Arithmetic & Geometric Series, Sum to Infinity & Recurrence Relations
Sequences and series bridge pure algebra, discrete modelling, and calculus. At A-Level, you need more than just plugging numbers into the arithmetic and geometric sum formulas: examiners regularly ask for the formal first-principles derivations of S_n for both series, test convergence criteria (|r| < 1) for infinite geometric series, and combine sigma notation (Σ) with logarithms to find when a sum first exceeds a threshold. You must also classify recurrence relations u_(n+1) = f(u_n) as increasing, decreasing, or periodic with period k.
- Arithmetic Sequences and Series (S_n Proof)
- An arithmetic sequence has first term a and common difference d, with nth term u_n = a + (n - 1)d. The sum of the first n terms is S_n = (n / 2)[2a + (n - 1)d] = (n / 2)(a + L), where L is the last term. To prove this from first principles, write S_n forwards: S_n = a + (a + d) + . + [a + (n - 1)d], then write S_n in reverse beneath it: S_n = [a + (n - 1)d] + [a + (n - 2)d] + . + a. Adding the two equations term-by-term gives 2S_n = n × [2a + (n - 1)d], and dividing by 2 completes the proof.
- Geometric Sequences and Series (S_n Proof)
- A geometric sequence has first term a and common ratio r, with nth term u_n = ar^(n-1). The sum of the first n terms is S_n = a(1 - r^n) / (1 - r) (for r ≠ 1). To prove this from first principles, write out S_n = a + ar + ar² + . + ar^(n-1) (Equation 1). Multiply every term by r to get rS_n = ar + ar² + . + ar^(n-1) + ar^n (Equation 2). Subtracting Equation 2 from Equation 1 cancels all middle terms, leaving S_n - rS_n = a - ar^n. Factorising both sides gives S_n(1 - r) = a(1 - r^n), so S_n = a(1 - r^n) / (1 - r).
- Convergence and Sum to Infinity (S_∞)
- An infinite geometric series converges to a finite limit if and only if |r| < 1 (that is, -1 < r < 1). When |r| < 1, as n → ∞, the power term r^n → 0. Substituting r^n → 0 into the finite sum formula gives the sum to infinity: S_∞ = lim_(n→∞) [a(1 - r^n) / (1 - r)] = a / (1 - r). If |r| ≥ 1, r^n does not tend to zero and the series diverges.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Arithmetic Sequences and Series (S_n Proof)
- An arithmetic sequence has first term a and common difference d, with nth term u_n = a + (n - 1)d. The sum of the first n terms is S_n = (n / 2)[2a + (n - 1)d] = (n / 2)(a + L), where L is the last term. To prove this from first principles, write S_n forwards: S_n = a + (a + d) + . + [a + (n - 1)d], then write S_n in reverse beneath it: S_n = [a + (n - 1)d] + [a + (n - 2)d] + . + a. Adding the two equations term-by-term gives 2S_n = n × [2a + (n - 1)d], and dividing by 2 completes the proof.
- Geometric Sequences and Series (S_n Proof)
- A geometric sequence has first term a and common ratio r, with nth term u_n = ar^(n-1). The sum of the first n terms is S_n = a(1 - r^n) / (1 - r) (for r ≠ 1). To prove this from first principles, write out S_n = a + ar + ar² + . + ar^(n-1) (Equation 1). Multiply every term by r to get rS_n = ar + ar² + . + ar^(n-1) + ar^n (Equation 2). Subtracting Equation 2 from Equation 1 cancels all middle terms, leaving S_n - rS_n = a - ar^n. Factorising both sides gives S_n(1 - r) = a(1 - r^n), so S_n = a(1 - r^n) / (1 - r).
- Convergence and Sum to Infinity (S_∞)
- An infinite geometric series converges to a finite limit if and only if |r| < 1 (that is, -1 < r < 1). When |r| < 1, as n → ∞, the power term r^n → 0. Substituting r^n → 0 into the finite sum formula gives the sum to infinity: S_∞ = lim_(n→∞) [a(1 - r^n) / (1 - r)] = a / (1 - r). If |r| ≥ 1, r^n does not tend to zero and the series diverges.
