BTEC Level 3 · Further Engineering MathsSequences & Series
🏠 Sequences & Series
A huge amount of engineering maths comes down to spotting a pattern, describing it precisely, and then summing it up. This resource works through the four building blocks in order — arithmetic progressions, geometric progressions, binomial expansion, and power series — each with worked examples before you try problems yourself.
A progression is a sequence with a fixed rule linking each term to the next
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A series is what you get when you add the terms together
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The first term is usually called a, the last terml
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Key Words You'll Meet Throughout
SequenceAn ordered list of numbers
SeriesThe sum of a sequence's terms
Common DifferenceFixed gap between AP terms
Common RatioFixed multiplier between GP terms
ConvergenceA series settling towards a finite total
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Visual: Sequence vs Series
🗺️ How these topics connect
Arithmetic progressions (AP) — terms that change by a constant amount each step (add the same number every time).
Geometric progressions (GP) — terms that change by a constant factor each step (multiply by the same number every time). This is where things like compound depreciation and gear ratios live.
Binomial expansion — a fast way to expand (a + b)ⁿ without multiplying it out term by term, built from Pascal's triangle.
Power series — writing a function as an infinite sum of powers of x, which is how calculators and computers actually evaluate things like eˣ, sin x, cos x, and ln(1+x).
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Arithmetic Progressions
A sequence where you get from one term to the next by adding the same number every time — the common difference, d.
📋 AP — Quick Reference
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General term: a, a+d, a+2d, a+3d, …
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nth term: l = a + (n − 1)d
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Sum to n terms: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
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Key Words — What Do They Mean?
aThe first term
dThe common difference
lThe last (nth) term
SₙSum of the first n terms
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Visual: AP terms increase by a fixed step
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AP Term & Sum Calculator
adn
Enter a, d and n, then click Calculate.
The nth term formula
Starting from a and adding d repeatedly, the nth term is:
l = a + (n − 1)d
✏️ Worked example
Problem: A sequence starts at 7 and has common difference 4. Find the 12th term.
Given: a = 7, d = 4, n = 12
Substitute: l = 7 + (12 − 1)(4) = 7 + 44
Answer: l = 51
The sum of an AP (arithmetic series)
Adding all n terms together gives:
Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
✏️ Worked example
Problem: Find the sum of the first 20 terms of the AP with a = 3, d = 5.
Both versions of the sum formula give the same answer — use whichever one you have the information for. If you know l already, the second form is faster; if you only know d, use the first.
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Geometric Progressions
A sequence where you get from one term to the next by multiplying by the same number every time — the common ratio, r.
📋 GP — Quick Reference
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General term: a, ar, ar², ar³, …
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nth term: arⁿ⁻¹
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Sum to n terms: Sₙ = a(1 − rⁿ)/(1 − r)
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Sum to infinity (only if |r| < 1): S∞ = a/(1 − r)
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Key Words — What Do They Mean?
rThe common ratio
S∞Sum to infinity
ConvergeApproach a fixed finite value
DivergeGrow without settling to a value
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Visual: GP terms shrinking toward the sum to infinity
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GP Term, Sum & Convergence Calculator
arn
Enter a, r and n, then click Calculate.
The nth term formula
nth term = a rⁿ⁻¹
✏️ Worked example
Problem: A GP has first term 5 and common ratio 3. Find the 6th term.
Given: a = 5, r = 3, n = 6
Substitute: T₆ = 5 × 3⁵ = 5 × 243
Answer: T₆ = 1215
The sum of a GP (geometric series)
Sₙ = a(1 − rⁿ) / (1 − r)
✏️ Worked example
Problem: Find the sum of the first 8 terms of the GP with a = 2, r = 1.5.
If you keep adding terms of a GP forever, one of two things happens:
If |r| < 1 (r is between −1 and 1), each term gets smaller and smaller, and the running total settles towards a fixed value — the series converges.
If |r| ≥ 1, the terms don't shrink, and the running total either grows forever or oscillates without settling — the series diverges, and there is no sum to infinity.
S∞ = a / (1 − r) (valid only when |r| < 1)
✏️ Worked example
Problem: Find the sum to infinity of the GP with a = 8, r = 0.25.
Check convergence: |0.25| < 1 ✅ so a sum to infinity exists
Substitute: S∞ = 8 / (1 − 0.25) = 8 / 0.75
Answer: S∞ = 10.67 (2 d.p.)
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Engineering Applications of Sequences & Series
AP and GP thinking shows up constantly on the shop floor and in design calculations — often disguised as a completely different-sounding problem.
