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JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Binomial Theorem

Let the coefficients of three consecutive terms T_r, T_ r+1 and T_...

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 :

Options

A $283$
B $287$
C $295$
D $299$

Correct answer

A. $283$

Step-by-step 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}

$

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