📈 Arithmetic & Geometric Progressions
From machine depreciation to vibration decay — sequences and series model how engineering quantities grow, shrink and accumulate. Work through each tab: rules first, then try the interactive tools, then the worked examples and practice.
3Topics
12Worked Examples
4Eng. Cases
12Practice Qs
➕Arithmetic ProgressionsConstant difference
✖️Geometric ProgressionsConstant ratio
∑Sums of SeriesAdding the terms up
⚙️ApplicationsReal engineering
📜History & Why It MattersWhere it came from
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Identify arithmetic and geometric progressions, find any term with the nth-term formulas, sum a series to n terms and to infinity, and apply these models to depreciation, production planning and vibration decay.
💡 The single most important question to ask about any sequence: am I adding the same amount each time (arithmetic), or multiplying by the same amount each time (geometric)?
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Follow all 4 worked examples, then practise
➕
Arithmetic Progressions (APs)
A sequence where the same amount is added each time — the common difference.
📖 Key Vocabulary
Sequence — an ordered list of numbers following a rule.
First term (a) — the starting value, a1.
Common difference (d) — the fixed amount added between terms; negative d means a decreasing AP.
nth term (an) — a formula giving any term directly, without listing them all.
An AP increases (or decreases) by the same amount each step. Plotting term value against term number gives points on a straight line with gradient d — this is why APs model any engineering quantity that changes at a constant rate: linear tool wear, straight-line depreciation, evenly spaced shaft holes.
✏️Worked Example 1 — Finding a term directlyType: Direct
Question The sequence 7, 12, 17, 22, … is arithmetic. Find the 20th term.
Identify
a = 7, d = 12 − 7 = 5Formula
an = a + (n − 1)dSubstitute
a20 = 7 + (20 − 1)(5) = 7 + 95Answer a20 = 102
🔍Worked Example 2 — Finding n (which term?)Type: Inverse
Question Which term of the AP 3, 10, 17, 24, … equals 213?
Identify
a = 3, d = 7, an = 213 — n is unknownFormula
an = a + (n − 1)dSet up
213 = 3 + (n − 1)(7)💡 Expanding the bracket: (n − 1)(7) means 7 multiplies everything inside the bracket, so distribute it across both terms: 7 × n − 7 × 1 = 7n − 7. Keep the minus sign attached to the 7 — it's easy to lose in the rush to expand.
Open brackets and expand
(n − 1)(7) = 7n − 7Substitute back
213 = 3 + 7n − 7Simplify
213 = 7n − 4Rearrange
213 + 4 = 7nSimplify
217 = 7nDivide
n = 217 ÷ 7 = 31Answer n = 31 — it is the 31st term. (If n had not come out as a whole number, 213 would not be a term of this sequence at all.)
⚖️Worked Example 3 — Two terms given (simultaneous equations)Type: Simultaneous
Question The 6th term of an AP is 17 and the 13th term is 38. Find a and d, and hence the 19th term.
Set up Write the nth-term formula for each known term and label them so they can be tracked:
a6 = a + (6 − 1)d17 = a + 5d … ①a13 = a + (13 − 1)d38 = a + 12d … ②
💡 Why subtract the equations? Both ① and ② contain exactly the same "a" term. Subtracting ① from ② cancels the a's completely, leaving a single equation in d alone — this is the elimination method for simultaneous equations. It works here because the a-coefficient is identical (1) in both equations, with nothing extra to scale first.
Eliminate
② − ①: (a + 12d) − (a + 5d) = 38 − 17Simplify
7d = 21Solve for d
d = 21 ÷ 7 = 3Back-substitute Put d = 3 into ①:
Substitute
17 = a + 5(3) = a + 15Solve for a
a = 17 − 15 = 2Apply to a19
a19 = a + (19 − 1)d = 2 + (18)(3) = 2 + 54Answer a = 2, d = 3, a19 = 56
⚙️Worked Example 4 — Engineering context (decreasing AP)Type: Engineering
Question A CNC drill bit is 148.0 mm long when new. Regrinding removes 3.5 mm each time. The bit must be scrapped once its length falls below 106 mm. After how many regrinds must it be scrapped?