📋 Spotting Which Type Applies
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Changes by a fixed amount each step → AP
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Changes by a fixed percentage or factor each step → GP
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"Depreciates by X% per year" is always a GP with r = (1 − X/100)
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Depreciation Calculator (GP model)
Initial value (£)Rate (% per year)Years
Enter values and click Calculate.
Lathe spindle speeds
Machine tool spindle speeds are often set up in a geometric progression, so that each speed step gives a roughly constant percentage change in cutting speed — this keeps the "step size" feeling consistent across the whole range, from slow to fast.
✏️ Worked example
Problem: A lathe has 6 spindle speeds forming a GP. The slowest speed is 40 rpm and the fastest (6th) speed is 1250 rpm. Find the common ratio and the 3rd speed.
Set up: a = 40, T₆ = 1250 = 40 × r⁵
Solve for r: r⁵ = 1250/40 = 31.25 → r = 31.25^(1/5) ≈ 2
3rd speed: T₃ = 40 × 2² = 40 × 4
Answer: r ≈ 2, 3rd speed = 160 rpm
Depreciation of capital equipment
When a machine's value falls by a fixed percentage each year (rather than a fixed amount), its value over time forms a GP, with common ratio r = 1 − (percentage as a decimal).
✏️ Worked example
Problem: A CNC machine costs £80,000 and depreciates at 15% per year. Find its value after 4 years.
Set up: a = 80000, r = 1 − 0.15 = 0.85
Value after 4 years = 5th term (year 0 is the 1st term): T₅ = 80000 × 0.85⁴
Evaluate: 0.85⁴ ≈ 0.5220
Answer: Value ≈ £41,760
Gearbox ratios
A multi-speed gearbox often steps its ratios geometrically for the same reason as lathe spindle speeds — an even "feel" between gears. Deep drilling costs and manufacturing batch estimation problems tend to use AP instead, since setup costs and per-unit costs usually add in fixed amounts.
✏️ Worked example
Problem: Drilling a batch of holes costs a £25 setup fee plus £3.50 per hole. Find the total cost of drilling 40 holes, treating the running total after each hole as an AP.
Set up: This is really just a linear cost — but written as a series: cost after n holes = 25 + 3.5n, so as an AP: a = 28.5 (cost after 1 hole), d = 3.5
Direct approach: Total = 25 + 3.5 × 40 = 25 + 140
Answer: Total = £165
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Binomial Expansion
A fast, systematic way to expand (a + b)ⁿ without multiplying it out term by term — essential for approximating errors and small changes in engineering measurements.
📋 Binomial — Quick Reference
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Coefficients come from Pascal's triangle for positive integer n
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Powers of a decrease, powers of b increase, across the expansion
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For non-integer or negative n, use the general binomial theorem — valid only for −1 < x < 1
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Key Words — What Do They Mean?
BinomialAn expression with two terms, e.g. (a+b)
CoefficientThe number multiplying each term
Pascal's TriangleA triangle of binomial coefficients
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Visual: How the powers shift across the expansion
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Binomial Coefficient Finder
nk
Enter n and k to find the binomial coefficient.
Pascal's triangle
Each row starts and ends with 1; every other number is the sum of the two numbers above it. Row n (starting from row 0) gives the coefficients for (a+b)ⁿ:
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1
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1
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1
1
3
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1
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4
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1
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Rows 0 to 5 — row 4 (1, 4, 6, 4, 1) gives the coefficients for (a+b)⁴
Expanding (a + b)ⁿ for positive integer n
✏️ Worked example
Problem: Expand (2x + 3)⁴.
Coefficients (row 4): 1, 4, 6, 4, 1
Set up terms: 1(2x)⁴(3)⁰ + 4(2x)³(3)¹ + 6(2x)²(3)² + 4(2x)¹(3)³ + 1(2x)⁰(3)⁴
Engineering application: small errors and approximate percentage changes
If a quantity like area or volume depends on a measured length, a small error in the length produces a predictable small error in the result — and the binomial theorem gives a fast way to estimate it without recalculating from scratch.
✏️ Worked example
Problem: The volume of a sphere is V = (4/3)πr³. If r is measured with a possible error of +1%, estimate the percentage error in V.
Set up: New radius = r(1 + 0.01). New volume ∝ r³(1+0.01)³
Answer: A 1% error in r gives roughly a 3% error in V — the power (3) multiplies the percentage error.
This "multiply the percentage error by the power" shortcut comes directly from the nx term in the binomial expansion, and is used constantly for quick error estimates in engineering measurement.
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Power Series
Writing a function as an infinite sum of powers of x — the method calculators actually use internally to evaluate things like eˣ.
