7.Binomial Theorem
normal

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 $p + q$ is equal to :

A$283$
B$295$
C$287$
D$299$
(JEE MAIN-2025)

Solution

$( a + b )^{\frac{1}{2}}$
$T_{ r }, T _{ r +1}, T_{ r +2} \rightarrow GP$
$\text { So, } \frac{ T _{ r +1}}{T_{ r }}=\frac{ T _{ r +2}}{T_{ r +1}}$
$\frac{{ }^{12} C _{ r }}{{ }^{12} C _{ r -1}}=\frac{{ }^{12} C _{ r +1}}{{ }^{12} C _{ r }}$
$\frac{12-r+1}{r}=\frac{12-(r+1)+1}{r+1}$
$(13-r)(r+1)=(12-r)(r)$
$-r+12 r+13=12 r-r^2$
$13=0$
No value of r possible
So $P=0$
$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C _{ r }\left(3^{\frac{1}{4}}\right)^{12- r }\left(4^{\frac{1}{3}}\right)^{ r }$
Exponent of $\left(3^{\frac{1}{4}}\right)$ exponent of $\left(4^{\frac{1}{3}}\right)$ term
$\begin{array}{llr}
12 & 0 & 27 \\
0 & 12 & 256
\end{array}$
$q=27+256=283$
$p+q=0+283=283$
Standard 11
Mathematics

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