Lesson 13.2: Arithmetic Sequences

New Concept

An arithmetic sequence is a set of numbers where each term differs from the last by a constant amount, called the common difference, $d$. You'll learn to identify these sequences, find terms, and write both recursive and explicit formulas.

What’s next

Now that you have the basic idea, you'll put it into practice with interactive examples for finding the common difference and writing the terms of a sequence.

Property

An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If $a_1$ is the first term of an arithmetic sequence and $d$ is the common difference, the sequence will be:

Examples

• Is the sequence $\{4, 10, 16, 22, \dots\}$ arithmetic? Yes, because subtracting any term from the next term gives a common difference of $6$. For example, $10 - 4 = 6$ and $16 - 10 = 6$.
• Is the sequence $\{10, 7, 4, 1, \dots\}$ arithmetic? Yes, the common difference is $-3$. For example, $7 - 10 = -3$ and $4 - 7 = -3$.
• Is the sequence $\{1, 3, 6, 10, \dots\}$ arithmetic? No, because the difference between terms is not constant. $3 - 1 = 2$, but $6 - 3 = 3$.

Explanation

Think of an arithmetic sequence as taking steps of the exact same size. Each term is just the previous term plus a fixed amount, the common difference. When graphed, these points always form a straight line.

Property

To find the terms of an arithmetic sequence, you can start with the first term, $a_1$, and add the common difference, $d$, repeatedly. To find any specific term without listing all previous terms, use the explicit formula:

Examples

• Write the first five terms of the sequence with $a_1 = 20$ and $d = -4$. The sequence is $\{20, 16, 12, 8, 4\}$.
• Given $a_1 = 5$ and $a_4 = 20$, find $a_5$. First, find $d$: $20 = 5 + (4-1)d$, so $15 = 3d$ and $d=5$. Then, $a_5 = a_4 + d = 20 + 5 = 25$.
• Given $a_3 = 10$ and $a_5 = 18$, find $a_1$. The difference over two steps is $18 - 10 = 8$, so $d = 4$. Working backward, $a_2 = 10 - 4 = 6$, and $a_1 = 6 - 4 = 2$.

Explanation

This formula is a shortcut to find any term directly. Instead of adding the common difference $n-1$ times, you can multiply it by $n-1$ and add it to the first term, getting you to your destination in one step.

Property

The recursive formula for an arithmetic sequence with common difference $d$ is:

Examples

• Write a recursive formula for the sequence $\{-5, -2, 1, 4, \dots\}$. The common difference is $d = -2 - (-5) = 3$. The formula is $a_1 = -5$; $a_n = a_{n-1} + 3$ for $n \geq 2$.
• Write the first five terms for the sequence with $a_1 = 50$ and $a_n = a_{n-1} - 8$. The terms are $\{50, 42, 34, 26, 18\}$.
• Write a recursive formula for the sequence $\{1.5, 3, 4.5, 6, \dots\}$. The common difference is $d = 3 - 1.5 = 1.5$. The formula is $a_1 = 1.5$; $a_n = a_{n-1} + 1.5$ for $n \geq 2$.

Explanation

A recursive formula defines each term based on the one right before it. It's like a set of dominoes: to know where one falls, you only need to look at the one that knocked it over, plus the fixed distance between them.

Property

An explicit formula for the $n$th term of an arithmetic sequence is given by

Examples

• Write an explicit formula for $\{3, 8, 13, 18, \dots\}$. The common difference is $d = 5$ and $a_1 = 3$. The formula is $a_n = 3 + 5(n-1)$, which simplifies to $a_n = 5n - 2$.
• Write an explicit formula for $\{100, 92, 84, 76, \dots\}$. The common difference is $d = -8$ and $a_1 = 100$. The formula is $a_n = 100 - 8(n-1)$, which simplifies to $a_n = 108 - 8n$.
• Find the 15th term of the sequence with the explicit formula $a_n = 7 + 3(n-1)$. $a_{15} = 7 + 3(15-1) = 7 + 3(14) = 7 + 42 = 49$.

Explanation

The explicit formula acts like a direct map to any term in the sequence. You don't need to know the previous term; just plug in the term number ($n$) you want, and the formula calculates its value for you.

Property

To find the total number of terms in a finite arithmetic sequence:

1. Find the common difference $d$.
2. Substitute the common difference, the first term $a_1$, and the last term $a_n$ into the explicit formula $a_n = a_1 + d(n-1)$.
3. Solve the resulting equation for $n$.

Examples

• Find the number of terms in the sequence $\{4, 7, 10, \dots, 34\}$. Here, $a_1=4$, $d=3$, and $a_n=34$. So, $34 = 4 + 3(n-1)$, which gives $30 = 3(n-1)$, so $10 = n-1$, and $n=11$. There are 11 terms.
• Find the number of terms in $\{22, 18, 14, \dots, -10\}$. Here, $a_1=22$, $d=-4$, and $a_n=-10$. So, $-10 = 22 + (-4)(n-1)$, which gives $-32 = -4(n-1)$, so $8 = n-1$, and $n=9$. There are 9 terms.
• Find the number of terms in $\{2, 3.5, 5, \dots, 17\}$. Here, $a_1=2$, $d=1.5$, and $a_n=17$. So, $17 = 2 + 1.5(n-1)$, which gives $15 = 1.5(n-1)$, so $10 = n-1$, and $n=11$. There are 11 terms.

Explanation

To count the terms in a sequence, you can use the explicit formula in reverse. By plugging in the last term's value, you solve for its position, $n$, which tells you the total number of terms in the sequence.

Property

In many application problems, it makes sense to use an initial term of $a_0$ instead of $a_1$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula:

Examples

• A seed is planted and is 5 cm tall. It grows 2 cm per week. What is its height after 8 weeks? Let $a_0 = 5$ and $d = 2$. After 8 weeks ($n=8$), its height is $A_8 = 5 + 2(8) = 21$ cm.
• A new car is purchased for 28,000 dollars and depreciates by 3,000 dollars per year. What is its value after 5 years? Let $a_0 = 28000$ and $d = -3000$. After 5 years, its value is $A_5 = 28000 - 3000(5) = 13000$ dollars.
• A person starts with 100 dollars in savings and adds 20 dollars each month. How much money is saved after one year (12 months)? Let $a_0 = 100$ and $d = 20$. After 12 months, the savings are $A_{12} = 100 + 20(12) = 340$ dollars.

Explanation

When a problem gives a starting value at time zero (like an initial deposit or starting height), it's easier to use this modified formula. It directly connects the value after $n$ steps to the starting point $a_0$.