📋 Power Series — Quick Reference
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Power series: f(x) = a₀ + a₁x + a₂x² + a₃x³ + …
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Maclaurin series = Taylor series centred at a = 0
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A series only gives useful answers where it converges
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eˣ, sin x and cos x converge for every x; ln(1+x) only converges for −1 < x ≤ 1
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Key Words — What Do They Mean?
Taylor SeriesA series built from a function's derivatives at a point
Maclaurin SeriesA Taylor series centred at x = 0
n!n factorial: n×(n−1)×…×1
Alternating SeriesTerms switch between + and − signs
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Visual: sin x and its Taylor approximations
Using more terms pulls the polynomial closer to the true curve — this is what "convergence" looks like.
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Power Series Approximator
Function
xNumber of terms
Choose a function, enter x and the number of terms, then click Approximate.
Taylor and Maclaurin series
The Taylor series builds up a function from its value and derivatives at a chosen point a:
Every derivative of eˣ is eˣ itself, and e⁰ = 1, so every coefficient in the Maclaurin series is 1/n!:
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + …
✏️ Worked example
Problem: Use the first 5 terms of the series to estimate e (i.e. e¹).
Substitute x = 1: e ≈ 1 + 1 + 1/2 + 1/6 + 1/24
Evaluate: = 1 + 1 + 0.5 + 0.1667 + 0.04167
Answer: e ≈ 2.7083 (true value 2.71828 — already accurate to 2 d.p. with just 5 terms)
Deriving the series for sin x
Differentiating sin x repeatedly cycles through cos x, −sin x, −cos x, sin x, … Evaluating each derivative at x = 0 kills every even-power term (they all evaluate to sin 0 = 0), leaving only odd powers with alternating signs:
sin x = x − x³/3! + x⁵/5! − x⁷/7! + …
✏️ Worked example
Problem: Use the first 3 terms of the series to estimate sin(0.5) (x in radians).
Answer: sin(0.5) ≈ 0.479427 (true value 0.479426 — accurate to 6 d.p. with just 3 terms)
This is why calculators and microcontrollers can evaluate sin and cos without a lookup table — a handful of terms already gives near-exact results for small to moderate x.
Deriving the series for cos x
cos x follows the same derivative cycle as sin x, but starting one step earlier, so it keeps only the even-power terms:
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …
✏️ Worked example
Problem: Use the first 3 terms of the series to estimate cos(0.3) (x in radians).
Answer: cos(0.3) ≈ 0.955338 (true value 0.955336 — accurate to 5 d.p. with just 3 terms)
Deriving the series for ln(1 + x)
Unlike eˣ, sin x and cos x, the natural log series does not converge everywhere — it only works for −1 < x ≤ 1, so it's only useful for values of x close to zero:
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … (valid for −1 < x ≤ 1)
✏️ Worked example
Problem: Use the first 4 terms of the series to estimate ln(1.2), using x = 0.2.
Answer: ln(1.2) ≈ 0.182267 (true value 0.182322 — accurate to 4 d.p. with just 4 terms)
Engineering note: this series links engineering strain to true (logarithmic) strain, ε_true = ln(1 + ε_eng) — for the small strains typical of elastic loading, the first term or two of this series is often accurate enough to use directly.
Convergence and divergence
An infinite series is only useful if it converges — meaning the partial sums settle towards a fixed value as you add more terms. The series for eˣ, sin x and cos x converge for every value of x, which is unusually generous; the series for ln(1+x), like the binomial series from the previous tab, only converges for a restricted range, −1 < x ≤ 1.
In practice, this matters because using a series outside its range of convergence gives a meaningless or wildly inaccurate answer — always check the stated validity range before using a series approximation.
Engineering application: error propagation
Power series are the standard tool for estimating how a small measurement error in one quantity affects a calculated result — exactly the same idea as the binomial shortcut, but applicable to any function, not just powers.
✏️ Worked example
Problem: The resonant frequency of an oscillator is f = 1/(2π√(LC)). If C has a small error of +2%, estimate the resulting percentage error in f.
Set up: f ∝ C^(−1/2), so using the small-error shortcut, %error in f ≈ (−1/2) × %error in C
Substitute: %error in f ≈ (−1/2)(2%)
Answer: f decreases by approximately 1%
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Practice
A mixed set of problems across all four topics — working from straightforward substitution through to multi-step engineering scenarios.
💡 Show your working for every problem. Click "Show answer" to check once you've had a genuine attempt — the value is in the attempt, not the answer.
