Arithmetic Sequences

In an arithmetic sequence, you add the same number to get from each term to the next, like $5, 9, 13, 17, \dots$ Rows of seats, weekly savings that grow by a fixed amount, and taxi fares all follow this pattern. One formula lets you find any term without listing all the ones before it.

Key ideas

The common difference

A sequence is arithmetic if the difference between consecutive terms is always the same. That difference is the common difference, $d$:

$$d = t_2 - t_1 = t_3 - t_2 = \dots$$

The first term is called $a$. For $5, 9, 13, 17, \dots$: $a = 5$ and $d = 4$. A decreasing arithmetic sequence has a negative $d$: for $10, 7, 4, \dots$, $d = -3$.

The general term

To reach the $n$th term, you start at $a$ and add $d$ a total of $n - 1$ times:

$$t_n = a + (n - 1)d$$

The recursion formula is $t_1 = a$, $t_n = t_{n - 1} + d$.

Arithmetic sequences are linear

Expanding gives $t_n = dn + (a - d)$, a linear function of $n$. So the points of an arithmetic sequence lie on a straight line with slope $d$.

Solving problems

Most problems give you some of $a$, $d$, $n$, and $t_n$ and ask for the rest. Substitute what you know into $t_n = a + (n - 1)d$ and solve. Remember that $n$ must be a positive whole number.

Worked examples

Example 1: The general term

For $5, 9, 13, 17, \dots$, find the general term and $t_{30}$.

Solution. $a = 5$ and $d = 4$:

$$t_n = 5 + (n - 1)(4) = 4n + 1$$ $$t_{30} = 4(30) + 1 = 121$$

Example 2: Which term?

Which term of $7, 10, 13, \dots$ is $250$?

Solution. $a = 7$, $d = 3$:

$$\begin{aligned} 7 + (n - 1)(3) &= 250 \\ 3(n - 1) &= 243 \\ n - 1 &= 81 \\ n &= 82 \end{aligned}$$

It’s the $82$nd term.

Example 3: From two terms

In an arithmetic sequence, $t_4 = 17$ and $t_{10} = 41$. Find the general term.

Solution. Going from $t_4$ to $t_{10}$ adds $d$ six times:

$$6d = 41 - 17 = 24 \quad\Rightarrow\quad d = 4$$

Then $t_4 = a + 3d$ gives $17 = a + 12$, so $a = 5$:

$$t_n = 5 + (n - 1)(4) = 4n + 1$$

Example 4: A real-world sequence

A theatre has $18$ seats in the first row, and each row has $2$ more seats than the row in front. How many seats are in row $15$?

Solution. $a = 18$, $d = 2$, $n = 15$:

$$t_{15} = 18 + (15 - 1)(2) = 18 + 28 = 46$$

Row $15$ has $46$ seats.

Common mistakes

Using $n$ instead of $n - 1$. $t_n = a + (n - 1)d$. The first term has had $d$ added zero times, not once.

Getting the sign of $d$ wrong. For a decreasing sequence like $10, 7, 4$, $d = 7 - 10 = -3$. Always subtract in order: later term minus earlier term.

Accepting a non-whole $n$. If solving for $n$ gives a fraction, the number isn’t a term of the sequence.

Confusing arithmetic with geometric. Arithmetic sequences add the same amount. If you multiply by the same amount, it’s geometric.

Practice

1. (Warm-up) Is each sequence arithmetic? If so, give $d$.

• (a) $3, 8, 13, 18, \dots$
• (b) $2, 4, 8, 16, \dots$
• (c) $10, 7, 4, 1, \dots$

Solution

(a) Yes, $d = 5$.

(b) No. The differences are $2, 4, 8$, which aren’t constant.

(c) Yes, $d = -3$.

2. (Warm-up) Write the general term of the arithmetic sequence with $a = 6$ and $d = -2$, and find $t_{12}$.

Solution

$t_n = 6 + (n - 1)(-2) = 8 - 2n$, so $t_{12} = 8 - 24 = -16$.

3. (Warm-up) Find $t_{25}$ for $2, 9, 16, \dots$

Solution

$t_{25} = 2 + 24(7) = 170$

4. (Core) How many terms are in the sequence $11, 15, 19, \dots, 299$?

Solution

$$11 + (n - 1)(4) = 299 \quad\Rightarrow\quad 4(n - 1) = 288 \quad\Rightarrow\quad n = 73$$

There are $73$ terms.

5. (Core) In an arithmetic sequence, $t_5 = 22$ and $t_{12} = 57$. Find the general term.

Solution

From $t_5$ to $t_{12}$ is $7$ steps: $7d = 35$, so $d = 5$. Then $22 = a + 4(5)$ gives $a = 2$.

$$t_n = 2 + (n - 1)(5) = 5n - 3$$

6. (Core) Is $100$ a term of $4, 11, 18, \dots$? Explain.

Solution

$$4 + (n - 1)(7) = 100 \quad\Rightarrow\quad 7(n - 1) = 96 \quad\Rightarrow\quad n - 1 = \tfrac{96}{7}$$

That isn’t a whole number, so $100$ is not a term.

7. (Core) A taxi charges $4.25 plus $1.80 per kilometre. Show that the costs for $1, 2, 3, \dots$ km form an arithmetic sequence, and find the cost of a $12$ km trip.

Solution

The costs are $6.05, $7.85, $9.65, \dots Each extra kilometre adds $1.80, so it’s arithmetic with $a = 6.05$ and $d = 1.80$.

$$t_{12} = 6.05 + 11(1.80) = 6.05 + 19.80 = 25.85$$

A $12$ km trip costs $25.85.

8. (Challenge) The terms $x + 1$, $3x - 2$, $2x + 7$ form an arithmetic sequence. Find $x$ and the three terms.

Solution

The two differences must be equal:

$$\begin{aligned} (3x - 2) - (x + 1) &= (2x + 7) - (3x - 2) \\ 2x - 3 &= -x + 9 \\ 3x &= 12 \\ x &= 4 \end{aligned}$$

The terms are $5, 10, 15$ (common difference $5$).

9. (Challenge) Show that the general term $t_n = a + (n - 1)d$ is a linear function of $n$, and give its slope and the value it would have at $n = 0$.

Solution

$$t_n = a + dn - d = dn + (a - d)$$

That’s in the form $mn + b$, so it’s linear with slope $d$. At $n = 0$ it would equal $a - d$, the “term before the first term”. (There’s no actual term $t_0$, but that’s where the line through the points crosses the vertical axis.)