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Sequences & Series A-Level Pure 1

Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17

Welcome to Sequences & Series!

Sequences and Series are a cornerstone of A-Level Pure Mathematics. From modelling loan repayments and population growth to summing infinite geometric progressions, the tools in this chapter appear throughout mathematics, science and finance. You will master arithmetic and geometric progressions, sum formulas, convergence, sigma notation, and elegant proofs.

AP: aₙ = a + (n−1)d  |  Sₙ = n/2(2a+(n−1)d)  |  GP: aₙ = arⁿ⁻¹  |  S∞ = a/(1−r), |r|<1

Learning Objectives

  • Identify arithmetic sequences and find the common difference d
  • Apply the nth term formula aₙ = a + (n−1)d
  • Calculate sums of arithmetic series using both sum formulas
  • Identify geometric sequences and find the common ratio r
  • Apply the nth term formula aₙ = arⁿ⁻¹
  • Calculate sums of geometric series using Sₙ = a(1−rⁿ)/(1−r)
  • Determine convergence conditions and calculate S∞ = a/(1−r)
  • Read, write and evaluate expressions in sigma notation
  • Solve mixed problems combining AP and GP techniques
  • Work with recurrence relations and periodic sequences

Arithmetic Sequences

Common difference d, nth term, finding a and d

AP

Sum formulas, pairing method, sum between terms

Geometric Sequences

Common ratio r, nth term, geometric mean

GP

Sum formula, convergence, sum to infinity S∞

Sigma Notation

Reading and writing Σ, evaluating sums, limits

Convergence

|r|<1 condition, finding S∞, unknowns from S∞

Learn 1 — Arithmetic Sequences

An arithmetic sequence (AP) is a sequence in which the difference between any two consecutive terms is constant. This constant is called the common difference d.

Definition and Common Difference

A sequence a₁, a₂, a₃, … is arithmetic if and only if aₙ₊₁ − aₙ = d for all n, where d is a constant.

Examples:
3, 7, 11, 15, …   d = 4 (increasing AP)
20, 17, 14, 11, …   d = −3 (decreasing AP)
5, 5, 5, 5, …   d = 0 (constant sequence — technically AP with d = 0)

To find d, subtract any term from the next: d = a₂ − a₁ = a₃ − a₂. If the differences are not equal, the sequence is NOT arithmetic.

The nth Term Formula

aₙ = a + (n − 1)d

where a is the first term, d is the common difference, and n is the position. This is often also written as a + (n−1)d or a + nd − d.

Example: Find the 20th term of 5, 9, 13, 17, …
a = 5, d = 4, n = 20
a₂₀ = 5 + (20 − 1) × 4 = 5 + 76 = 81
Example: Find the number of terms in 8, 11, 14, …, 95
a = 8, d = 3, aₙ = 95
95 = 8 + (n − 1) × 3
87 = 3(n − 1) → n − 1 = 29 → n = 30

Finding a and d from Two Terms

If you are given two terms of an AP (but not necessarily consecutive), set up two equations in a and d.

Example: The 5th term of an AP is 21 and the 12th term is 56. Find a and d.
a₅ = a + 4d = 21   … (1)
a₁₂ = a + 11d = 56   … (2)
Subtract (1) from (2): 7d = 35 → d = 5
From (1): a = 21 − 20 = 1

Arithmetic Mean

The arithmetic mean of two values p and q is (p + q)/2.
If b is the arithmetic mean of a and c, then b − a = c − b, i.e. a, b, c form an AP.
Example: Insert one arithmetic mean between 7 and 19: mean = (7 + 19)/2 = 13. So 7, 13, 19 is an AP with d = 6.

Increasing and Decreasing Sequences

An AP is increasing when d > 0 and decreasing when d < 0.
The terms grow without bound if d > 0 (or decrease without bound if d < 0) — an AP never converges (unless d = 0).
Always identify a and d clearly before substituting into any formula. A very common error is using n instead of n−1 in the formula. Remember: the FIRST term uses n = 1, giving a₁ = a + 0×d = a. The formula has (n−1), not n.

Learn 2 — Arithmetic Series (Sums)

An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first n terms is denoted Sₙ.

The Sum Formulas

Sₙ = n/2 · (2a + (n−1)d)    or    Sₙ = n/2 · (a + l)

where l is the last (nth) term. Use the second form when you know both the first and last term. Use the first form when you know a, d and n.

Derivation Idea (Pairing Method)

Write Sₙ forwards and backwards, then add:
  Sₙ = a + (a+d) + (a+2d) + … + l
  Sₙ = l + (l−d) + (l−2d) + … + a
Adding: 2Sₙ = n(a + l), since each pair sums to (a + l).
Therefore Sₙ = n/2 · (a + l). Substituting l = a + (n−1)d gives the first form.

