Lesson 2 of 12
16% complete
1. Arithmetic Sequences and Series P1
Arithmetic Sequence (AP): A sequence where consecutive terms differ by a fixed constant d, called the common difference. The general term and sum are determined entirely by the first term a and the common difference d.
Arithmetic Sequence Formulae
nth term: uₙ = a + (n−1)d
Sum of n terms: Sₙ = n/2 [2a + (n−1)d] = n/2 (a + l)
where a = first term, d = common difference, l = last term
📐 Worked Example 1 — Arithmetic Series
The 5th term of an AP is 23 and the 12th term is 58. Find the first term, common difference, and the sum of the first 20 terms.
1
u₅ = a+4d = 23 ...(1) u₁₂ = a+11d = 58 ...(2)
Subtract (1) from (2): 7d = 35 → d = 5
Subtract (1) from (2): 7d = 35 → d = 5
2
From (1): a + 20 = 23 → a = 3
3
S₂₀ = 20/2 [2(3) + 19(5)] = 10[6 + 95] = 10 × 101 = 1010
📐 Worked Example 2 — Sum Greater Than a Value
An AP has first term 4 and common difference 3. Find the least value of n for which Sₙ > 500.
1
Sₙ = n/2[8+(n−1)×3] = n/2(3n+5) > 500
3n²+5n > 1000 → 3n²+5n−1000 > 0
3n²+5n > 1000 → 3n²+5n−1000 > 0
2
Using quadratic formula: n = (−5+√(25+12000))/6 = (−5+√12025)/6 = (−5+109.7)/6 ≈ 17.4
Since n must be a positive integer and the inequality is strict: n = 18
Since n must be a positive integer and the inequality is strict: n = 18
2. Geometric Sequences and Series P1
Geometric Sequence (GP): A sequence where each term is obtained by multiplying the previous term by a fixed constant r, called the common ratio. uₙ₊₁/uₙ = r for all n.
Geometric Sequence Formulae
nth term: uₙ = arⁿ⁻¹
Sum of n terms: Sₙ = a(1−rⁿ)/(1−r) for r ≠ 1
Sum of n terms: Sₙ = a(rⁿ−1)/(r−1) for r > 1 (alternative form)
Sum to infinity: S∞ = a/(1−r) valid only when |r| < 1
Condition for Sum to Infinity: The series converges (S∞ exists) if and only if |r| < 1. If |r| ≥ 1, the series diverges — no sum to infinity. This is a key condition that must be stated in exam answers.
📐 Worked Example 3 — Geometric Series
A GP has first term 12 and common ratio 2/3. (a) Find the 6th term. (b) Find S∞. (c) Find the smallest n such that Sₙ > 95% of S∞.
1
(a) u₆ = 12 × (2/3)⁵ = 12 × 32/243 = 384/243 = 128/81
2
(b) |r|=2/3 < 1, so S∞ exists.
S∞ = 12/(1−2/3) = 12/(1/3) = 36
S∞ = 12/(1−2/3) = 12/(1/3) = 36
3
(c) Need Sₙ > 0.95 × 36 = 34.2
Sₙ = 12(1−(2/3)ⁿ)/(1/3) = 36(1−(2/3)ⁿ) > 34.2
1−(2/3)ⁿ > 0.95 → (2/3)ⁿ < 0.05
n ln(2/3) < ln(0.05) → n > ln(0.05)/ln(2/3) = −2.996/(−0.405) ≈ 7.40
n = 8
Sₙ = 12(1−(2/3)ⁿ)/(1/3) = 36(1−(2/3)ⁿ) > 34.2
1−(2/3)ⁿ > 0.95 → (2/3)ⁿ < 0.05
n ln(2/3) < ln(0.05) → n > ln(0.05)/ln(2/3) = −2.996/(−0.405) ≈ 7.40
n = 8
📐 Worked Example 4 — Finding a and r
The second and fifth terms of a GP are 6 and −48 respectively. Find the first term, common ratio, and the sum of the first 8 terms.
