2021 AMC 10A Problem 4

1. Write out the first few seconds to see the pattern:

1. To reach second $4$ we added $7$ three times: $5 + 3 \times 7 = 26$. So in second $n$ the cart has had $n - 1$ increases of $7$.
2. In second $30$ it travels $5 + 29 \times 7 = 5 + 203 = 208$ inches.
3. Now add $5 + 12 + 19 + \cdots + 208$. Pair the first with the last: $5 + 208 = 213$. Pair the second with the second-to-last: $12 + 201 = 213$. Every pair adds to $213$, because one number goes up by $7$ exactly when the other goes down by $7$.
4. Thirty numbers make $15$ pairs, so the total is $15 \times 213 = 3195$ inches.

The answer is $\boxed{\textbf{(D)}\ 3195}$.

A list that grows by the same amount each step is called an arithmetic sequence. Its total is the number of terms times the average of the first and last terms, which is exactly what the pairing shows. Remember that the $n$th term has only $n - 1$ jumps; choice (A), $215$, is what you get if you count $30$ jumps.

Estimate as a check. The cart goes about halfway between $5$ and $208$ inches per second on average, roughly $106$, for $30$ seconds: about $3200$ inches. That rules out $215$, $360$ and $2992$ at once, and the exact pairing above decides between $3195$ and $3242$.

The distance covered in each second is an arithmetic sequence with first term $5$ and common difference $7$. The $30$th term is
$$5 + 29 \cdot 7 = 5 + 203 = 208.$$
The total distance is the sum of these $30$ terms, which equals the number of terms times the average of the first and last:
$$30 \cdot \frac{5 + 208}{2} = 15 \cdot 213 = 3195.$$

The answer is $\boxed{\textbf{(D)}\ 3195}$.