Binomial theorem – General term – JEE Main 24 January 2025 Shift 2

Solution:

Given $A = \text{Coefficient of } 30^{th} \text{ term} = {}^{2n-1}C_{29}$
And $B = \text{Coefficient of } 12^{th} \text{ term} = {}^{2n-1}C_{11}$

According to the given condition $2A = 5B$:
$$ 2 \cdot {}^{2n-1}C_{29} = 5 \cdot {}^{2n-1}C_{11} $$

Expanding the combinations:
$$ 2 \frac{(2n-1)!}{29!(2n-30)!} = 5 \frac{(2n-1)!}{(2n-12)!11!} $$

Canceling $(2n-1)!$ and rearranging terms:
$$ \therefore\frac{2}{29 \cdot 28 \dots 12 \cdot 11! (2n-30)!} = \frac{5}{(2n-12)(2n-13)\dots(2n-29)(2n-30)! 11!} $$

$$ \therefore \frac{1}{29 \dots 12 \cdot 5} = \frac{1}{(2n-12)(2n-13)\dots(2n-29) \cdot 2} $$
$$ \therefore \frac{1}{29 \dots 12 \cdot 5 \times 6} = \frac{1}{(2n-12)(2n-13)\dots(2n-29) \cdot 2 \times 6} $$
$$ \therefore \frac{1}{30 \cdot 29 \dots 12 } = \frac{1}{(2n-12)(2n-13)\dots(2n-29) \cdot 12} $$

By comparing the factorial structures and terms on both sides:
$$ 2n – 12 = 30 $$

$$ 2n = 42 \Rightarrow n = 21 $$

Ans. (2)