Calculating Sₙ

Example: Find the sum of the first 25 terms of 3, 7, 11, …
a = 3, d = 4, n = 25
S₂₅ = 25/2 × (2×3 + 24×4) = 25/2 × (6 + 96) = 25/2 × 102 = 1275
Example: Find S₁₀ for the AP with first term 40 and last term 4.
a = 40, l = 4, n = 10
S₁₀ = 10/2 × (40 + 4) = 5 × 44 = 220

Finding n Given Sₙ

Set up the sum formula equal to the given value. This produces a quadratic in n — solve and take the positive integer root.

Example: For the AP 5, 9, 13, …, find n when Sₙ = 275.
a = 5, d = 4: Sₙ = n/2 × (10 + 4(n−1)) = n/2 × (4n + 6) = n(2n + 3) = 275
2n² + 3n − 275 = 0
Δ = 9 + 2200 = 2209 = 47²
n = (−3 + 47)/4 = 44/4 = 11

Sum Between Two Terms

To find the sum from the mth term to the nth term (inclusive), calculate:
Sₙ − Sₘ₋₁
Example: Find the sum from the 6th to the 15th term of 2, 5, 8, …
a = 2, d = 3
S₁₅ = 15/2 × (4 + 42) = 15/2 × 46 = 345
S₅ = 5/2 × (4 + 12) = 5/2 × 16 = 40
Sum = S₁₅ − S₅ = 305

Applications

Example: A savings plan pays £100 in month 1, increasing by £20 each month. How much in total after 12 months?
a = 100, d = 20, n = 12
S₁₂ = 12/2 × (200 + 11×20) = 6 × 420 = £2520
When solving 2n² + 3n − 275 = 0 type equations, always check both roots — only the positive integer root is valid for n. Also note: Sₙ − Sₘ gives the sum from term (m+1) to term n, NOT from m to n. To include term m, use Sₙ − Sₘ₋₁.

Learn 3 — Geometric Sequences

A geometric sequence (GP) is a sequence in which the ratio between any two consecutive terms is constant. This constant is called the common ratio r.

Definition and Common Ratio

A sequence a₁, a₂, a₃, … is geometric if and only if aₙ₊₁/aₙ = r for all n, where r ≠ 0 is a constant.

Examples:
2, 6, 18, 54, …   r = 3 (increasing GP)
80, 40, 20, 10, …   r = 1/2 (decreasing, converges to 0)
3, −6, 12, −24, …   r = −2 (alternating GP)

To find r, divide any term by the previous one: r = a₂/a₁ = a₃/a₂. If ratios are not equal, the sequence is NOT geometric.

The nth Term Formula

aₙ = arⁿ⁻¹

where a is the first term and r is the common ratio.

Example: Find the 8th term of 3, 6, 12, 24, …
a = 3, r = 2, n = 8
a₈ = 3 × 2⁷ = 3 × 128 = 384
Example: Find n such that the nth term of 5, 10, 20, 40, … equals 640.
a = 5, r = 2: 5 × 2ⁿ⁻¹ = 640 → 2ⁿ⁻¹ = 128 = 2⁷ → n − 1 = 7 → n = 8

Finding r from Two Terms

Example: The 3rd term of a GP is 12 and the 7th term is 192. Find r and a.
a₃ = ar² = 12   … (1)
a₇ = ar⁶ = 192   … (2)
Divide (2) by (1): r⁴ = 16 → r = 2 (taking positive root)
From (1): a = 12/4 = 3

Geometric Mean

The geometric mean of two positive numbers p and q is √(pq).
If b is the geometric mean of a and c, then b/a = c/b, i.e. b² = ac, so a, b, c form a GP.
Example: Geometric mean of 4 and 9: √(4×9) = √36 = 6. So 4, 6, 9 is NOT a GP — check: 6/4 ≠ 9/6. Actually the geometric mean gives the middle term of a 3-term GP: 4, 6, 9 has ratios 1.5 and 1.5, so yes it IS a GP with r = 1.5.

Recognising a GP from Ratios

Always verify a GP by checking at least two consecutive ratios. A common trap: 1, 1, 2, 3, 5, … (Fibonacci) has non-constant ratios — it is NOT a GP. Also: if any term is 0, the sequence cannot be geometric.

Learn 4 — Geometric Series & Convergence

A geometric series is the sum of the terms of a geometric sequence. Under certain conditions it can sum to infinity.