Q1
An AP has first term 4 and common difference 6. Find the 15th term.
l = a + (n−1)d = 4 + (14)(6) = 4 + 84 = 88
Q2
Find the sum of the first 25 terms of the AP: 2, 9, 16, 23, …
a = 2, d = 7, n = 25 S₂₅ = 25/2 [2(2) + 24(7)] = 12.5 × [4 + 168] = 12.5 × 172 = 2150
Q3
A GP has first term 3 and common ratio 2. Find the 10th term and the sum of the first 10 terms.
Find the sum to infinity of the GP with a = 12, r = 0.4. State why a sum to infinity exists.
|r| = 0.4 < 1, so the series converges and a sum to infinity exists. S∞ = 12/(1−0.4) = 12/0.6 = 20
Q5
A tool's usable value falls by 12% each year. It cost £45,000 new. Find its value after 5 years, to the nearest £100.
a = 45000, r = 0.88, value after 5 years = T₆ = 45000 × 0.88⁵ 0.88⁵ ≈ 0.5277 Value ≈ 45000 × 0.5277 ≈ £23,700
Q6
A production line's daily output forms an AP as new staff are trained in: 40 units on day 1, increasing by 5 units per day. On what day does daily output first exceed 100 units?
a = 40, d = 5. Need a + (n−1)d > 100 40 + 5(n−1) > 100 → 5(n−1) > 60 → n−1 > 12 → n > 13 First day output exceeds 100 is day 14 (giving 40+13×5 = 105)
Expand (1 + x)⁻¹ up to the term in x³ using the binomial theorem, and use it to estimate 1/1.05 with x = 0.05.
(1+x)⁻¹ = 1 − x + x² − x³ + … Substitute x = 0.05: 1 − 0.05 + 0.0025 − 0.000125 = 0.952375 (true value 1/1.05 = 0.952381 — a very close match)
Q9
The kinetic energy of a rotating part is E = ½Iω². If ω (angular velocity) is measured with a possible error of +3%, estimate the percentage error in E.
E ∝ ω², so %error in E ≈ 2 × %error in ω = 2 × 3% = 6%
Q10
Write the Maclaurin series for eˣ up to the term in x⁴, then use it to estimate e^0.5.
eˣ ≈ 1 + x + x²/2 + x³/6 + x⁴/24 Substitute x = 0.5: 1 + 0.5 + 0.125 + 0.02083 + 0.00260 ≈ 1.6484 (true value e^0.5 = 1.64872 — accurate to 3 d.p.)
Q11
A 9-speed gearbox has speeds forming a GP. The lowest speed is 60 rpm and the highest (9th) is 3840 rpm. Find the common ratio and all speeds between the lowest and highest.
A company invests in a machine expected to save £2,000 in its first year of use, with savings increasing by £150 each subsequent year (an AP). Find the total savings over a 10-year period.
a = 2000, d = 150, n = 10 S₁₀ = 10/2 [2(2000) + 9(150)] = 5 × [4000 + 1350] = 5 × 5350 = £26,750
Q13
Use the first 3 terms of the Maclaurin series for sin x to estimate sin(0.2) (x in radians). Compare with the true value.
sin x ≈ x − x³/3! + x⁵/5! Substitute x = 0.2: 0.2 − 0.008/6 + 0.00032/120 = 0.2 − 0.001333 + 0.0000027 ≈ 0.198669 (true value 0.198669 — matches to 6 d.p.)
Q14
Use the first 3 terms of the Maclaurin series for cos x to estimate cos(0.4) (x in radians).
cos x ≈ 1 − x²/2! + x⁴/4! Substitute x = 0.4: 1 − 0.16/2 + 0.0256/24 = 1 − 0.08 + 0.001067 ≈ 0.921067 (true value 0.921061 — accurate to 4 d.p.)
Q15
Use the first 4 terms of the series for ln(1+x) to estimate ln(0.9), using x = −0.1. State why this series is valid to use here.
Valid since −1 < −0.1 ≤ 1. ln(1+x) ≈ x − x²/2 + x³/3 − x⁴/4 Substitute x = −0.1: −0.1 − 0.01/2 − 0.001/3 − 0.0001/4 = −0.1 − 0.005 − 0.000333 − 0.000025 ≈ −0.105358 (true value −0.105361 — accurate to 4 d.p.)
🎯 Working through this set: Q1–4 build the core AP/GP formulas; Q5–6 and Q11–12 apply them to realistic engineering and business scenarios; Q7–10 cover binomial expansion and power series, including error-approximation reasoning that goes beyond simple substitution; Q13–15 extend this to the sin, cos and ln(1+x) Maclaurin series.