- Sigma Notation (Σ) and Splitting Series
- The notation Σ_(r=1)^n f(r) represents the sum f(1) + f(2) + . + f(n). Always write out the first three terms and the last term of a sigma expression to identify whether it is arithmetic, geometric, or a combination of both. If a sum starts at r = k (where k > 1), contains a number of terms equal to (top limit, bottom limit + 1), or you can evaluate it as Σ_(r=1)^n f(r) - Σ_(r=1)^(k-1) f(r).
- Recurrence Relations and Periodic Sequences
- A recurrence relation defines each term as a function of preceding terms, u_(n+1) = f(u_n) with a given starting term u₁. A sequence is strictly increasing if u_(n+1) > u_n for all n, strictly decreasing if u_(n+1) < u_n for all n, and periodic if the terms repeat in a regular cycle such that u_(n+k) = u_n for all n. The smallest positive integer k for which this holds is the order (or period) of the sequence. To sum a large number of terms (e.g., Σ_(n=1)^100 u_n) of a periodic sequence, divide 100 by the period k, multiply the sum of one full cycle by the number of complete cycles, and add any remaining terms.
- Modelling with Series and Logarithmic Inequalities
- Real-world problems involving fixed annual increments (e.g., adding £500 per year to a salary) follow arithmetic models, whereas percentage changes or compound interest (e.g., increasing output by 4% per year, where r = 1.04, or decreasing by 6% per year, where r = 0.94) follow geometric models. When solving inequalities like r^n < 0.05 by taking logs, remember that if 0 < r < 1, then ln(r) is negative, so dividing both sides of the inequality by ln(r) reverses the inequality sign.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How many terms are there in the sum Σ_(r=5)^42 u_r?
What is the condition for an infinite geometric series to converge, and why?
How can you relate the nth term u_n of any series to its partial sums S_n and S_(n-1)?
If three consecutive terms x, y, z form an arithmetic sequence vs a geometric sequence, what equations link them?
Why does the inequality sign flip when solving 0.85^n < 0.01 using logarithms?
What does it mean if a sequence is periodic with period (order) 3?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Prove from first principles that the sum of the first n terms of a geometric series with first term a and common ratio r (r ≠ 1) is given by S_n = a(1 - r^n) / (1 - r). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write out the sum of the first n terms explicitly: S_n = a + ar + ar² + . + ar^(n-2) + ar^(n-1) (Equation 1) [1 mark]. 2. Multiply every term of Equation 1 by r: rS_n = ar + ar² + ar³ + . + ar^(n-1) + ar^n (Equation 2) [1 mark]. 3. Subtract Equation 2 from Equation 1 so all intermediate terms cancel: S_n - rS_n = a - ar^n [1 mark]. 4. Factorise both sides as S_n(1 - r) = a(1 - r^n) and divide by (1 - r) (since r ≠ 1) to obtain S_n = a(1 - r^n) / (1 - r) [1 mark].
- Q2
The 14th, 5th, and 2nd terms of an arithmetic sequence with non-zero common difference d are, in that order, the first three consecutive terms of a convergent geometric sequence. Find the common ratio r of the geometric sequence, and given that the sum to infinity of the geometric sequence is 36, find the first term a and common difference d of the arithmetic sequence. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write the 14th, 5th, and 2nd terms of the arithmetic sequence in terms of a and d: u₁₄ = a + 13d, u₅ = a + 4d, and u₂ = a + d [1 mark]. 2. Since they form a geometric sequence in that order, equate the ratio of consecutive terms: (a + 4d) / (a + 13d) = (a + d) / (a + 4d) [1 mark]. 3. Cross-multiply and expand: (a + 4d)² = (a + 13d)(a + d) gives a² + 8ad + 16d² = a² + 14ad + 13d², which simplifies to 3d² = 6ad [1 mark]. 4. Since d ≠ 0, divide by 3d to obtain d = 2a; substitute d = 2a into the common ratio r = (a + d) / (a + 4d) = 3a / 9a = 1/3 [1 mark]. 5. The first term of the geometric sequence is u₁₄ = a + 13(2a) = 27a; substitute this and r = 1/3 into S_∞ = u₁₄ / (1 - r) = 36: 27a / (2/3) = 36 [1 mark]. 6. Solve for a and d: (81/2)a = 36 gives a = 8/9, and d = 2a = 16/9 [1 mark].
