2022 AMC 10B Problem 15

Let the first term be $a$. With common difference $2$,
$$S_n = \frac{n}{2}\bigl(2a + 2(n-1)\bigr) = n(a + n - 1).$$
Therefore
$$\frac{S_{3n}}{S_n} = \frac{3n(a + 3n - 1)}{n(a + n - 1)} = 3 \cdot \frac{(a-1) + 3n}{(a-1) + n}.$$

Write $c = a - 1$. The fraction $\frac{c + 3n}{c + n}$ equals $3$ when $c = 0$, and otherwise changes with $n$ (for $n = 1$ and $n = 2$ it gives $\frac{c+3}{c+1}$ and $\frac{c+6}{c+2}$, which are equal only when $c = 0$). So $a = 1$.

The sequence is $1, 3, 5, \dots$, the odd numbers, and $S_n = n \cdot n = n^2$. Hence $S_{20} = 400$.

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

A ratio of two linear expressions in $n$ is independent of $n$ exactly when the numerator is a constant multiple of the denominator; here that forces the constant term $a - 1$ to vanish. The result is the classic fact that the sum of the first $n$ odd numbers is $n^2$, which is what makes $S_{3n}/S_n = 9$ for every $n$.

Impose the condition for $n = 1$ and $n = 2$: $\frac{S_3}{S_1} = \frac{S_6}{S_2}$ gives $\frac{3a + 6}{a} = \frac{6a + 30}{2a + 2}$. Cross-multiplying: $6a^2 + 18a + 12 = 6a^2 + 30a$, so $a = 1$. Then $S_{20} = 20 \cdot 1 + 2 \cdot \frac{20 \cdot 19}{2} = 400$.

• Solving for a but then computing S_20 with the wrong formula, e.g. 20 · (first term) + 2 · 20 or forgetting to use d = 2.
• Assuming the first term is $2$ (the common difference) and getting $S_{20} = 420$.