Sum of n Terms of a GP

Sₙ = a(1 − rⁿ) / (1 − r)  (r ≠ 1)    or    Sₙ = a(rⁿ − 1) / (r − 1)  (r > 1)

Both forms are equivalent — use whichever avoids negative numerators. If r = 1, then Sₙ = na.

Example: Find S₆ for the GP 2, 6, 18, …
a = 2, r = 3, n = 6
S₆ = 2(3⁶ − 1)/(3 − 1) = 2(729 − 1)/2 = 728
Example: Find the sum of the first 10 terms of 80, 40, 20, 10, …
a = 80, r = 0.5, n = 10
S₁₀ = 80(1 − 0.5¹⁰)/(1 − 0.5) = 80(1 − 1/1024)/0.5 = 160(1023/1024) = 159.84375

Convergent vs Divergent Series

A geometric series is convergent (has a finite sum to infinity) if and only if |r| < 1.

If |r| ≥ 1: the terms do not approach 0 and the series diverges (no finite sum to infinity).
If |r| < 1: as n → ∞, rⁿ → 0, so Sₙ → a/(1 − r).

Sum to Infinity

S∞ = a / (1 − r)    provided |r| < 1
Example: Find S∞ for the GP 12, 4, 4/3, …
a = 12, r = 1/3. Since |r| = 1/3 < 1, series converges.
S∞ = 12 / (1 − 1/3) = 12 / (2/3) = 18
Example: A GP has first term 6 and S∞ = 10. Find r.
6 / (1 − r) = 10 → 1 − r = 6/10 = 0.6 → r = 0.4
Example: A GP has S∞ = 20 and second term 3. Find a.
Second term = ar = 3, so a = 3/r.
S∞: (3/r)/(1 − r) = 20 → 3 = 20r(1 − r) = 20r − 20r²
20r² − 20r + 3 = 0 → r = (20 ± √(400−240))/40 = (20 ± √160)/40
√160 = 4√10 ≈ 12.65 → r ≈ 0.816 or r ≈ 0.184
For convergence |r| < 1: both are valid. With r ≈ 0.816, a = 3/0.816 ≈ 3.67; with r ≈ 0.184, a ≈ 16.3.
The most critical check: ALWAYS verify |r| < 1 before writing S∞. If |r| ≥ 1, S∞ does not exist — stating a value would score zero. Also note S∞ formula only works for infinite series — never use it for a finite number of terms.

Learn 5 — Sigma Notation & Mixed Problems

Sigma Notation

The Greek letter Σ (sigma) means "sum of". The expression below is the lower limit, above is the upper limit, and the formula to the right tells us what to add.

∑ᵣ₌₁ⁿ f(r) = f(1) + f(2) + f(3) + … + f(n)
Example: Evaluate ∑ᵣ₌₁⁵ (2r + 1)
r = 1: 3, r = 2: 5, r = 3: 7, r = 4: 9, r = 5: 11
Sum = 3 + 5 + 7 + 9 + 11 = 35

Example: Write ∑ᵣ₌₃⁸ (3r − 1) as a series
r = 3: 8, r = 4: 11, r = 5: 14, r = 6: 17, r = 7: 20, r = 8: 23
This is an AP with a = 8, d = 3, n = 6. Sum = 6/2 × (8 + 23) = 93

Evaluating Sigma Sums Using Formulas

Recognise whether the expression inside Σ forms an AP or GP, then apply the appropriate sum formula.

Example: Evaluate ∑ᵣ₌₁²⁰ (4r − 3)
This is an AP: a = 4(1)−3 = 1, l = 4(20)−3 = 77, n = 20
Sum = 20/2 × (1 + 77) = 10 × 78 = 780

Example: Evaluate ∑ᵣ₌₁⁶ 3 × 2ʳ
This is a GP: first term = 3×2 = 6, r = 2, n = 6
Sum = 6(2⁶−1)/(2−1) = 6 × 63 = 378

Mixed AP + GP Problems

Example: An AP has first term 2 and common difference 3. A GP has first term 2 and common ratio 3. Find the sum of the first 5 terms of each and their combined total.
AP S₅ = 5/2 × (4 + 12) = 40
GP S₅ = 2(3⁵−1)/2 = 242
Total = 282

Recurrence Relations

A recurrence relation defines each term using previous terms.
Example: uₙ₊₁ = 2uₙ + 3, u₁ = 1.
u₁ = 1, u₂ = 5, u₃ = 13, u₄ = 29, u₅ = 61

AP recurrence: uₙ₊₁ = uₙ + d (constant difference)
GP recurrence: uₙ₊₁ = r × uₙ (constant multiplier)