1
u₂ = ar = 6 ...(1) u₅ = ar⁴ = −48 ...(2)
Divide (2) by (1): r³ = −8 → r = −2
Divide (2) by (1): r³ = −8 → r = −2
2
From (1): a(−2) = 6 → a = −3
3
S₈ = −3(1−(−2)⁸)/(1−(−2)) = −3(1−256)/3 = −3×(−255)/3 = 255
Increasing, Decreasing, and Periodic Sequences
| Type | AP Condition | GP Condition |
|---|---|---|
| Increasing | d > 0 | a > 0 and r > 1, or a < 0 and 0 < r < 1 |
| Decreasing | d < 0 | a > 0 and 0 < r < 1, or a < 0 and r > 1 |
| Periodic | d = 0 (constant) | r = 1 (constant) or r = −1 (alternating) |
| Convergent (S∞) | Never (diverges) | |r| < 1 |
3. The Binomial Theorem P1
Binomial Expansion — Positive Integer n
(a+b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ (r = 0 to n)
= aⁿ + n·aⁿ⁻¹b + n(n−1)/2! · aⁿ⁻²b² + ... + bⁿ
General term: T_{r+1} = ⁿCᵣ · aⁿ⁻ʳ · bʳ
ⁿCᵣ = n! / [r!(n−r)!]
📐 Worked Example 5 — Binomial Expansion
Expand (2x − 3)⁴ fully. Hence find the coefficient of x³.
1
a=2x, b=−3, n=4. T_{r+1} = ⁴Cᵣ(2x)⁴⁻ʳ(−3)ʳ
2
r=0: ⁴C₀(2x)⁴(−3)⁰ = 16x⁴
r=1: ⁴C₁(2x)³(−3)¹ = 4×8x³×(−3) = −96x³
r=2: ⁴C₂(2x)²(−3)² = 6×4x²×9 = 216x²
r=3: ⁴C₃(2x)¹(−3)³ = 4×2x×(−27) = −216x
r=4: ⁴C₄(2x)⁰(−3)⁴ = 81
r=1: ⁴C₁(2x)³(−3)¹ = 4×8x³×(−3) = −96x³
r=2: ⁴C₂(2x)²(−3)² = 6×4x²×9 = 216x²
r=3: ⁴C₃(2x)¹(−3)³ = 4×2x×(−27) = −216x
r=4: ⁴C₄(2x)⁰(−3)⁴ = 81
3
(2x−3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81
Coefficient of x³ = −96
Coefficient of x³ = −96
📐 Worked Example 6 — Finding a Specific Term
Find the term independent of x in the expansion of (x² + 2/x)⁹.
1
General term: T_{r+1} = ⁹Cᵣ(x²)⁹⁻ʳ(2/x)ʳ = ⁹Cᵣ · 2ʳ · x^(18−2r) · x^(−r) = ⁹Cᵣ · 2ʳ · x^(18−3r)
2
For term independent of x: power of x = 0
18 − 3r = 0 → r = 6
18 − 3r = 0 → r = 6
3
T₇ = ⁹C₆ · 2⁶ = 84 × 64 = 5376
📐 Worked Example 7 — Using Two Expansions
The coefficient of x² in the expansion of (1+ax)⁶ + (1−2x)⁴ is 93. Find a.
1
Coefficient of x² in (1+ax)⁶: ⁶C₂(ax)² = 15a²
2
Coefficient of x² in (1−2x)⁴: ⁴C₂(−2x)² = 6×4 = 24
3
15a² + 24 = 93 → 15a² = 69 → a² = 69/15 = 23/5
a = ±√(23/5)
a = ±√(23/5)
4. Radians, Arcs, and Sectors P1
Radian: The angle subtended at the centre of a circle by an arc equal in length to the radius. 2π radians = 360°. All A Level calculus with trigonometry requires radians.
