JEE Main 2025 — Binomial Theorem Question with Solution

Coefficient of $\begin{aligned} & T_r, T_{r+1}, T_{r+2} \rightarrow G P \\ & \Rightarrow\left({ }^{12} C_r\right)^2={ }^{12} C_{r-1} \cdot{ }^{12} C_{r+1} \end{aligned}$ $\Rightarrow {\left({}^{12}{C}_r\right)}^2={}^{12}{C}_{r-1}\cdot {}^{12}{C}_{r+1}$ but no three consecutive binomial coefficient are in GP $\Rightarrow P=0$ Now for ${\left(3^{1/4}+4^{1/3}\right)}^{12},T_{r+1}={}^{12}{C}_r(4{)}^{K/3}(3{)}^{\frac{12-K}{4}}$ for rational terms $\mathrm{K}=0,12$ sum of rational terms $\begin{aligned} & ={ }^{12} \mathrm{C}_0 4^0 \cdot 3^3+{ }^{12} \mathrm{C}_{12} \cdot 4^4 \cdot 3^0 \\ & =27+256=283=q \\ & \therefore p+q=283 \end{aligned}$