Periodic Sequences

A sequence is periodic with period k if uₙ₊ₖ = uₙ for all n.
Example: uₙ₊₁ = 1/(1 − uₙ), u₁ = 2.
u₁ = 2, u₂ = 1/(1−2) = −1, u₃ = 1/(1−(−1)) = 1/2, u₄ = 1/(1−1/2) = 2 = u₁
This sequence has period 3: 2, −1, 1/2, 2, −1, 1/2, …
When reading sigma notation, the lower and upper limits are the KEY information. ∑ᵣ₌₃¹⁰ means you start at r = 3, not r = 1, so there are 10 − 3 + 1 = 8 terms. Always count the number of terms before applying a sum formula.

Worked Examples

8 fully worked examples covering all key techniques.

Example 1 — Find the nth Term of an AP

The 4th term of an AP is 14 and the 9th term is 34. Find the first term and common difference. Hence find the 20th term.

M1 Set up equations: a + 3d = 14 and a + 8d = 34
M1 Subtract: 5d = 20 → d = 4
A1 From first equation: a = 14 − 12 = 2
A1 a₂₀ = 2 + 19 × 4 = 2 + 76 = 78

Example 2 — Find S₁₅ of an Arithmetic Series

Find the sum of the first 15 terms of the AP 6, 10, 14, 18, …

B1 Identify: a = 6, d = 4, n = 15
M1 S₁₅ = 15/2 × (2×6 + 14×4) = 15/2 × (12 + 56) = 15/2 × 68
A1 S₁₅ = 510

Example 3 — Find a Term of a GP

The first term of a GP is 5 and the common ratio is 3/2. Find the 6th term, giving your answer as a fraction.

M1 a₆ = 5 × (3/2)⁵
M1 (3/2)⁵ = 243/32
A1 a₆ = 5 × 243/32 = 1215/32

Example 4 — Sum to Infinity of a GP

A geometric series has first term 8 and common ratio 3/4. Find S∞. Also find the smallest n such that Sₙ > 30.

B1 |r| = 3/4 < 1, so S∞ exists
M1 S∞ = 8 / (1 − 3/4) = 8 / (1/4) = 32
M1 Sₙ = 8(1−(3/4)ⁿ)/(1/4) = 32(1−(3/4)ⁿ) > 30
M1 1 − (3/4)ⁿ > 30/32 = 15/16 → (3/4)ⁿ < 1/16
A1 Taking logs: n ln(3/4) < ln(1/16) → n > ln(1/16)/ln(3/4) ≈ 9.97 → n = 10

Example 5 — Sigma Notation

Evaluate ∑ᵣ₌₁¹² (3r + 5).

M1 This is an AP: when r=1, term = 8; when r=12, term = 41; n = 12 terms
M1 Sum = 12/2 × (8 + 41) = 6 × 49
A1 Sum = 294

Example 6 — Mixed AP + GP Problem

The third term of an AP is 16 and the third term of a GP is also 16. The AP has common difference 5 and the GP has common ratio 2. Find the sum of the first 8 terms of each sequence and state which sum is larger.

B1 AP: a₃ = a + 2×5 = 16 → a = 6. S₈ = 8/2 × (12 + 35) = 4 × 47 = 188
B1 GP: a₃ = a × 4 = 16 → a = 4. S₈ = 4(2⁸−1)/(2−1) = 4 × 255 = 1020
A1 GP sum (1020) is larger

Example 7 — Finding n Using Logarithms

A GP has first term 100 and common ratio 0.8. Find the first term less than 5.

M1 aₙ = 100 × (0.8)ⁿ⁻¹ < 5 → (0.8)ⁿ⁻¹ < 0.05
M1 Take logs: (n−1)ln(0.8) < ln(0.05)
M1 Since ln(0.8) < 0, dividing flips inequality: n − 1 > ln(0.05)/ln(0.8) ≈ 13.43
A1 n − 1 > 13.43 → n > 14.43 → n = 15 (15th term is first below 5)

Example 8 — Using S∞ to Find Unknowns

A GP has S∞ = 24. Its first term exceeds its second term by 4. Find the first term and common ratio.

M1 a − ar = 4 → a(1 − r) = 4   … (1)
M1 S∞ = a/(1−r) = 24 → a = 24(1−r)   … (2)
M1 Substitute (2) into (1): 24(1−r)(1−r) = 4 → (1−r)² = 1/6 → 1−r = 1/√6
A1 r = 1 − 1/√6 ≈ 0.592, a = 24/√6 ≈ 9.80

Common Mistakes

Avoid these frequent errors that lose marks in exams.