Radian Measure — Key Conversions
π radians = 180° → 1 radian = 180°/π ≈ 57.3°
Key values: π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°, π = 180°
Arc length: s = rθ (θ in radians)
Area of sector: A = ½r²θ (θ in radians)
Area of segment = Area of sector − Area of triangle = ½r²θ − ½r²sinθ = ½r²(θ − sinθ)
📐 Worked Example 8 — Arc, Sector, and Segment
A sector of a circle has radius 8 cm and area 20 cm². Find the angle θ (in radians), the arc length, and the area of the corresponding segment.
1
Area of sector = ½r²θ = ½×64×θ = 32θ = 20
θ = 20/32 = 5/8 radians
θ = 20/32 = 5/8 radians
2
Arc length s = rθ = 8 × 5/8 = 5 cm
3
Area of segment = ½r²(θ−sinθ) = ½×64×(5/8−sin(5/8))
sin(5/8) = sin(0.625) ≈ 0.5847
= 32×(0.625−0.5847) = 32×0.0403 ≈ 1.29 cm²
sin(5/8) = sin(0.625) ≈ 0.5847
= 32×(0.625−0.5847) = 32×0.0403 ≈ 1.29 cm²
5. Trigonometric Identities and Equations P1
Exact Values — Must Know
sin 0 = 0, sin π/6 = ½, sin π/4 = √2/2, sin π/3 = √3/2, sin π/2 = 1
cos 0 = 1, cos π/6 = √3/2, cos π/4 = √2/2, cos π/3 = ½, cos π/2 = 0
tan 0 = 0, tan π/6 = 1/√3, tan π/4 = 1, tan π/3 = √3, tan π/2 = undefined
Fundamental Trigonometric Identities
sin²θ + cos²θ ≡ 1
tan θ ≡ sin θ / cos θ
1 + tan²θ ≡ sec²θ (divide sin²θ+cos²θ=1 by cos²θ)
1 + cot²θ ≡ cosec²θ (divide sin²θ+cos²θ=1 by sin²θ)
Reciprocals: sec θ = 1/cosθ, cosec θ = 1/sinθ, cot θ = cosθ/sinθ
Graphs of Trigonometric Functions
| Function | Period | Amplitude | Domain | Range |
|---|---|---|---|---|
| y = sin x | 2π | 1 | ℝ | [−1, 1] |
| y = cos x | 2π | 1 | ℝ | [−1, 1] |
| y = tan x | π | ∞ | x ≠ π/2+nπ | ℝ |
| y = a sin(bx+c)+d | 2π/b | a | ℝ | [d−a, d+a] |
Solving Trigonometric Equations
Strategy for Solving Trig Equations in [0, 2π]:
1. Use identities to reduce to a single trig function.
2. Find the principal value α using inverse trig.
3. Use CAST or symmetry to find all solutions in the given interval:
• sin θ = k: θ = α and θ = π − α (in [0, 2π])
• cos θ = k: θ = α and θ = 2π − α (in [0, 2π])
• tan θ = k: θ = α and θ = π + α (in [0, 2π])
4. Adjust for compound angles — solve for the inner angle, then find x.
1. Use identities to reduce to a single trig function.
2. Find the principal value α using inverse trig.
3. Use CAST or symmetry to find all solutions in the given interval:
• sin θ = k: θ = α and θ = π − α (in [0, 2π])
• cos θ = k: θ = α and θ = 2π − α (in [0, 2π])
• tan θ = k: θ = α and θ = π + α (in [0, 2π])
4. Adjust for compound angles — solve for the inner angle, then find x.
📐 Worked Example 9 — Trig Equation Using Identity
Solve 2sin²θ − cosθ − 1 = 0 for 0 ≤ θ ≤ 2π.
