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2015 AMC 10A Problem 7

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2015 AMC 10A Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 1 of 5 · about 1 min · position P1-10

How many terms are there in the arithmetic sequence $13$ , $16$ , $19$ , . . ., $70$ , $73$ ?

A)
$20$
B)
$21$
C)
$24$
D)
$60$
E)
$61$

Answer

Key insight

The number of terms is (last - first)/(common difference) + 1 = (73 - 13)/3 + 1 = 21.

Solution

The common difference is $3$ . The $n$ -th term is $13 + 3(n - 1)$ , and it equals $73$ when
$3(n - 1) = 60 \quad\Longrightarrow\quad n - 1 = 20 \quad\Longrightarrow\quad n = 21.$

Equivalently, there are $20$ jumps of size $3$ from $13$ to $73$ , and $20$ jumps connect $21$ terms.

The answer is $\boxed{\textbf{(B)}\ 21}$ .

Why this works

Counting terms of an arithmetic sequence is counting fenceposts: the gaps number $\frac{\text{last} - \text{first}}{d}$ , and the terms number one more. Subtracting a constant and dividing by $d$ turns the list into $1, 2, \ldots, n$ , which is the cleanest way to see it.

Alternative approach

Subtract $10$ from every term to get $3, 6, 9, \ldots, 63$ , then divide by $3$ to get $1, 2, \ldots, 21$ . This bijection with $\{1, \ldots, 21\}$ counts the terms without any formula.

The trap

Forgetting the +1 (fencepost error) and answering 20.

Common mistakes

Forgetting the $+1$ (fencepost error) and answering $20$ .
Dividing only the last term by $3$ ( $73/3 \approx 24$ ) and picking choice (C).

Techniques

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