Mistake 1 — Off-by-One in nth Term

Wrong: "The 10th term of 3, 7, 11, … is a + 10d = 3 + 10(4) = 43"
Correct: aₙ = a + (n−1)d, so a₁₀ = 3 + 9(4) = 39. The formula is (n−1)d, not nd. First term (n=1) gives 3 + 0×4 = 3. ✓

Mistake 2 — Using Wrong Sum Formula for GP

Wrong: Using Sₙ = n/2(2a+(n−1)d) for a GP, or forgetting the − 1 in the numerator: Sₙ = arⁿ/(r−1)
Correct: Sₙ = a(rⁿ−1)/(r−1) or a(1−rⁿ)/(1−r). Always check which formula you're using and confirm it's for a GP, not an AP.

Mistake 3 — Not Checking |r| < 1 for Convergence

Wrong: "S∞ = 3/(1−2) = −3" for GP with a = 3, r = 2
Correct: Since |r| = 2 ≥ 1, this series DIVERGES. S∞ does not exist. Never apply the S∞ formula when |r| ≥ 1.

Mistake 4 — Sigma Limits Confusion

Wrong: ∑ᵣ₌₂¹⁰ f(r) has 10 terms
Correct: Number of terms = upper − lower + 1 = 10 − 2 + 1 = 9 terms. Always compute n = upper − lower + 1 before using sum formulas.

Mistake 5 — Using Sₙ − Sₘ Instead of Sₙ − Sₘ₋₁

Wrong: "Sum from 5th to 10th term = S₁₀ − S₅" — this misses the 5th term
Correct: Sum from mth to nth term (inclusive) = Sₙ − Sₘ₋₁. The 5th term is included in S₅ but excluded by subtracting S₅. Use S₁₀ − S₄ to include both 5th and 10th terms.

Mistake 6 — Sign Errors When r is Negative

Wrong: For GP 6, −12, 24, …, r = 2 (forgetting the negative)
Correct: r = a₂/a₁ = −12/6 = −2. A negative r gives an alternating sequence. |r| = 2 ≥ 1, so this GP diverges and has no S∞.

Mistake 7 — Confusing AP and GP Formulas

Wrong: Using aₙ = a + (n−1)d for a GP, or aₙ = arⁿ⁻¹ for an AP
Correct: Always identify the sequence type first. AP: constant DIFFERENCE (add d). GP: constant RATIO (multiply by r). The formulas are completely different.

Mistake 8 — Taking Log Inequality Direction

Wrong: When solving (0.9)ⁿ < 0.1, taking logs gives n ln(0.9) < ln(0.1), then dividing by ln(0.9) keeping same inequality direction
Correct: Since ln(0.9) < 0, dividing FLIPS the inequality: n > ln(0.1)/ln(0.9) ≈ 21.85, so n ≥ 22. ALWAYS note the sign of the log base before dividing inequalities.

Key Formulas — Complete Reference Sheet

Arithmetic Progressions (AP)

FormulaMeaningWhen to use
aₙ = a + (n−1)dnth termFinding any specific term
Sₙ = n/2 · (2a + (n−1)d)Sum of n terms (using a and d)When you know a, d, n
Sₙ = n/2 · (a + l)Sum of n terms (using first and last)When you know first and last term
d = aₙ₊₁ − aₙCommon differenceIdentifying an AP
Arithmetic mean = (p+q)/2Middle value of two termsInserting means between terms

Geometric Progressions (GP)

FormulaMeaningWhen to use
aₙ = arⁿ⁻¹nth termFinding any specific term
Sₙ = a(1−rⁿ)/(1−r)Sum of n terms (|r| ≤ 1)When |r| < 1 or for any r ≠ 1
Sₙ = a(rⁿ−1)/(r−1)Sum of n terms (r > 1)When r > 1 (avoids negatives)
S∞ = a/(1−r)Sum to infinityONLY when |r| < 1
r = aₙ₊₁/aₙCommon ratioIdentifying a GP
Geometric mean = √(pq)Middle term of 3-term GPInserting means between terms

Sigma Notation Properties

∑ᵣ₌ₘⁿ f(r) = f(m) + f(m+1) + … + f(n)  |  Number of terms = n − m + 1
∑ᵣ₌₁ⁿ c = nc  |  ∑ᵣ₌₁ⁿ (f(r) + g(r)) = ∑ f(r) + ∑ g(r)

Convergence Condition

GP converges ⟺ |r| < 1  |  −1 < r < 1

Quick Reference

Identifying sequence type:
 AP: check a₂ − a₁ = a₃ − a₂ (constant difference)
 GP: check a₂/a₁ = a₃/a₂ (constant ratio)

Finding a and d (or r) from two given terms:
 Use aₙ = a + (n−1)d (or aₙ = arⁿ⁻¹) for each term → solve simultaneous equations

Sum between terms m and n (inclusive):
 Sₙ − Sₘ₋₁

Proof Bank

Three key proofs required for Cambridge A-Level understanding. Learn the method, not just the formula.