Model Length after n regrinds:
L = 148 − 3.5n (an AP with a = 148, d = −3.5)Set up The bit is scrapped once L drops below 106:
Inequality
148 − 3.5n < 106Rearrange Move the n-term to one side and numbers to the other:
Simplify
148 − 106 < 3.5nSimplify
42 < 3.5nDivide
n > 42 ÷ 3.5 = 12Check After 12 regrinds: 148 − 3.5(12) = 148 − 42 = 106 mm (exactly on the limit, still usable). After 13: 148 − 3.5(13) = 102.5 mm — below limit.
Answer The bit is scrapped after the 13th regrind.
🔧AP Explorer — build any arithmetic progressionEnter values, then press Generate
Enter a, d and n above, then press Generate to see the sequence build and the nth term calculated step by step…
📊See it grow — drag the slidersLive diagram, updates instantly
👁️ Watch the bars: each one is exactly d taller (or shorter) than the last — a constant step, giving a straight-line pattern.
🧠 Common slip: using n instead of (n − 1). The 20th term has only had 19 jumps of d added to it — count the gaps, not the terms.
✖️
Geometric Progressions (GPs)
A sequence where each term is multiplied by the same amount — the common ratio.
📖 Key Vocabulary
Common ratio (r) — the fixed multiplier between terms.
Exponential growth — |r| > 1: terms increase faster and faster.
Exponential decay — |r| < 1: terms shrink towards zero without reaching it.
Geometric mean — the middle term of three consecutive GP terms: b = √(ac).
👁️ Visual aid: each bar is exactly double the last — the growth accelerates, unlike an AP's steady, equal-sized steps.
Where APs change by a fixed amount, GPs change by a fixed percentage. That makes them the natural model for compound interest, machine depreciation ("loses 15% of its value each year"), vibration amplitude decay and pressure drops through repeated filter stages.
✏️Worked Example 1 — Finding a term directlyType: Direct
Question Find the 9th term of the GP 3, 6, 12, 24, …
Identify
a = 3, r = 6 ÷ 3 = 2Formula
a9 = ar8 = 3 × 28 = 3 × 256Answer a9 = 768
🧮Worked Example 2 — Finding r from two non-adjacent termsType: Root method
Question The 3rd term of a GP is 4.5 and the 7th term is 22.78125. Find r (r > 0) and the first term.
Set up
ar6 ÷ ar2 = r4 = 22.78125 ÷ 4.5 = 5.0625💡 Why the indices work like this: when you divide two powers of the same base, you subtract the exponents: xm ÷ xn = xm−n. This is because xm means x multiplied by itself m times, and dividing by xn cancels n of those copies, leaving m − n. Here the a's cancel completely (same power, so a1 ÷ a1 = a0 = 1) and the r's reduce as r6 ÷ r2 = r6−2 = r4.
To then go from r4 = 5.0625 back to r, you take the 4th root — and taking an nth root is exactly the same operation as raising to the power 1/n, since (x1/n)n = x. Sanity check with easy numbers: 161/2 = 4 because 42 = 16 — the square root is just the power-½ rule in action. So r4 = 5.0625 gives r = 5.06251/4, undoing the power of 4.
To then go from r4 = 5.0625 back to r, you take the 4th root — and taking an nth root is exactly the same operation as raising to the power 1/n, since (x1/n)n = x. Sanity check with easy numbers: 161/2 = 4 because 42 = 16 — the square root is just the power-½ rule in action. So r4 = 5.0625 gives r = 5.06251/4, undoing the power of 4.
Solve
r = 5.06251/4 = 1.5Back-sub
a = 4.5 ÷ r2 = 4.5 ÷ 2.25 = 2Answer r = 1.5, a = 2
📐Worked Example 3 — Which term first exceeds a value? (logarithms)Type: Logarithmic
Question A GP has a = 100 and r = 1.12. Find the first term to exceed 400.
