Grade by Grade
C. Discrete Functions · Sequences and Series

Arithmetic Sequences and the General Term

C2.1 C2.2 C1.4About 75 minutes
Learning goals
  • I can identify an arithmetic sequence and find its common difference.
  • I can explain where the formula tₙ = a + (n − 1)d comes from.
  • I can use the formula to find any term, the number of terms, or the first term and common difference.
  • I can connect arithmetic sequences to linear functions.

Words to know

arithmetic sequence
A sequence in which each term is found by adding the same number to the previous term.
common difference (d)
The constant amount added each time: d = tₙ − tₙ₋₁. It can be positive, negative or a fraction.
first term (a)
t₁, the first term of the sequence.
general term of an arithmetic sequence
tₙ = a + (n − 1)d.

Same jump every time

Remember the concert chairs from the first lesson: 12, 14, 16, 18, …. Each row has 2 more chairs than the one before. Or think of a stack of plastic cups: each cup you add raises the stack by the same small amount. When a sequence grows (or shrinks) by the same amount every step, it is an arithmetic sequence. Arithmetic sequences are the simplest and most common sequences, and they have a neat formula for the general term.

Definition

A sequence is arithmetic if the difference between consecutive terms is constant. That constant is the common difference d:

d = t₂ − t₁ = t₃ − t₂ = … = tₙ − tₙ₋₁

  • 5, 9, 13, 17, … is arithmetic with d = 4.
  • 20, 17, 14, 11, … is arithmetic with d = −3.
  • 1, 4, 9, 16, … is not arithmetic: the differences are 3, 5, 7, ….

The recursion formula of an arithmetic sequence is t₁ = a, tₙ = tₙ₋₁ + d.

01234567891011121314151617181920+4+4+4
The arithmetic sequence 5, 9, 13, 17 on a number line. To reach the 4th term from the 1st, you make 3 jumps of 4.

Developing the general term

Let's connect the steps in a number example to the algebra. Use 5, 9, 13, 17, … (a = 5, d = 4) on the left and a general arithmetic sequence on the right:

  • t₁ = 5 ↔ t₁ = a
  • t₂ = 5 + 4 ↔ t₂ = a + d
  • t₃ = 5 + 4 + 4 = 5 + 2(4) ↔ t₃ = a + 2d
  • t₄ = 5 + 3(4) ↔ t₄ = a + 3d

The number of d's is always one less than the position, because the first term has no jumps yet. So

tₙ = a + (n − 1)d

For 5, 9, 13, 17, …: tₙ = 5 + (n − 1)(4) = 4n + 1. Then t₅₀ = 4(50) + 1 = 201.

t₁ = 33 + 0(2)t₂ = 53 + 1(2)t₃ = 73 + 2(2)t₄ = 93 + 3(2)a = 3 (first term)d = 2 (each jump)Tower n has the firstterm plus (n − 1)jumps of d:tₙ = a + (n − 1)d
Cube towers for 3, 5, 7, 9, …. Every tower starts with the first term a = 3 (the bottom three blocks) and adds one block of d = 2 (stacked on top) for each step after the first.

Arithmetic sequences are linear

Expand the general term: tₙ = a + (n − 1)d = dn + (a − d). That has the form y = mx + b with slope m = d. So an arithmetic sequence is a discrete linear function: its graph is a set of points on a straight line whose slope is the common difference.

012345651015202530ntₙ(1, 7)(2, 11)(3, 15)(4, 19)(5, 23)1d = 4The points lie on a linewith slope d = 4:tₙ = 4n + 3an arithmetic sequence is adiscrete linear function
The arithmetic sequence 7, 11, 15, 19, 23 graphed. Each step of 1 to the right goes up d = 4, so the points lie on a line with slope 4.
Worked example

Example 1: Finding the general term and a term

For 7, 11, 15, 19, …, find the general term and t₂₀.

  1. Check it is arithmetic: 11 − 7 = 15 − 11 = 19 − 15 = 4. So a = 7, d = 4.
  2. tₙ = 7 + (n − 1)(4) = 7 + 4n − 4 = 4n + 3.
  3. t₂₀ = 4(20) + 3 = 83.

tₙ = 4n + 3; t₂₀ = 83

Worked example

Example 2: Negative and decimal differences

For 10, 8.5, 7, 5.5, …, find t₁₅.

  1. d = 8.5 − 10 = −1.5 (check: 7 − 8.5 = −1.5 ✓). a = 10.
  2. t₁₅ = 10 + (15 − 1)(−1.5) = 10 + 14(−1.5) = 10 − 21 = −11.

t₁₅ = −11

Worked example

Example 3: How many terms?

How many terms are in the arithmetic sequence 5, 12, 19, …, 222?

  1. a = 5, d = 7, and the last term is tₙ = 222. Find n.
  2. 222 = 5 + (n − 1)(7)
  3. 217 = 7(n − 1)
  4. 31 = n − 1, so n = 32.
  5. A quick check: 5 + 31(7) = 5 + 217 = 222 ✓.