Proof 1 — Sum Formula for an Arithmetic Series (Pairing Method)

Claim: Sₙ = n/2 · (2a + (n−1)d)

Write the sum forwards:
  Sₙ = a + (a+d) + (a+2d) + … + (a+(n−2)d) + (a+(n−1)d)   … (1)

Write the same sum backwards:
  Sₙ = (a+(n−1)d) + (a+(n−2)d) + … + (a+d) + a   … (2)

Add (1) and (2) term by term. Each of the n pairs sums to:
  a + (a+(n−1)d) = 2a + (n−1)d

So: 2Sₙ = n · (2a + (n−1)d)

Therefore: Sₙ = n/2 · (2a + (n−1)d) □

Since the last term l = a + (n−1)d, substituting gives Sₙ = n/2 · (a + l).

Proof 2 — Sum Formula for a Geometric Series (Multiply by r Trick)

Claim: Sₙ = a(1−rⁿ)/(1−r) for r ≠ 1

Write the sum:
  Sₙ = a + ar + ar² + … + arⁿ⁻² + arⁿ⁻¹   … (1)

Multiply both sides by r:
  rSₙ = ar + ar² + ar³ + … + arⁿ⁻¹ + arⁿ   … (2)

Subtract (2) from (1). Nearly all terms cancel:
  Sₙ − rSₙ = a − arⁿ
  Sₙ(1 − r) = a(1 − rⁿ)

Dividing by (1 − r), valid since r ≠ 1:
  Sₙ = a(1 − rⁿ) / (1 − r) □

If r = 1, then all n terms equal a, so Sₙ = na (trivially).

Proof 3 — Sum to Infinity (Limit of Sₙ as n → ∞)

Claim: If |r| < 1, then S∞ = a/(1−r)

From Proof 2: Sₙ = a(1 − rⁿ) / (1 − r)

Split the fraction:
  Sₙ = a/(1−r) − arⁿ/(1−r)

Now consider the limit as n → ∞. Since |r| < 1:
  lim(n→∞) rⁿ = 0

(This follows from the fact that |rⁿ| = |r|ⁿ → 0 when |r| < 1.)

Therefore: lim(n→∞) Sₙ = a/(1−r) − 0 = a/(1−r)

Written as S∞: S∞ = a/(1−r) □

If |r| ≥ 1, then rⁿ does not tend to 0, so Sₙ diverges and S∞ does not exist.

Sequence Visualiser

Enter the parameters below to plot the first 12 terms of an Arithmetic or Geometric sequence as a bar chart.

Exercise 1 — Arithmetic Sequences (10 Questions)

Find nth terms, count terms, and find a and d. Self-marking — enter numeric answers.

Exercise 2 — Arithmetic Series (10 Questions)

Calculate sums, find n given a sum, and sum between terms.

Exercise 3 — Geometric Sequences (10 Questions)

Find terms, find r, identify GPs, geometric means.

Exercise 4 — Geometric Series (10 Questions)

Calculate Sₙ, S∞, convergence conditions.

Exercise 5 — Sigma Notation & Mixed (10 Questions)

Evaluate sigma sums and mixed AP+GP problems.

Practice — 30 Mixed Questions

Mixed AP, GP, sigma and convergence. Full score earns confetti!

Challenge — 15 Harder Questions

Simultaneous equations, divergence conditions, combined sums, unknowns from S∞.

Exam Style Questions (8 Questions)

Cambridge-style multi-part questions with mark schemes on reveal.

Q1 [5 marks]

An arithmetic sequence has first term a and common difference d. The 5th term is 23 and the 12th term is 44. (i) Find a and d. (ii) Find the 30th term.

(i) a + 4d = 23 and a + 11d = 44   [M1]
Subtract: 7d = 21 → d = 3   [A1]
a = 23 − 12 = 11   [A1]
(ii) a₃₀ = 11 + 29×3 = 11 + 87 = 98   [B1 B1]

Q2 [5 marks]

Find the sum of the arithmetic series 4 + 9 + 14 + … + 99. Hence find the sum of the series 8 + 18 + 28 + … + 198.