1
Replace sin²θ = 1−cos²θ:
2(1−cos²θ)−cosθ−1=0 → 2−2cos²θ−cosθ−1=0
−2cos²θ−cosθ+1=0 → 2cos²θ+cosθ−1=0
2(1−cos²θ)−cosθ−1=0 → 2−2cos²θ−cosθ−1=0
−2cos²θ−cosθ+1=0 → 2cos²θ+cosθ−1=0
2
Factorise: (2cosθ−1)(cosθ+1)=0
cosθ = ½ or cosθ = −1
cosθ = ½ or cosθ = −1
3
cosθ = ½: θ = π/3 and θ = 5π/3
cosθ = −1: θ = π
θ = π/3, π, 5π/3
cosθ = −1: θ = π
θ = π/3, π, 5π/3
📐 Worked Example 10 — Compound Angle Equation
Solve sin(2x + π/6) = 0.5 for 0 ≤ x ≤ 2π.
1
Let φ = 2x + π/6. When x ∈ [0, 2π], φ ∈ [π/6, 4π + π/6] = [π/6, 25π/6].
2
sinφ = 0.5 → principal value = π/6.
In [π/6, 25π/6]: φ = π/6, 5π/6, π/6+2π=13π/6, 5π/6+2π=17π/6
In [π/6, 25π/6]: φ = π/6, 5π/6, π/6+2π=13π/6, 5π/6+2π=17π/6
3
x = (φ − π/6)/2:
φ=π/6: x=(π/6−π/6)/2=0
φ=5π/6: x=(5π/6−π/6)/2=π/3
φ=13π/6: x=(13π/6−π/6)/2=π
φ=17π/6: x=(17π/6−π/6)/2=4π/3
x = 0, π/3, π, 4π/3
φ=π/6: x=(π/6−π/6)/2=0
φ=5π/6: x=(5π/6−π/6)/2=π/3
φ=13π/6: x=(13π/6−π/6)/2=π
φ=17π/6: x=(17π/6−π/6)/2=4π/3
x = 0, π/3, π, 4π/3
📐 Worked Example 11 — Proving a Trig Identity
Prove: (1 + sinθ)/cosθ + cosθ/(1 + sinθ) ≡ 2secθ
1
LHS: Combine over a common denominator: cosθ(1+sinθ):
= [(1+sinθ)² + cos²θ] / [cosθ(1+sinθ)]
= [(1+sinθ)² + cos²θ] / [cosθ(1+sinθ)]
2
Expand numerator: 1+2sinθ+sin²θ+cos²θ = 1+2sinθ+1 = 2+2sinθ = 2(1+sinθ)
3
= 2(1+sinθ)/[cosθ(1+sinθ)] = 2/cosθ = 2secθ = RHS ✓
Inverse Trigonometric Functions
Restricted Domains for Inverse Trig: Since sin, cos, and tan are many-to-one, their inverses exist only on restricted domains.
| Function | Domain (input) | Range (output) |
|---|---|---|
| arcsin x (sin⁻¹x) | −1 ≤ x ≤ 1 | −π/2 ≤ y ≤ π/2 |
| arccos x (cos⁻¹x) | −1 ≤ x ≤ 1 | 0 ≤ y ≤ π |
| arctan x (tan⁻¹x) | x ∈ ℝ | −π/2 < y < π/2 |
6. Harder AP/GP Problems P1
📐 Worked Example 12 — Mixed AP and GP
The first three terms of an AP are also three consecutive terms of a GP. The AP has first term 1 and common difference d ≠ 0. The three terms are 1, 1+d, 1+2d. Find the common ratio of the GP.
1
For a GP, the ratio between consecutive terms must be constant:
(1+d)/1 = (1+2d)/(1+d)
(1+d)/1 = (1+2d)/(1+d)
2
(1+d)² = 1+2d → 1+2d+d² = 1+2d → d² = 0
This gives d=0, which contradicts d ≠ 0. Revisit: the problem means the AP terms (in some other positions) form a GP — typically stated as three specific terms. Standard version: if 1, 1+d, 4 are in GP:
This gives d=0, which contradicts d ≠ 0. Revisit: the problem means the AP terms (in some other positions) form a GP — typically stated as three specific terms. Standard version: if 1, 1+d, 4 are in GP:
3
For the standard problem type: three terms a, b, c in GP means b² = ac (geometric mean condition). Always apply b² = ac when told three quantities form a GP.
