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2021 AMC 10A Problem 4

2021 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P1-10

A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

A)
$215$
B)
$360$
C)
$2992$
D)
$3195$
E)
$3242$

Answer

Key insight

The distances per second form an arithmetic sequence 5, 12, 19, ... with 30 terms; sum it as 30 times the average of the first and last terms.

Solution

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 $\boxed{\textbf{(D)}\ 3195}$ .

Why this works

"Each interval adds a constant amount" is the definition of an arithmetic sequence, and the total distance is its partial sum. The $n$ th term uses $n - 1$ jumps, not $n$ ; choice (A) $= 215$ is exactly the value you get by miscounting.

Alternative approach

Estimate: the average speed over the trip is about the middle term, roughly $5 + 14.5 \cdot 7 \approx 106$ inches per second, times $30$ seconds gives about $3200$ . Only (D) and (E) are close, and the exact arithmetic settles it.

The trap

Using 30 as the last term's index incorrectly (last term 5 + 30*7 = 215) or reporting the last term instead of the total.

Common mistakes

Using 30 as the last term's index incorrectly (last term 5 + 30*7 = 215) or reporting the last term instead of the total.
Multiplying $30 \cdot 213 = 6390$ and forgetting to halve.

Techniques

Set up the equation/formula and compute; no special trick needed