Lesson 08 · Arithmetic Sequences

Worksheet · Mia
Every equation you've solved so far described one moment in time. But many real patterns keep going — Copper's weight each week, Mia's savings each month, the number of jumps in each training set. When a list of numbers grows (or shrinks) by the same amount each time, it's called an arithmetic sequence. The fixed amount you add each time is the common difference.

Vocabulary

Term ($a_n$): one number in the sequence. The 1st term is $a_1$, the 5th is $a_5$.
Common difference ($d$): the fixed amount added each step. Can be positive (growing) or negative (shrinking).
Position ($n$): which term it is (1st, 2nd, 3rd …).
Formula for the $n$th term: $$a_n = a_1 + (n - 1) \cdot d$$ Start at $a_1$, then add $d$ exactly $(n-1)$ more times.

Worked Examples

Worked Example 1 — identify and extend

Copper's training schedule adds 3 extra minutes of trot each week. Week 1: 12 min. List the first 5 weeks and find the 10th term.

12 +3→ 15 +3→ 18 +3→ 21 +3→ 24

Common difference $d = 3$. For the 10th term:

$$a_{10} = 12 + (10 - 1) \cdot 3 = 12 + 27 = 39 \text{ minutes}$$

By week 10, Copper trots for 39 minutes.

Worked Example 2 — find the common difference

Mia's hay bills (in €) for the last 6 months: 85, 91, 97, 103, 109, 115.

Each month the bill rises by $91 - 85 = 6$ €. So $d = 6$.

Check: $103 - 97 = 6$ ✓    $115 - 109 = 6$ ✓

Formula: $a_n = 85 + (n-1) \cdot 6$

Month 12: $a_{12} = 85 + 11 \cdot 6 = 85 + 66 = $ 151 €

Worked Example 3 — find a missing term

A show-jumping course has a sequence of obstacle heights (in cm): 65, __, 83, 92, …

The difference between $a_3 = 83$ and $a_4 = 92$ gives $d = 9$. So $a_2 = 83 - 9 = $ 74 cm. Check: $65 + 9 = 74$ ✓.

Worked Example 4 — write the expression

A sequence starts at 7 and has common difference $-5$. Write a formula for the $n$th term and find the first negative term.

$$a_n = 7 + (n-1)(-5) = 7 - 5n + 5 = 12 - 5n$$

When does $a_n < 0$?   $12 - 5n < 0 \;\Rightarrow\; n > 2.4$, so the 3rd term ($n=3$): $12 - 15 = -3$. ✓

Practice Problems

Problems 1–3: identify the common difference and fill in the missing term(s).

  1. Feeding routine. Mia measures how many kg of hay Copper eats per day during different seasons. The amounts follow a sequence: 7, 9, 11, __, 15, __
    $d = $ Missing terms:
  2. Stable rent. The monthly stable rent increases each year: $\$520,\ \$547,\ \__,\ \$601,\ \$628$
    $d = $ Missing term:
  3. Competition scores. Mia's dressage scores over 6 rounds: 58, 61, 64, __, 70, __
    $d = $ Missing terms:

Problems 4–6: use the $n$th term formula.

  1. Training load. Copper starts with 18 minutes of canter per session. Each week the trainer adds 2.5 minutes.
    (a) Write the formula for $a_n$.
    (b) How long will the canter session be in week 11?
  2. Prize money. A local show offers prize money in arithmetic sequence by placing: 1st place 120 €, 2nd place 103 €, 3rd place 86 €, …
    (a) What is $d$?
    (b) What does 8th place receive?
    (c) Which placing first receives $0$ € or less?
  3. Savings plan. Mia has $\$35$ saved. She adds $\$22$ each week.
    (a) Write the formula for $a_n$ (week $n$, counting week 1 as when she already has $\$35$).
    (b) After how many weeks does she first have at least $\$200$?
From here — write the sequence and expression yourself.
  1. Bale countdown. Mia starts the month with 53 kg of hay. Copper eats 7 kg per day.
    Write out the first 5 terms of the sequence (day 1 = start of day, so $a_1 = 53$).
    Write the formula for $a_n$:
    On which day does she fall below 10 kg and need to reorder?
  2. Jump height progression. At a show, the 1st fence is 55 cm. Each fence is 8 cm higher than the previous one.
    Write the expression $a_n = $ ____
    The maximum safe height for Copper is 135 cm. What is the last fence number he can safely attempt?
  3. Vet bills. Routine vet visits cost 45 €, 62 €, 79 €, 96 € over 4 visits.
    Is this arithmetic? Show how you know.
    Predict the 9th visit cost.
  4. Challenge. The 3rd term of an arithmetic sequence is 29, and the 7th term is 53. Find $d$ and $a_1$, then write the $n$th term formula. (Hint: how many steps separate the 3rd and 7th terms?)
What's next → Arithmetic sequences add a fixed amount each step. But what if you multiply by a fixed amount instead? Doubling every day, halving every week — these are geometric sequences. Lesson 09 covers how to spot them, find the ratio, and write the formula.
Show answers
  1. $d = 2$; missing terms: 13 and 17
  2. $d = 27$; missing term: $\$574$
  3. $d = 3$; missing terms: 67 and 73
  4. (a) $a_n = 18 + (n-1) \cdot 2.5 = 15.5 + 2.5n$; (b) $a_{11} = 18 + 10 \cdot 2.5 = 43$ minutes
  5. (a) $d = -17$; (b) $a_8 = 120 + 7 \cdot (-17) = 120 - 119 = 1$ €; (c) $a_n \leq 0$: $120 - 17(n-1) \leq 0 \Rightarrow n \geq 8.06$, so 9th place receives $120 - 8 \cdot 17 = -16$ € — i.e., nothing from 9th place onward
  6. (a) $a_n = 35 + (n-1) \cdot 22 = 13 + 22n$; (b) $13 + 22n \geq 200 \Rightarrow n \geq 8.5$, so after 9 weeks ($a_9 = 13 + 198 = 211$)
  7. Terms: 53, 46, 39, 32, 25; formula $a_n = 53 + (n-1)(-7) = 60 - 7n$; $60 - 7n < 10 \Rightarrow n > 7.1$, so on day 8 ($a_8 = 60 - 56 = 4$ kg)
  8. $a_n = 55 + (n-1) \cdot 8 = 47 + 8n$; max fence: $47 + 8n \leq 135 \Rightarrow n \leq 11$, so fence 11 ($47 + 88 = 135$ cm exactly)
  9. Yes — differences: $62-45=17$, $79-62=17$, $96-79=17$ ✓; $a_9 = 45 + 8 \cdot 17 = 45 + 136 = 181$ €
  10. From $a_3$ to $a_7$ is 4 steps: $53 - 29 = 24$, so $d = 6$. $a_1 = 29 - 2 \cdot 6 = 17$. Formula: $a_n = 17 + (n-1) \cdot 6 = 11 + 6n$
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