📐 Worked Example 13 — Geometric Mean Condition
x−1, x+1, and 2x+3 are the first three terms of a GP. Find x and the common ratio.
1
For a GP: b² = ac → (x+1)² = (x−1)(2x+3)
2
x²+2x+1 = 2x²+x−3 → x²−x−4 = 0... wait:
x²+2x+1 = 2x²+3x−2x−3 = 2x²+x−3
0 = x²−x−4 → x = (1±√17)/2
x²+2x+1 = 2x²+3x−2x−3 = 2x²+x−3
0 = x²−x−4 → x = (1±√17)/2
3
r = (x+1)/(x−1). For x=(1+√17)/2: r = ((3+√17)/2)/((√17−1)/2) = (3+√17)/(√17−1)
Rationalise: = (3+√17)(√17+1)/16 = (3√17+3+17+√17)/16 = (20+4√17)/16 = (5+√17)/4
Rationalise: = (3+√17)(√17+1)/16 = (3√17+3+17+√17)/16 = (20+4√17)/16 = (5+√17)/4
📝 Exam Practice Questions
Q1 [4 marks] — An AP has third term 11 and eighth term 31. Find the first term, common difference, and the sum of the first 15 terms.
u₃=a+2d=11 ...(1), u₈=a+7d=31 ...(2)
(2)−(1): 5d=20 → d=4
From (1): a=11−8=3
S₁₅=15/2[6+14×4]=15/2×62=465
(2)−(1): 5d=20 → d=4
From (1): a=11−8=3
S₁₅=15/2[6+14×4]=15/2×62=465
Q2 [4 marks] — A GP has first term 5 and common ratio r. The sum of the first four terms is 4 times the first term. Find r (r ≠ 1) and the sum to infinity if it exists.
S₄ = 5(1−r⁴)/(1−r) = 20
(1−r⁴)/(1−r) = 4 → (1−r²)(1+r²)/(1−r) = 4
(1+r)(1+r²) = 4 → r³+r²+r+1=4 → r³+r²+r−3=0
Try r=1: 1+1+1−3=0 ✓ but r≠1. Factor: (r−1)(r²+2r+3)=0
r²+2r+3=0: Δ=4−12<0, no real roots besides r=1.
Re-examine: S₄=5+5r+5r²+5r³=4×5=20 → 1+r+r²+r³=4
Try r=1 excluded. Try r=−1: 1−1+1−1=0≠4. Try negatives more carefully.
Clean version: r³+r²+r=3; (r−1)(r²+2r+3)=0 gives only r=1 real.
The condition implies r = 1 is the only real solution (excluded) — the problem as stated has no other real solution. Check: likely intended S₄ = 5 times first term → gives r = (real cubic root).
(1−r⁴)/(1−r) = 4 → (1−r²)(1+r²)/(1−r) = 4
(1+r)(1+r²) = 4 → r³+r²+r+1=4 → r³+r²+r−3=0
Try r=1: 1+1+1−3=0 ✓ but r≠1. Factor: (r−1)(r²+2r+3)=0
r²+2r+3=0: Δ=4−12<0, no real roots besides r=1.
Re-examine: S₄=5+5r+5r²+5r³=4×5=20 → 1+r+r²+r³=4
Try r=1 excluded. Try r=−1: 1−1+1−1=0≠4. Try negatives more carefully.
Clean version: r³+r²+r=3; (r−1)(r²+2r+3)=0 gives only r=1 real.
The condition implies r = 1 is the only real solution (excluded) — the problem as stated has no other real solution. Check: likely intended S₄ = 5 times first term → gives r = (real cubic root).
Exam Tip: When a GP problem yields only r=1 as a real root, re-read the problem — the multiplier given may need re-checking. Always verify your r value by substituting back.
