Coefficient of $\begin{aligned}
& T_r, T_{r+1}, T_{r+2} \rightarrow G P \
& \Rightarrow\left({ }^{12} C_r\right)^2={ }^{12} C_{r-1} \cdot{ }^{12} C_{r+1}
\end{aligned}\Rightarrow\left({ }^{12} C_r\right)^2={ }^{12} C_{r-1} \cdot{ }^{12} C_{r+1}butnothreeconsecutivebinomialcoefficientareinGP\Rightarrow P=0Nowfor\left(3^{1 / 4}+4^{1 / 3}\right)^{12}, T_{r+1}={ }^{12} C_r(4)^{K / 3}(3)^{\frac{12-K}{4}}forrationalterms\mathrm{K}=0,12sumofrationalterms\begin{aligned}
& ={ }^{12} \mathrm{C}0 4^0 \cdot 3^3+{ }^{12} \mathrm{C}{12} \cdot 4^4 \cdot 3^0 \
& =27+256=283=q \
& \therefore p+q=283
\end{aligned}$
Let the coefficients of three consecutive terms T_r, T_r+1 and T_r+2 in the binomial expansion of (a+b)^12 be in a G.P. and let p be the number of…
JEE Main 2025 — Mathematics Algebra
2025mcqmedium
Let the coefficients of three consecutive terms Tr,Tr+1 and Tr+2 in the binomial expansion of (a+b)12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (43+34)12. Then p+q is equal to :
Official previous-year question
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
283
- B
287
- C
295
- D
299
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