- Q3
A convergent geometric series has first term a and common ratio r. The sum of the first 4 terms is 15, and the sum to infinity is 16. Find the two possible values of r and the corresponding values of a. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write the two equations from the given information: S₄ = a(1 - r⁴) / (1 - r) = 15 and S_∞ = a / (1 - r) = 16 [1 mark]. 2. Substitute a / (1 - r) = 16 directly into the S₄ equation: 16(1 - r⁴) = 15 [1 mark]. 3. Rearrange to solve for r⁴: 1 - r⁴ = 15 / 16 → r⁴ = 1 / 16 [1 mark]. 4. Take the fourth root, keeping both positive and negative real roots (since both satisfy |r| < 1): r = +1/2 or r = -1/2 [1 mark]. 5. Substitute back into a = 16(1 - r): when r = 1/2, a = 16(1 - 1/2) = 8; when r = -1/2, a = 16(1 - (-1/2)) = 24 [1 mark].
- Q4
A sequence u₁, u₂, u₃, . is defined by u₁ = 4 and the recurrence relation u_(n+1) = (u_n - 3) / (u_n - 1) for n ≥ 1. Show that the sequence is periodic, state its period, and calculate Σ_(n=1)^100 u_n. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate u₂: u₂ = (4 - 3) / (4 - 1) = 1 / 3 [1 mark]. 2. Calculate u₃: u₃ = (1/3 - 3) / (1/3 - 1) = (-8/3) / (-2/3) = 4 [1 mark]. 3. Since u₃ = u₁ = 4, the sequence repeats as 4, 1/3, 4, 1/3, . so it is periodic with period (order) 2 [1 mark]. 4. Find the sum of one complete period of 2 terms: u₁ + u₂ = 4 + 1/3 = 13/3 [1 mark]. 5. Since 100 terms contain 100 ÷ 2 = 50 complete cycles of 2 terms: Σ_(n=1)^100 u_n = 50 × (13/3) = 650 / 3 (or 216⅔) [1 mark].
Common Exam Questions (4)
How do I evaluate a combined sigma expression like Σ_(r=1)^20 (4r + 3 × 2^r)?
Split the sigma into two separate sums using linearity: 4 Σ_(r=1)^20 r + Σ_(r=1)^20 (3 × 2^r). The first part is an arithmetic series (a = 4, d = 4, n = 20) and the second part is a geometric series (a = 6, r = 2, n = 20). Calculate each separately and add the results.
In compound interest or depreciation problems, how do I avoid being off by one year on n?
Write out Year 1, Year 2, and Year 3 explicitly with actual numbers. For example, if £1000 is invested at the start of Year 1 at 5% interest, the value at the start of Year 1 is 1000, at the end of Year 1 is 1000(1.05)¹, and at the end of Year n is 1000(1.05)^n.
Why must I keep both positive and negative roots when solving r² = 0.25 or r⁴ = 1/16?
An even power produces two real roots (e.g., r = ±0.5). Unless the question specifies that all terms of the geometric series are positive, the negative common ratio gives a valid alternating geometric series and earns half the marks.
Can I use S_n = a(r^n - 1) / (r - 1) instead of a(1 - r^n) / (1 - r)?
Yes, both formulas are algebraically identical (the top and bottom are just multiplied by -1). Usually a(r^n - 1) / (r - 1) is neater when r > 1 because both numerator and denominator stay positive, while a(1 - r^n) / (1 - r) is better when |r| < 1.
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