32 terms

Worked example

Example 4: Two terms given

In an arithmetic sequence, t₄ = 17 and t₁₀ = 41. Find a, d and the general term.

  1. Going from t₄ to t₁₀ is 10 − 4 = 6 jumps of d: 6d = 41 − 17 = 24, so d = 4.
  2. t₄ = a + 3d: 17 = a + 12, so a = 5.
  3. tₙ = 5 + (n − 1)(4) = 4n + 1. Check t₁₀: 41 ✓.
  4. (You could also solve the system a + 3d = 17 and a + 9d = 41 by elimination.)

a = 5, d = 4, tₙ = 4n + 1

Worked example

Example 5: Algebraic terms

Show that x, 2x + 1, 3x + 2, … is arithmetic, and find t₁₀.

  1. (2x + 1) − x = x + 1 and (3x + 2) − (2x + 1) = x + 1. Constant, so it is arithmetic with d = x + 1.
  2. t₁₀ = a + 9d = x + 9(x + 1) = 10x + 9.

d = x + 1; t₁₀ = 10x + 9

Just for fun

An arithmetic sequence's favourite saying?

"Same difference."

Hands-on activity

The stacking-cup sequence

You need: 5 or more identical plastic or paper cups that nest inside each other, and a ruler.

  1. Measure the height of 1 cup, then a stack of 2 nested cups, then 3, 4 and 5. Record the heights in a table.
  2. Find the differences. Is the sequence (close to) arithmetic? What is d, and what part of the cup does d measure?
  3. Write tₙ = a + (n − 1)d for your cups.
  4. Predict the height of a stack of 20 cups, then the number of cups needed to reach 1 m (100 cm).

What to look for: example data: 1 cup 12 cm, then 13.5, 15, 16.5, 18 cm, so a = 12 and d = 1.5 (d is the height of the rim that sticks out above the cup below). tₙ = 12 + (n − 1)(1.5). For 20 cups: 12 + 19(1.5) = 40.5 cm. For 100 cm: 12 + (n − 1)(1.5) = 100 gives n − 1 ≈ 58.7, so you need 60 cups (59 cups reach only 99 cm). Your numbers will differ; the method is the same.

Your turn

1

Is 3, 8, 13, 18, … arithmetic? If so, give a and d.

2

Find the general term of 3, 8, 13, 18, … and find t₃₀.

3

Find t₁₂ for 40, 34, 28, ….

4

Find t₂₁ for 1/2, 1, 3/2, 2, ….

5

How many terms are in 2, 9, 16, …, 191?

6

In an arithmetic sequence, t₃ = 14 and t₈ = 39. Find a, d and tₙ.

7

Is 500 a term of 3, 10, 17, …? If so, which one?

More practice: warm-up

1

What is the common difference of 9, 15, 21, 27, …?

2

What is the common difference of 4, 1, −2, −5, …?

3

Write the first four terms of the arithmetic sequence with a = 6 and d = 5.

4

For tₙ = 2 + (n − 1)(3), find t₁₀.

5

Is 2, 4, 8, 16, … arithmetic? Explain.

More practice: on your own

1

The 1st term of an arithmetic sequence is 11 and the 25th term is 155. Find d.

2

How many terms are in −8, −5, −2, …, 97?

3

Find the general term of the arithmetic sequence with t₆ = 30 and t₁₅ = 3.

4

Write the arithmetic sequence tₙ = 3n + 8 as a recursion formula and in function notation.

5

Show that 2a + b, 3a + 2b, 4a + 3b, … is arithmetic and find t₁₀.

Challenge

1

Find x so that x + 2, 3x − 1 and 4x + 1 are consecutive terms of an arithmetic sequence.

2

Three numbers form an arithmetic sequence. Their sum is 27 and their product is 504. Find them. (Hint: call them a − d, a, a + d.)

3

How many multiples of 7 are there between 100 and 1000?

Check your understanding

Quiz

1.

What is the common difference of 12, 7, 2, −3, …?

2.

Which is the general term of 6, 10, 14, 18, …?

3.

For a = 20 and d = −3, what is t₁₁?

4.

How many terms are in 4, 7, 10, …, 64?

5.

In an arithmetic sequence, t₂ = 9 and t₇ = 34. What is d?

6.

The graph of an arithmetic sequence is:

For parents and teachers

Big idea: Expectation C2.1 asks students to identify arithmetic sequences, and C2.2 asks them to develop the formula tₙ = a + (n − 1)d through investigation (the curriculum mentions linking cubes, diagrams and connecting the steps of a number example to the algebra) and apply it to find any term.

How to help: The key idea is "n − 1 jumps": to get to the 10th term you make 9 jumps from the 1st. Use stairs, fence posts or stacked cups to make it concrete.

Common difficulties:

  • Using n instead of n − 1: t₁₀ = a + 9d, not a + 10d.
  • Getting the sign of d wrong for decreasing sequences: d = (later term) − (earlier term).
  • Getting a non-whole n when counting terms. That usually means an arithmetic slip, or that the last number is not actually a term.
  • In two-term problems, dividing by the wrong number of jumps: from t₄ to t₁₀ is 6 jumps, not 10.