Q3 [4 marks] — Find the coefficient of x³ in the expansion of (3−2x)⁷.
T_{r+1} = ⁷Cᵣ(3)⁷⁻ʳ(−2x)ʳ = ⁷Cᵣ · 3⁷⁻ʳ · (−2)ʳ · xʳ
For x³: r=3
T₄ = ⁷C₃ · 3⁴ · (−2)³ = 35 × 81 × (−8) = −22680
For x³: r=3
T₄ = ⁷C₃ · 3⁴ · (−2)³ = 35 × 81 × (−8) = −22680
Q4 [4 marks] — A sector has perimeter 30 cm and area 54 cm². Find the radius and the angle in radians.
Perimeter = 2r + rθ = r(2+θ) = 30 ...(1)
Area = ½r²θ = 54 ...(2)
From (1): θ = 30/r − 2. Sub into (2):
½r²(30/r − 2) = 54 → 15r − r² = 54 → r²−15r+54=0
(r−6)(r−9)=0 → r=6 or r=9
r=6: θ=30/6−2=3 rad r=9: θ=30/9−2=10/9 rad
Both valid. Two solutions: (r=6, θ=3) and (r=9, θ=10/9).
Area = ½r²θ = 54 ...(2)
From (1): θ = 30/r − 2. Sub into (2):
½r²(30/r − 2) = 54 → 15r − r² = 54 → r²−15r+54=0
(r−6)(r−9)=0 → r=6 or r=9
r=6: θ=30/6−2=3 rad r=9: θ=30/9−2=10/9 rad
Both valid. Two solutions: (r=6, θ=3) and (r=9, θ=10/9).
Q5 [4 marks] — Solve 3tan²θ − tanθ − 2 = 0 for 0 ≤ θ ≤ 2π. Give answers in radians to 3 s.f. where not exact.
Factorise: (3tanθ + 2)(tanθ − 1) = 0
tanθ = 1 or tanθ = −2/3
tanθ = 1: θ = π/4 and π/4+π = 5π/4
tanθ = −2/3: α = arctan(−2/3) = −0.588 rad (principal value, negative → 2nd or 4th quad)
θ = π+(−0.588) = 2.55 rad and θ = 2π+(−0.588) = 5.70 rad
θ = π/4, 2.55, 5π/4, 5.70 radians
tanθ = 1 or tanθ = −2/3
tanθ = 1: θ = π/4 and π/4+π = 5π/4
tanθ = −2/3: α = arctan(−2/3) = −0.588 rad (principal value, negative → 2nd or 4th quad)
θ = π+(−0.588) = 2.55 rad and θ = 2π+(−0.588) = 5.70 rad
θ = π/4, 2.55, 5π/4, 5.70 radians
Q6 [4 marks] — A GP has first term a and common ratio r where r > 0. The sum of the first two terms is 7.2 and the sum to infinity is 12. Find a and r.
S∞ = a/(1−r) = 12 → a = 12(1−r) ...(1)
S₂ = a + ar = a(1+r) = 7.2 ...(2)
Sub (1) into (2): 12(1−r)(1+r) = 7.2
12(1−r²) = 7.2 → 1−r² = 0.6 → r² = 0.4 → r = √0.4 = √(2/5)
Since r>0: r = √(2/5) = √10/5 ≈ 0.632
a = 12(1−√(2/5)) ≈ 12(1−0.632) = a ≈ 4.42
S₂ = a + ar = a(1+r) = 7.2 ...(2)
Sub (1) into (2): 12(1−r)(1+r) = 7.2
12(1−r²) = 7.2 → 1−r² = 0.6 → r² = 0.4 → r = √0.4 = √(2/5)
Since r>0: r = √(2/5) = √10/5 ≈ 0.632
a = 12(1−√(2/5)) ≈ 12(1−0.632) = a ≈ 4.42
Exam Tip: S∞ problems almost always pair with S₂ or Sₙ conditions. Form two equations, substitute to eliminate a, and solve for r first. Then substitute back to find a.