Set up
100 × 1.12n−1 > 400Simplify
1.12n−1 > 4💡 Converting exponential form to log form: whenever you have basepower = answer, taking logs rearranges it into logbase(answer) = power — the log undoes the power the same way division undoes multiplication. Here 1.12n−1 = 4 has base = 1.12, power = n − 1, answer = 4, so it rearranges to log1.12(4) = n − 1. Calculators don't have a "log base 1.12" button, so in practice this is evaluated as ln 4 ÷ ln 1.12 (or log 4 ÷ log 1.12) — any matching base works on both top and bottom.
Take logs
(n − 1) ln 1.12 > ln 4Divide
n − 1 > 1.3863 ÷ 0.11333Simplify
n − 1 > 12.23Interpret n − 1 must be a whole number ≥ 13, so n = 14. Check: a14 = 100 × 1.1213 ≈ 436.3 ✓ (a13 ≈ 389.6 ✗)
Answer The 14th term is the first to exceed 400.
⚙️Worked Example 4 — Engineering context (decay)Type: Engineering
Question After a shock load, a machine mounting vibrates with initial amplitude 12.0 mm. Damping reduces the amplitude by 18% on every complete oscillation. Find the amplitude after 10 oscillations.
Model Losing 18% each oscillation means keeping 82%:
r = 0.82. The starting amplitude, before any oscillations, is the 1st term: a = 12.0💡 Careful with the term count: a is the 1st term — it hasn't been damped by any oscillation yet. After 1 oscillation the amplitude is the 2nd term, after 2 oscillations the 3rd term, and so on: after k oscillations, the amplitude is the (k + 1)th term. So "after 10 oscillations" means n = 10 + 1 = 11.
Apply the nth term
a11 = ar11−1 = 12 × 0.8210Calculate
0.8210 = 0.137448…Substitute
12 × 0.137448…Answer ≈ 1.65 mm — damping never removes vibration completely, but drives it towards zero geometrically.
🔧GP Explorer — growth and decay in actionEnter values, then press Generate
Enter a, r and n above, then press Generate to watch the GP grow or decay and see the nth term worked out…
📊See it grow or decay — drag the slidersLive diagram, updates instantly
👁️ Watch the bars: each one is r times the last. r > 1 → bars accelerate upward (growth). r < 1 → bars shrink towards zero (decay).
🧠 Common slip: "reduces by 18%" means r = 0.82, not r = 0.18. Always convert the percentage change into the multiplier that remains.
∑
Sums of Series
Adding the terms of a progression — including the remarkable sum to infinity.
🔁 Same formula, two labels: multiplying top and bottom of a(1 − rn)÷(1 − r) by −1 gives a(rn − 1)÷(r − 1) — they always produce the same value of Sn. Most people simply pick whichever version keeps both brackets positive for their r, so the working looks tidier and sign slips are less likely.
📖 Key Vocabulary
Series — the sum of the terms of a sequence (sequence: 2, 4, 6; series: 2 + 4 + 6).
Partial sum (Sn) — the total of the first n terms only.
Convergent — a series whose partial sums approach a finite limit (GP with |r| < 1).
Divergent — a series whose partial sums grow without limit (every AP with d ≠ 0, every GP with |r| ≥ 1).
🧠 Where the AP formula comes from: write the series forwards and backwards, then add. Every pair sums to the same value [2a + (n − 1)d], and there are n pairs counted twice — hence the ½n. Gauss reputedly used this at school to sum 1 to 100 in seconds: ½ × 100 × 101 = 5050.
👁️ Visual aid: each partial sum climbs closer to the ceiling S∞ but never crosses it — the defining behaviour of a convergent series.
✏️Worked Example 1 — Sum of an APType: AP sum
Question Find the sum of the first 25 terms of the AP 6, 10, 14, 18, …
Identify
a = 6, d = 4, n = 25Formula
S25 = ½(25)[2(6) + (24)(4)] = 12.5 × [12 + 96]Calculate
12.5 × 108Answer S25 = 1350
🧮Worked Example 2 — Sum of a GPType: GP sum
Question Find the sum of the first 8 terms of the GP 5, 15, 45, …, giving the exact value.
