Pearson Edexcel International Advanced Level Mathematics: Sequences and series: arithmetic, geometric and the binomial expansion (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4 (topics 4.1 to 4.5) of Unit P2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. Unit P2 is compulsory for IAS Mathematics and IAL Mathematics (and Pure Mathematics). Teaching is in the study guide and the revision notes. The mark totals shown are Marlbridge’s own allocations.

Questions

1. A sequence is defined by x₁ = 2 and xₙ₊₁ = 5xₙ − 7.

(a) Find x₂, x₃ and x₄. [2] (b) Find the value of x₁ for which every term of the sequence is equal. [1]

2. An arithmetic sequence has fifth term 23 and twelfth term 51.

(a) Find the first term and the common difference. [3] (b) Find the sum of the first 15 terms. [2]

3. Evaluate (a) Σ (r = 1 to 40) of (3r + 2) [3], and (b) Σ (r = 11 to 40) of (3r + 2). [2]

4. Prove that the sum of the first n terms of an arithmetic series with first term a and common difference d is (n/2)(2a + (n − 1)d). [4] Hence write down 1 + 2 + … + n. [1]

5. Ria saves £20 in month 1, and each later month she saves £6 more than the month before.

(a) How much does she save in month 24? [2] (b) How much has she saved in total after 24 months? [2] (c) Find the first month at the end of which her total savings exceed £1200. [3]

6. (a) For uₙ = n² − 10n, show that uₙ₊₁ − uₙ = 2n − 9. [2] (b) Hence describe where the sequence decreases and where it increases. [2] (c) A sequence has x₁ = 2 and xₙ₊₁ = 6/xₙ. Show it is periodic, state its order, and find the sum of its first 51 terms. [3]

7. A geometric sequence has third term 20 and sixth term 160.

(a) Find the common ratio and the first term. [3] (b) Write down an expression for the nth term. [1] (c) Find the sum of the first 10 terms. [2]

8. A geometric series has first term 30 and sum to infinity 75.

(a) Find the common ratio. [2] (b) Find the third term. [1] (c) Find the sum of all terms from the fourth term onwards. [2]

9. Prove that the sum of the first n terms of a geometric series with first term a and common ratio r (r ≠ 1) is a(1 − rⁿ)/(1 − r). [4] State the condition under which the series has a sum to infinity, and give its value. [1]

10. (a) £5000 is invested at 4% compound interest per year. Find the smallest whole number of years after which the value exceeds £8000. [4] (b) A ball is dropped from 3 m. After each bounce it rises to 4/5 of its previous height. Find the total distance it travels before coming to rest, assuming the pattern continues indefinitely. [3]

11. (a) Expand (5 − 2x)⁴ in ascending powers of x. [3] (b) Find the coefficient of x² in the expansion of (1 + 3x)(5 − 2x)⁴. [3] (c) The coefficient of x² in the expansion of (1 + 2x)ⁿ is 180, where n is a positive integer. Find n. [3]

Answers

1. (a) x₂ = 5(2) − 7 = 3 [1], x₃ = 8 and x₄ = 33 [1] (b) All terms equal means x = 5x − 7, so 4x = 7 and x₁ = 7/4 [1] [3] Examiner insight: in a recurrence, show each substitution. A bare list of terms with one slip loses the later values, and the working lets a marker see which step went wrong.

2. (a) a + 4d = 23 and a + 11d = 51 [1]. Subtract: 7d = 28 [1], so d = 4 and a = 7 [1] (b) S₁₅ = (15/2)(2×7 + 14×4) [1] = (15/2)(70) = 525 [1] [5] Examiner insight: write both simultaneous equations before solving. If you slip in the arithmetic, the correct equations still show the method.

3. (a) Σ(3r + 2) = 3Σr + 2×40 [1]; Σr = 40×41/2 = 820 [1]; 2460 + 80 = 2540 [1] (b) Σ (r = 1 to 10) = 3×55 + 20 = 185 [1]; 2540 − 185 = 2355 [1] [5] Examiner insight: for a sum starting at r = 11, subtract the sum to 10, not the sum to 11. That off-by-one choice is the usual way to lose the final mark.

4. Sₙ = a + (a + d) + … + (a + (n − 1)d) [1]. Reverse the order and add the two lines [1]: each pair totals 2a + (n − 1)d, and there are n pairs [1]. So 2Sₙ = n(2a + (n − 1)d) and Sₙ = (n/2)(2a + (n − 1)d) [1]. With a = 1, d = 1: n(n + 1)/2 [1] [5] Examiner insight: a proof is marked on its reasoning. Say that the reversed series is the same series, and show the pairing, rather than writing the result and stopping.