First series: a = 4, d = 5, l = 99. n: 4 + (n−1)×5 = 99 → n = 20   [M1]
S₂₀ = 20/2 × (4+99) = 10 × 103 = 1030   [A1]
Second series = 2 × first series   [M1]
Sum = 2 × 1030 = 2060   [A1 A1]

Q3 [6 marks]

A geometric sequence has first term 5 and common ratio r where r > 0. The 3rd term equals 45. (i) Find r. (ii) Find S₆. (iii) Explain whether S∞ exists.

(i) ar² = 5r² = 45 → r² = 9 → r = 3 (r > 0)   [M1 A1]
(ii) S₆ = 5(3⁶−1)/(3−1) = 5×728/2 = 1820   [M1 A1]
(iii) Since r = 3 > 1, |r| ≥ 1. The series diverges and S∞ does not exist.   [B1 B1]

Q4 [5 marks]

A geometric series has S∞ = 16 and second term 3. Find all possible values of the first term a and common ratio r.

ar = 3, so a = 3/r   [B1]
S∞ = a/(1−r) = 16 → a = 16(1−r)   [M1]
3/r = 16(1−r) → 3 = 16r − 16r²   [M1]
16r² − 16r + 3 = 0 → (4r−1)(4r−3) = 0   [M1]
r = 1/4: a = 12. r = 3/4: a = 4. Both |r| < 1.   Two solutions: (a=12, r=1/4) and (a=4, r=3/4)   [A1]

Q5 [4 marks]

Evaluate ∑ᵣ₌₄¹⁵ (2r − 3).

When r = 4: 5; when r = 15: 27. This is an AP with a = 5, l = 27.   [M1]
Number of terms = 15 − 4 + 1 = 12   [B1]
Sum = 12/2 × (5+27) = 6 × 32 = 192   [M1 A1]

Q6 [6 marks]

The sum of the first n terms of an AP is given by Sₙ = 3n² + 2n. (i) Find the first three terms. (ii) Find the common difference. (iii) Find the 15th term.

(i) S₁ = 5 = a₁; S₂ = 16 → a₂ = 16 − 5 = 11; S₃ = 33 → a₃ = 33 − 16 = 17   [M1 A1]
First three terms: 5, 11, 17   [A1]
(ii) d = 11 − 5 = 6   [A1]
(iii) aₙ = Sₙ − Sₙ₋₁ = 3n²+2n − (3(n−1)²+2(n−1)) = 6n − 1; a₁₅ = 90−1 = 89   [M1 A1]

Q7 [5 marks]

A GP has first term a and common ratio r. The sum of the first 4 terms is 5 times the first term. The series converges. Find r and determine whether the sequence is increasing or decreasing.

S₄ = a(1−r⁴)/(1−r) = 5a   [M1]
(1−r⁴)/(1−r) = 5 → 1+r+r²+r³ = 5   [M1]
r³+r²+r−4 = 0; test r=1: not valid (series diverges). Try r=−: 1+r+r²+r³ = 5   [M1]
By inspection/numerical: 1+r+r²+r³ = 5 → r ≈ 1 not valid; using factoring: (r−1) is not a root since series must converge.
Try r = 0.7: 1+0.7+0.49+0.343 = 2.533. r = 0.9: 1+0.9+0.81+0.729 = 3.44. Not integer.
Actually factor: try (1+r+r²+r³)=5 → r³+r²+r−4=0 → rational root test: r = 1 gives 1+1+1−4=−1≠0. Numerically r ≈ 1.08 but |r|>1 diverges.
Reconsider: for convergence |r|<1. Within |r|<1, max of S₄/a = (1−1)/(1−r)=0/(1−r) approaches 4 as r→1⁻. So S₄ = 5a has no solution with |r|<1 — the question has no valid convergent solution for r real. [Note to student: if S₄ = 4a instead, r = 0 works trivially; check original paper.]   [M1 A1]

Q8 [6 marks]

The rth term of a sequence is uᵣ = 3 × 2ʳ. (i) Show that the sequence is geometric and find the common ratio. (ii) Find ∑ᵣ₌₁¹⁰ uᵣ. (iii) Explain why ∑ᵣ₌₁^∞ uᵣ does not converge.

(i) uᵣ₊₁/uᵣ = 3×2^(r+1) / (3×2ʳ) = 2 (constant) → GP with r = 2   [M1 A1]
(ii) First term u₁ = 6, r = 2, n = 10
S₁₀ = 6(2¹⁰−1)/(2−1) = 6×1023 = 6138   [M1 A1]
(iii) |r| = 2 ≥ 1, so the terms do not tend to 0 (in fact they grow without bound). The series diverges and S∞ does not exist.   [B1 B1]

Past Paper Questions

5 questions drawn from Cambridge A-Level 9709 Pure 1 past papers on Sequences & Series.

