IB Maths
Paper 1 & 2
20 min read
Applications of Sequences & Series
Real-life problems often hide a sequence inside. The trick: spot whether it’s arithmetic or geometric, then plug into the right formula. Let’s go.
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What you need to know
- How to spot arithmetic vs geometric in word problems
- How to identify u1, d or r from the question
- How to choose between un and Sn based on what the question wants
- 5 worked exam-style examples covering both types
The “spot the type” rule
Read the question carefully. Look for one of these two patterns:
Arithmetic (uses + or −)
Same amount added or subtracted each step.
Real-life examples:
- Saving the same amount each month
- Simple interest on savings
- Stacking rows of seats in a theatre
- Building blocks in a tower (one less per row)
Geometric (uses × or ÷)
Same amount multiplied or divided each step.
Real-life examples:
- Population growth (e.g. 5% each year)
- Bacteria doubling
- Bouncing ball (each bounce = fraction of last)
- Radioactive decay
Quick test: calculate the difference and the ratio between the first two terms. If the difference is constant → arithmetic. If the ratio is constant → geometric.
The 4 formulas to remember
All from earlier lessons (and all in the formula booklet):
Arithmetic
un = u1 + (n − 1)d
Sn = n2 [2u1 + (n − 1)d]
Geometric
un = u1 rn − 1
Sn = u1(rn − 1)r − 1
The “what does the question want?” rule
Always ask: am I finding one specific term or a total?
- One term? → use un
e.g. “What does she earn in year 5?” / “How many bacteria on day 10?” - Total? → use Sn
e.g. “How much has she earned over 5 years?” / “Total infections after 10 days?” - Forever? → use S∞ (only if geometric and |r| < 1)
Worked Examples
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Example 1 — Arithmetic: monthly saving
Sara saves $50 in January. Every month she saves $15 more than the previous month.
(a) How much does she save in December (month 12)?
(b) How much has she saved in total by the end of December?
Answer:
Spot the type: same amount added each month → arithmetic.
u1 = 50, d = 15
Part (a): single month → use un.
u12 = 50 + (12 − 1)(15)
= 50 + 165 = 215
Part (b): total → use Sn.
S12 = (12/2)[2(50) + 11(15)]
= 6[100 + 165] = 6 × 265
Dec: $215 | Total: $1590
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Example 2 — Geometric: population growth
A small town has a population of 8000. The population grows by 3% each year.
(a) Find the population after 10 years.
(b) After how many full years will the population first exceed 12000?
Answer:
Spot the type: 3% growth = multiply by 1.03 → geometric.
u1 = 8000, r = 1.03
Year 1 = 8000, Year 2 = 8000(1.03), …, Year n = u1rn−1
Part (a): population in year 10.
u10 = 8000 × (1.03)9
= 8000 × 1.30477…
= 10438.2…
Part (b): solve 8000(1.03)n−1 > 12000.
(1.03)n−1 > 1.5
(n − 1) ln(1.03) > ln(1.5)
n − 1 > 13.72…
n > 14.72…
(a) ≈ 10 438 | (b) Year 15
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Example 3 — Arithmetic: theatre seating
A theatre has 18 seats in the front row. Each row behind has 4 more seats than the one in front. The theatre has 25 rows in total.
(a) How many seats are in the back row?
(b) How many seats are in the theatre?
Answer:
Spot the type: same amount added each row → arithmetic.
u1 = 18, d = 4, n = 25
Part (a): back row = u25.
u25 = 18 + (25 − 1)(4)
= 18 + 96 = 114
Part (b): total seats = S25. Use the (u1 + un) version since we know both.
S25 = (25/2)(18 + 114)
= 12.5 × 132
Back row: 114 | Total seats: 1650
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Example 4 — Geometric: bouncing ball (S∞)
A ball is dropped from a height of 2 m. After each bounce, it reaches 70% of its previous height. Find the total vertical distance the ball travels before coming to rest.
Answer:
Spot the type: × 0.7 each bounce → geometric.
Total = first drop + sum of all up-and-down bounces.
First drop = 2 m
After bouncing, ball goes UP 1.4, DOWN 1.4, UP 0.98, DOWN 0.98, …
Each bounce contributes (up + down) = 2 × height.
Bounce heights: 1.4, 0.98, 0.686, …
u1 = 1.4, r = 0.7
Sum of bounce heights = S∞ (since |r| < 1).
S∞ = 1.4 / (1 − 0.7) = 1.4 / 0.3 = 14/3
Total bounces (up + down) = 2 × 14/3 = 28/3 m
Total distance = first drop + bounces.
Total = 2 + 28/3 = 6/3 + 28/3 = 34/3
Total distance = 34/3 m ≈ 11.3 m
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Example 5 — Spot the type (mixed)
A gym membership has two payment plans:
Plan A: $50 in month 1, increasing by $10 each month.
Plan B: $30 in month 1, increasing by 8% each month.
Which plan is cheaper over 12 months?
Answer:
Plan A is arithmetic: u1 = 50, d = 10.
S12 = (12/2)[2(50) + 11(10)]
= 6[100 + 110] = 6 × 210 = 1260
Plan B is geometric: u1 = 30, r = 1.08.
S12 = 30(1.0812 − 1) / (1.08 − 1)
= 30(2.5182 − 1) / 0.08
= 30 × 1.5182 / 0.08
= 569.31…
Plan A: $1260 | Plan B: $569.32 → Plan B is cheaper
Geometric grew slower because it started lower and 8% < $10 per month early on.
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Tips for word problems
- Always identify u1 first. Read the question carefully — the “first term” is whatever happens at step 1, not step 0.
- “Increases by 3%” means r = 1.03 (not 0.03). For decrease of 3%, r = 0.97.
- “Total” always means sum → use Sn.
- “Eventually” or “in the long run” → think S∞.
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Common mistakes
- Picking the wrong type. Check both: difference AND ratio between first two terms. Only one will be constant.
- Using r = 0.05 for “5% growth”. Wrong! Growth means r = 1.05.
- Confusing “after 10 years” with u10. Read carefully — sometimes “after year 1” means u1, sometimes u2. Set up your starting term clearly.
- Using Sn when the question wants one term (or vice versa). Always re-read what’s being asked.
- Forgetting the first drop in bouncing-ball questions. The ball drops once, THEN bounces forever — so add the first drop separately.
Final word: the maths is the same as before — the only new skill is translating words into u1, d or r, and n. Once you’ve practised 5–10 word problems, you’ll spot the pattern instantly.
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