1. Identify the general term

In the expansion of $(1+x)^{2n-1},$ the general term is $T_{r+1} = \binom{2n-1}{r}x^r.$

So:

• the $30^{\text{th}}$ term corresponds to $r=29$
• the $12^{\text{th}}$ term corresponds to $r=11$

Hence, $A = \binom{2n-1}{29}, \qquad B = \binom{2n-1}{11}.$

2. Use the given relation

We are given: $2A = 5B.$ Substitute $A$ and $B$: $2\binom{2n-1}{29} = 5\binom{2n-1}{11}.$

Let $N = 2n-1.$ Then, $2\binom{N}{29} = 5\binom{N}{11}.$

3. Form the ratio

$\frac{\binom{N}{29}}{\binom{N}{11}} = \frac{5}{2}.$

Using factorial form, $\frac{\binom{N}{29}}{\binom{N}{11}} = \frac{N!}{29!(N-29)!}\cdot \frac{11!(N-11)!}{N!} = \frac{11!(N-11)!}{29!(N-29)!}.$

Now, $\frac{(N-11)!}{(N-29)!} = (N-11)(N-12)\cdots(N-28),$ and $\frac{29!}{11!} = 12\cdot 13\cdots 29.$

So, $\frac{\binom{N}{29}}{\binom{N}{11}} = \frac{(N-11)(N-12)\cdots(N-28)}{12\cdot 13\cdots 29}.$

Trying the options is quickest.

4. Check the options

Option A: $n=20$

Then $N=2n-1=39.$ So, $\binom{39}{29}=\binom{39}{10}, \qquad \binom{39}{11}=\binom{39}{28}.$ Then $\frac{\binom{39}{29}}{\binom{39}{11}} = \frac{\binom{39}{10}}{\binom{39}{11}} = \frac{11}{39-10} = \frac{11}{29} \neq \frac{5}{2}.$ Not correct.

Option B: $n=19$

Then $N=37.$ $\frac{\binom{37}{29}}{\binom{37}{11}} = \frac{\binom{37}{8}}{\binom{37}{11}},$ clearly much smaller than $1$, so cannot be $\frac{5}{2}$. Not correct.

Option C: $n=22$

Then $N=43.$ Using symmetry, $\binom{43}{29}=\binom{43}{14}.$ Then $\frac{\binom{43}{29}}{\binom{43}{11}}=\frac{\binom{43}{14}}{\binom{43}{11}}.$ Now, $\frac{\binom{43}{14}}{\binom{43}{11}} = \frac{32\cdot 31\cdot 30}{14\cdot 13\cdot 12}$ which is not equal to $\frac{5}{2}$. Not correct.

Option D: $n=21$

Then $N=41.$ Using symmetry, $\binom{41}{29}=\binom{41}{12}.$ Thus, $\frac{A}{B} = \frac{\binom{41}{12}}{\binom{41}{11}}.$ Using $\frac{\binom{N}{r+1}}{\binom{N}{r}} = \frac{N-r}{r+1},$ we get $\frac{\binom{41}{12}}{\binom{41}{11}} = \frac{41-11}{12} = \frac{30}{12} = \frac{5}{2}.$ Therefore, $\frac{A}{B} = \frac{5}{2} \implies 2A=5B,$ which satisfies the condition.

So, $n=21.$

5. Final answer

The correct option is $\boxed{21}.$