JEE Main 2025 — 4 April, Morning Shift — Question 27

In the expansion of $(a+b)^{n}$

$15^{\text {th }}$ term from beginning: $T_{15}={ }^{n} C_{14} a^{n-14} b^{14}$

$15^{\text {th }}$ term from end: $T_{15}^{\prime}={ }^{n} C_{14} b^{n-14} a^{14}$

$\therefore \quad \frac{T_{15}}{T_{15}^{\prime}}=\frac{1}{6}$

$\left(\frac{a}{b}\right)^{n-28}=\frac{1}{6}$

$\left(6^{\frac{1}{3}}\right)^{n-28}=6^{-1}$

$\Rightarrow \frac{n-28}{3}=-1$ $n=25$

$\therefore \quad{ }^{25} C_{3}=2300$