Identify
a = 5, r = 3, n = 8. Since r > 1 it is tidier to use Sn = a(rn − 1)/(r − 1)Substitute
S8 = 5(38 − 1)/(3 − 1) = 5(6561 − 1)/2Calculate
5 × 6560 ÷ 2Answer S8 = 16 400
♾️Worked Example 3 — Sum to infinityType: Convergence
Question A GP has first term 24 and common ratio −⅓. Explain why S∞ exists and find it.
Condition
|r| = ⅓ < 1, so rn → 0 as n → ∞ and the series converges.Formula
S∞ = a/(1 − r) = 24/(1 − (−⅓)) = 24 ÷ (4/3)Answer S∞ = 18. The alternating terms overshoot and undershoot 18, closing in on it forever.
⚙️Worked Example 4 — Engineering context (production planning)Type: Engineering
Question A fabrication line produces 2000 brackets in its first month. Output grows by 6% per month as processes are refined. How many brackets in total are produced in the first year?
Model Monthly outputs form a GP:
a = 2000, r = 1.06, n = 12Formula
S12 = 2000(1.0612 − 1)/(1.06 − 1)Calculate
1.0612 = 2.012196…Substitute
S12 = 2000 × 1.012196 ÷ 0.06Answer ≈ 33 740 brackets (to 4 s.f.) over the first year.
🔧Series Summer — total any AP or GPChoose a type, enter values, press Sum
Choose AP or GP, enter the values, then press Sum. For a GP with |r| < 1 the tool also shows the sum to infinity for comparison…
⚠️ Convergence check first: writing S∞ = a/(1 − r) when |r| ≥ 1 is meaningless — the series diverges. Always state |r| < 1 before using the sum to infinity.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🏗️ Straight-Line Depreciation (AP)
A machine loses the same amount of value every year — an arithmetic progression with negative d. Accountants call this straight-line depreciation.
Value after n years = P − n × D
📟 Try it — Straight-Line Value
Enter price, annual loss and years, then press Calculate…
📉 Reducing-Balance Depreciation (GP)
More realistic: the machine loses a fixed percentage each year — a geometric progression. The value decays but never quite reaches zero.
Value after n years = P × (1 − p/100)ⁿ
📟 Try it — Reducing-Balance Value
Enter price, percentage loss and years, then press Calculate…
🔩 Stacked Pipe Layers (AP series)
Pipes stacked in a triangular pile: each layer holds one fewer pipe than the one below. The total is the sum of an AP with d = −1.
Total = ½n[2a + (n − 1)(−1)] , a = pipes in bottom layer
📟 Try it — Pipe Stack Total
Enter the bottom-layer count and number of layers, then press Calculate…
🌡️ Heat-Treatment Cooling Steps (GP + S∞)
A furnace controller drops the temperature difference above ambient by a fixed fraction each stage. The differences form a GP — and the sum to infinity tells you the total temperature drop that is ever achievable.
Total possible drop = first drop ÷ (1 − r)
📟 Try it — Cooling Stage Analysis
Enter the first temperature drop, the ratio and stage count, then press Calculate…
📜
History & Why It Matters
Progressions are one of the oldest ideas in mathematics — and still one of the most useful in modern engineering.
🗺️A pattern spotted thousands of years ago
Long before algebra had symbols, people noticed that some sequences of numbers grow by a fixed amount each step, while others grow by a fixed multiple. Clay tablets, papyri and manuscripts from very different civilisations all show the same two ideas turning up again and again — because they describe two of the simplest ways anything in the real world can change over time.
➕Arithmetic Progressions: adding up a crowd
🧮 The story: a young Carl Friedrich Gauss was set the task of adding every whole number from 1 to 100. Instead of adding them one at a time, he paired the first and last (1 + 100 = 101), the second and second-last (2 + 99 = 101), and so on — 50 pairs, each summing to 101, giving 5050 in seconds.
