Mathematics · Binomial Theorem

JEE Main 2025 — 28 January, Evening Shift — Question 18

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 all possible values of $r$ . Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$ . Then $\mathrm{p}+\mathrm{q}$ is equal to :

Option A:
283
Correct
Option B:
295
Option C:
287
Option D:
299

Answer: A

Step-by-step solution

\begin{aligned} &\textbf{Given:} \quad (a+b)^{12} = \sum_{k=0}^{12} \binom{12}{k} a^{12-k} b^{k} \\[6pt] &\text{Let\; the\; coefficients\; of \; three\; consecutive\; terms\; } T_r,\, T_{r+1},\, T_{r+2}\; \text{ be\; in\; G.P.} \\[6pt] &\text{Then\; } \binom{12}{r+1}^2 = \binom{12}{r} \binom{12}{r+2}. \\[6pt] &\text{We\; know\; that\; } \frac{\binom{12}{r+1}}{\binom{12}{r}} = \frac{12-r}{r+1}, \quad \frac{\binom{12}{r+2}}{\binom{12}{r+1}} = \frac{11-r}{r+2}. \\[6pt] &\text{For\; G.P.,\; these\; ratios\; must\; be\; equal:} \\[4pt] &\frac{12-r}{r+1} = \frac{11-r}{r+2}. \\[6pt] &\text{Cross-multiplying:\; } (12-r)(r+2) = (11-r)(r+1). \\[6pt] &12r + 24 - r^2 - 2r = 11r + 11 - r^2 - r \\[4pt] &\Rightarrow 10r + 24 = 10r + 11 \Rightarrow 24 \ne 11. \\[4pt] &\text{Hence,\; there\; is\; no\; integer\; solution\; for\; } r. \\[4pt] &\therefore p = 0. \\[12pt] % &\textbf{Now,\; consider\; } (\sqrt[4]{3} + \sqrt[3]{4})^{12}. \\[6pt] &\text{General \; term:\; } T_{k+1} = \binom{12}{k} (\sqrt[4]{3})^{12-k} (\sqrt[3]{4})^{k} = \binom{12}{k} 3^{\frac{12-k}{4}} 4^{\frac{k}{3}}. \\[6pt] &\text{This\; term\; is\; rational\; } \iff \frac{12-k}{4} \text{ and } \frac{k}{3} \text{ are\; integers.} \\[6pt] &\Rightarrow 12-k \equiv 0 \pmod{4} \text{ and } k \equiv 0 \pmod{3}. \\[6pt] &\text{Possible } k \text{ values:} \\[4pt] &12-k \equiv 0 \pmod{4} \Rightarrow k = 0,4,8,12. \\[4pt] &k \equiv 0 \pmod{3} \Rightarrow k = 0,3,6,9,12. \\[4pt] &\text{Intersection:\; } k = 0, 12. \\[6pt] &\text{Corresponding\; rational\; terms:\; } \\[4pt] &T_1 = \binom{12}{0} 3^{\frac{12}{4}} = 3^{3} = 27, \\[4pt] &T_{13} = \binom{12}{12} 4^{\frac{12}{3}} = 4^{4} = 256. \\[6pt] &\Rightarrow q = 27 + 256 = 283. \\[8pt] &\text{Hence,\; } p + q = 0 + 283 = \boxed{283}. \end{aligned}

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Applications of Binomial Theorem