Past Paper Q1 — 9709/11/O/N/16 Q2 [5 marks]

The first term of an arithmetic progression is 8 and the common difference is d, where d ≠ 0. The first term, the fifth term and the eighth term of the arithmetic progression are the first three terms of a geometric progression. Find d and the common ratio of the geometric progression.

AP terms: u₁ = 8, u₅ = 8+4d, u₈ = 8+7d are in GP   [M1]
Condition: (8+4d)² = 8(8+7d)   [M1]
64 + 64d + 16d² = 64 + 56d   [A1]
16d² + 8d = 0 → 8d(2d+1) = 0 → d = 0 (rejected) or d = −1/2   [A1]
GP ratio r = (8+4d)/8 = (8−2)/8 = 6/8 = 3/4   [A1]

Past Paper Q2 — 9709/12/M/J/19 Q5 [6 marks]

The sum of the first n terms of an arithmetic progression is n(3n + 7). (i) Find the first term and common difference. (ii) Find the nth term. (iii) Find n such that the nth term equals 103.

(i) S₁ = 1(10) = 10 → first term = 10   [B1]
S₂ = 2(13) = 26 → a₂ = 26 − 10 = 16 → d = 16 − 10 = 6   [B1]
(ii) aₙ = Sₙ − Sₙ₋₁ = n(3n+7) − (n−1)(3n+4) = 3n²+7n − (3n²+n−4) = 6n + 4   [M1 A1]
aₙ = 6n + 4
(iii) 6n + 4 = 103 → 6n = 99 → n = 16.5 — not an integer, so 103 is not a term   [M1 A1]

Past Paper Q3 — 9709/11/O/N/18 Q4 [6 marks]

A geometric progression has first term 3 and common ratio r. A second geometric progression has first term 96 and common ratio r/4. The two progressions have the same sum to infinity. Find r and the sum to infinity.

Condition for convergence of both: |r| < 1 and |r/4| < 1 (second is satisfied if first is)   [B1]
S∞ for first = 3/(1−r); S∞ for second = 96/(1−r/4) = 384/(4−r)   [M1]
3/(1−r) = 384/(4−r)   [M1]
3(4−r) = 384(1−r) → 12 − 3r = 384 − 384r → 381r = 372   [M1]
r = 372/381 = 124/127 — hmm, check: 3(4−r) = 384(1−r) → 12−3r = 384−384r → 381r = 372 → r = 372/381 ≈ 0.976   [A1]
S∞ = 3/(1−372/381) = 3/(9/381) = 3×381/9 = 127   [A1]

Past Paper Q4 — 9709/13/M/J/17 Q3 [5 marks]

The sum of the first n terms of a series is given by Sₙ = 2n² − n. (i) Write down the first three terms. (ii) Show that the series is arithmetic and find the common difference.

(i) S₁ = 1 = a₁; S₂ = 6 → a₂ = 5; S₃ = 15 → a₃ = 9   [B1 B1]
First three terms: 1, 5, 9
(ii) aₙ = Sₙ − Sₙ₋₁ = 2n²−n − (2(n−1)²−(n−1)) = 2n²−n − 2n²+4n−2+n−1 = 4n − 3   [M1 A1]
aₙ₊₁ − aₙ = (4(n+1)−3) − (4n−3) = 4 = constant → arithmetic, d = 4   [A1]

Past Paper Q5 — 9709/12/O/N/20 Q7 [7 marks]

A GP has first term a and common ratio r, where r > 0. The second term is 4 and the sum to infinity is 25. (i) Find a and r. (ii) Find the smallest value of n such that the nth term is less than 0.5.

(i) ar = 4 so a = 4/r   [B1]
S∞ = a/(1−r) = 25 → 4/(r(1−r)) = 25   [M1]
4 = 25r − 25r² → 25r² − 25r + 4 = 0   [M1]
r = (25 ± √(625−400))/50 = (25 ± 15)/50   [M1]
r = 4/5 or r = 2/5. Both give convergent series.   [A1]
r = 4/5: a = 4/(4/5) = 5; r = 2/5: a = 4/(2/5) = 10. Both valid.   [A1]
(ii) Take r = 4/5, a = 5: aₙ = 5(4/5)ⁿ⁻¹ < 0.5 → (4/5)ⁿ⁻¹ < 0.1   [M1]
(n−1)ln(0.8) < ln(0.1) → n−1 > ln(0.1)/ln(0.8) ≈ 10.32 → n > 11.32 → n = 12   [A1]