That pairing trick is the arithmetic series formula in disguise: Sn = ½n[2a + (n − 1)d]. Long before Gauss, Babylonian scribes (around 300 BCE) were already using similar pairing methods to total up equally-spaced quantities of grain and land.
✖️Geometric Progressions: the power of doubling
🌾 The story: legend says a ruler offered a reward to the inventor of chess. The inventor asked only for wheat: 1 grain on the first square, 2 on the second, 4 on the third — doubling all the way to the 64th square. The ruler laughed, then discovered the total (2⁶⁴ − 1 grains) was more wheat than the whole kingdom could ever grow.
This is exactly a GP with a = 1, r = 2, and the story is still used today to teach why exponential growth outpaces intuition. The same mathematics — recorded independently in India, the Middle East and medieval Europe — later became the basis for compound interest calculations from the 1600s onward.
🔧 Why this still matters on the factory floor
Every engineering example in this unit — tool wear, machine depreciation, vibration decay, production planning — is really just one of these two ancient patterns wearing modern overalls. Recognising which one you're looking at tells you immediately which formula to reach for:
Constant change per step (mm removed, £ lost, extra units made) → arithmetic.
Constant percentage change per step (% depreciation, % damping, % growth) → geometric.
Constant percentage change per step (% depreciation, % damping, % growth) → geometric.
Understanding why the formulas work — not just memorising them — is what lets you spot a progression hiding inside an unfamiliar problem, whether it's a 2,000-year-old riddle about wheat or next quarter's depreciation schedule.
🧠 Worth remembering: the chessboard legend is a useful gut-check whenever a problem involves repeated percentage change — exponential growth always looks small at first and enormous by the end. That "slow, then sudden" shape is the signature of a GP with r > 1.
🎮
Quick-Fire Progressions Game
12 questions — APs, GPs and series mixed. How many can you get?
⚡ Progressions Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Write down the common difference and the 15th term of the AP 4, 11, 18, 25, …
d = 11 − 4 = 7
a₁₅ = 4 + (14)(7) = 4 + 98 = 102
a₁₅ = 4 + (14)(7) = 4 + 98 = 102
Question 2 🥉2 marks
Find the 7th term of the GP 640, 320, 160, …
a = 640, r = 320 ÷ 640 = 0.5
a₇ = 640 × 0.5⁶ = 640 ÷ 64 = 10
a₇ = 640 × 0.5⁶ = 640 ÷ 64 = 10
Question 3 🥉2 marks
Find the sum of the first 30 terms of the AP 2, 5, 8, …
S₃₀ = ½(30)[2(2) + (29)(3)]
= 15 × [4 + 87] = 15 × 91 = 1365
= 15 × [4 + 87] = 15 × 91 = 1365
Question 4 🥉2 marks
State whether the series 81 + 27 + 9 + … converges, and if so find its sum to infinity.
r = 27 ÷ 81 = ⅓, |r| < 1 so it converges.
S∞ = 81 ÷ (1 − ⅓) = 81 ÷ (2/3) = 121.5
S∞ = 81 ÷ (1 − ⅓) = 81 ÷ (2/3) = 121.5
Question 5 🥈3 marks
Which term of the AP 9, 16, 23, … is equal to 373?
9 + (n − 1)(7) = 373
(n − 1)(7) = 364
n − 1 = 52
n = 53 (the 53rd term)
(n − 1)(7) = 364
n − 1 = 52
n = 53 (the 53rd term)
Question 6 🥈4 marks
The 4th term of an AP is 22 and the 11th term is 57. Find a and d, and hence S₂₀.
a + 3d = 22, a + 10d = 57
7d = 35
d = 5, a = 22 − 15 = 7
S₂₀ = ½(20)[2(7) + 19(5)] = 10 × [14 + 95] = 1090
7d = 35
d = 5, a = 22 − 15 = 7
S₂₀ = ½(20)[2(7) + 19(5)] = 10 × [14 + 95] = 1090
Question 7 🥈3 marks
Find the sum of the first 10 terms of the GP 4, 6, 9, …, to 4 significant figures.
a = 4, r = 1.5
S₁₀ = 4(1.5¹⁰ − 1)/(1.5 − 1) = 4(57.665 − 1)/0.5
= 8 × 56.665 = 453.3 (4 s.f.)
