2022 AMC 10B Problem 15

The first term is not given, so call it $a$. Test small values of $n$ to find $a$.

1. The first few sums. The terms are $a, a + 2, a + 4, a + 6, a + 8, a + 10, \ldots$ So
- $S_1 = a$,
- $S_2 = 2a + 2$,
- $S_3 = 3a + 6$,
- $S_6 = 6a + (2 + 4 + 6 + 8 + 10) = 6a + 30$.
2. Use "does not depend on $n$". The ratio must be the same for $n = 1$ and $n = 2$:
$$\frac{S_3}{S_1} = \frac{S_6}{S_2}, \qquad\text{that is}\qquad \frac{3a + 6}{a} = \frac{6a + 30}{2a + 2}.$$
3. Solve. Cross-multiply: $(3a + 6)(2a + 2) = a(6a + 30)$, so $6a^2 + 18a + 12 = 6a^2 + 30a$. The $6a^2$ cancels, leaving $12 = 12a$, so $a = 1$.
4. The sequence is $1, 3, 5, 7, \ldots$, the odd numbers. Their running sums are $1, 4, 9, 16, \ldots$, the perfect squares: $S_n = n^2$.
5. Check the condition. $\frac{S_{3n}}{S_n} = \frac{9n^2}{n^2} = 9$ for every $n$. Good.
6. Finish. $S_{20} = 20^2 = 400$.

The answer is $\boxed{\textbf{(D)}\ 400}$.

If something is true for every $n$, it is true for the easiest ones, so plug in $n = 1$ and $n = 2$ to get an equation for the unknown. Here it pins the sequence down as the odd numbers, and the sum of the first $n$ odd numbers is always $n^2$. That is why the ratio is always $9$.

Use the sum formula for all $n$ at once. With first term $a$ and gap $2$, $S_n = \frac{n}{2}\bigl(2a + 2(n - 1)\bigr) = n(a + n - 1)$. Then
$$\frac{S_{3n}}{S_n} = \frac{3n(a + 3n - 1)}{n(a + n - 1)} = 3 \cdot \frac{3n + (a - 1)}{n + (a - 1)}.$$
The fraction $\frac{3n + (a - 1)}{n + (a - 1)}$ stays fixed as $n$ changes only when $a - 1 = 0$, which gives $a = 1$ again.