T_4={ }^n C_3 y^3 x^{n-3}=\frac{10}{3}\end{aligned}$

$\begin{aligned}\Rightarrow \frac{135}{30} & =\left(\frac{x}{y}\right) \frac{n \cdot 2}{n(n-1)}=\left(\frac{2}{n-1}\right)\left(\frac{x}{y}\right) \quad \text{... (i)}\\\frac{30}{\frac{10}{3}} & =\frac{n(n-1)}{2} \frac{3!}{n(n-1)(n-2)}\left(\frac{x}{y}\right) \\9 & =\left(\frac{3}{n-2}\right)\left(\frac{x}{y}\right)\end{aligned}$

$\begin{aligned}\Rightarrow & 3(n-2)=\frac{135}{60}(n-1) \Rightarrow n=5 \\\Rightarrow & x=9 y \quad \text{.... (i)}\\& y \cdot x^4=27 \Rightarrow \frac{x}{9} \cdot x^4=3^3 \\\Rightarrow & x^5=3^5 \Rightarrow x=3 y=\frac{1}{3} \\\Rightarrow & 6\left(5^3+3^2+\frac{1}{3}\right)=6\left(125+9+\frac{1}{3}\right)\end{aligned}$

$=6(134)+2=806$