S₁₀ = 4(1.5¹⁰ − 1)/(1.5 − 1) = 4(57.665 − 1)/0.5
= 8 × 56.665 = 453.3 (4 s.f.)
Question 8 🥈4 marks
A lathe costing £48 000 depreciates at 12% per year (reducing balance). Find its value after 6 years, to the nearest £100.
Multiplier per year = 0.88
Value = 48000 × 0.88⁶ = 48000 × 0.464404…
= £22 291.41 ≈ £22 300
Value = 48000 × 0.88⁶ = 48000 × 0.464404…
= £22 291.41 ≈ £22 300
Question 9 🥇4 marks
The 2nd term of a GP is 12 and the 5th term is 324. Find r, a and the 8th term.
ar⁴ ÷ ar = r³ = 324 ÷ 12 = 27
r = 3
a = 12 ÷ 3 = 4
a₈ = 4 × 3⁷ = 4 × 2187 = 8748
r = 3
a = 12 ÷ 3 = 4
a₈ = 4 × 3⁷ = 4 × 2187 = 8748
Question 10 🥇5 marks
A compressor's output grows by 8% per year from 5000 units in year 1. Using logarithms, find the first year in which annual output exceeds 10 000 units.
5000 × 1.08ⁿ⁻¹ > 10000
1.08ⁿ⁻¹ > 2
(n − 1) ln 1.08 > ln 2
n − 1 > 0.6931 ÷ 0.07696 = 9.006
n − 1 = 10
n = 11
Check: year 10: 5000 × 1.08⁹ ≈ 9995 ✗; year 11: ≈ 10 794 ✓
Year 11
1.08ⁿ⁻¹ > 2
(n − 1) ln 1.08 > ln 2
n − 1 > 0.6931 ÷ 0.07696 = 9.006
n − 1 = 10
n = 11
Check: year 10: 5000 × 1.08⁹ ≈ 9995 ✗; year 11: ≈ 10 794 ✓
Year 11
Question 11 🥇4 marks
Drill sizes in a metric set form a GP from 1.00 mm to 8.00 mm in 10 sizes (10 terms). Find the common ratio to 4 d.p. and the 6th drill size to 3 s.f.
ar⁹ = 8 with a = 1
r⁹ = 8
r = 8^(1/9) = 2^(3/9) = 2^(1/3) = 1.2599
a₆ = 1 × r⁵ = 2^(5/3) = 3.17 mm
r⁹ = 8
r = 8^(1/9) = 2^(3/9) = 2^(1/3) = 1.2599
a₆ = 1 × r⁵ = 2^(5/3) = 3.17 mm
Question 12 🥇5 marks
A damped system's successive swing distances form a GP: first swing 500 mm, each swing 72% of the previous one. Find (a) the total distance travelled in the first 8 swings, (b) the total distance it can ever travel.
(a) S₈ = 500(1 − 0.72⁸)/(1 − 0.72)
0.72⁸ = 0.072222…
S₈ = 500 × 0.927778 ÷ 0.28 = 1656.7 mm ≈ 1660 mm (3 s.f.)
(b) |r| < 1 so S∞ = 500 ÷ 0.28 = 1785.7 mm ≈ 1790 mm (3 s.f.)
0.72⁸ = 0.072222…
S₈ = 500 × 0.927778 ÷ 0.28 = 1656.7 mm ≈ 1660 mm (3 s.f.)
(b) |r| < 1 so S∞ = 500 ÷ 0.28 = 1785.7 mm ≈ 1790 mm (3 s.f.)