5. (a) Arithmetic, a = 20, d = 6 [1]: u₂₄ = 20 + 23×6 = £158 [1] (b) S₂₄ = 12(40 + 23×6) [1] = 12 × 178 = £2136 [1] (c) Sₙ = (n/2)(40 + 6(n − 1)) = n(3n + 17) [1]. Solve 3n² + 17n − 1200 > 0, positive root (−17 + √14689)/6 ≈ 17.4 [1]. S₁₇ = 1156 and S₁₈ = 1278, so month 18 [1] [7] Examiner insight: in a context question, state which model you are using and answer in the context’s units. Checking the two neighbouring values of n protects against rounding the root the wrong way.

6. (a) uₙ₊₁ = (n + 1)² − 10(n + 1) = n² − 8n − 9 [1]; subtract uₙ = n² − 10n to get 2n − 9 [1] (b) 2n − 9 < 0 for n ≤ 4, so u₁ > u₂ > … > u₅ (decreasing up to u₅ = −25) [1]; 2n − 9 > 0 for n ≥ 5, so the sequence increases from u₅ onwards [1] (c) Terms 2, 3, 2, 3, … so x₃ = x₁, periodic with order 2 [1]. 51 terms = 25 pairs plus x₅₁ = 2 [1]. Sum = 25 × 5 + 2 = 127 [1] [7] Examiner insight: to claim a sequence is increasing or decreasing you need the sign of uₙ₊₁ − uₙ for all n, not a few terms. Say clearly where the behaviour changes.

7. (a) ar² = 20 and ar⁵ = 160 [1]. Divide: r³ = 8, so r = 2 [1] and a = 5 [1] (b) uₙ = 5 × 2ⁿ⁻¹ [1] (c) S₁₀ = 5(2¹⁰ − 1)/(2 − 1) [1] = 5115 [1] [6] Examiner insight: dividing the two equations is the efficient route. Write the powers as ar² and ar⁵, not ar³ and ar⁶, which is a common indexing slip.

8. (a) 30/(1 − r) = 75 [1], so 1 − r = 2/5 and r = 3/5 [1] (b) 30 × (3/5)² = 10.8 [1] (c) The sum from term 4 is ar³/(1 − r) = 30 × (27/125)/(2/5) [1] = 16.2 [1] [5] Examiner insight: either S∞ − S₃ or the tail formula can earn full credit if each step is shown. State |r| < 1 in (a) to justify using S∞.

9. Sₙ = a + ar + … + arⁿ⁻¹ [1]. Multiply by r: rSₙ = ar + … + arⁿ [1]. Subtract: Sₙ − rSₙ = a − arⁿ [1]. Factor: Sₙ(1 − r) = a(1 − rⁿ), so Sₙ = a(1 − rⁿ)/(1 − r) [1]. The sum to infinity needs |r| < 1, since then rⁿ → 0 and S∞ = a/(1 − r) [1] [5] Examiner insight: lay the two lines out one under the other so the cancelling is visible. Mention why r ≠ 1 before dividing.

10. (a) 5000 × 1.04ⁿ > 8000 [1], so 1.04ⁿ > 1.6 [1]. Take logs: n > ln 1.6/ln 1.04 = 11.98… [1], so n = 12 years [1] (b) Total = 3 + 2(2.4 + 1.92 + …) [1], and the bracket is a geometric series with a = 2.4, r = 0.8 whose sum is 2.4/0.2 = 12 [1]. Total = 3 + 24 = 27 m [1] [7] Examiner insight: give a whole number of years, because 11.98 does not answer the question as asked. In (b), counting each bounce as an upward and a downward journey is the step to show.

11. (a) Coefficients 1, 4, 6, 4, 1 [1]. 625 − 4×125×2x + 6×25×4x² − 4×5×8x³ + 16x⁴ [1] = 625 − 1000x + 600x² − 160x³ + 16x⁴ [1] (b) x² terms: 1 × 600x² and 3x × (−1000x) [1], so 600 + (−3000) [1] = −2400 [1] (c) ⁿC₂ × 2² = 180, so n(n − 1)/2 × 4 = 180 [1]. n² − n − 90 = 0, (n − 10)(n + 9) = 0 [1]. n is a positive integer, so n = 10 [1] [9] Examiner insight: in (b) take every pair of terms that makes x²; using only 1 × 600x² is the usual error. In (c), say why you reject n = −9.

Where marks are usually lost

• Using n in place of n − 1 in the nth-term formulas.
• Not putting brackets round (−2x) before raising it to a power, which loses signs and the factor of 4.
• Missing a cross term when multiplying a binomial expansion by a bracket.
• Leaving n as a decimal when the question asks for a whole number.
• Not reversing the inequality when dividing by a negative logarithm.
• Quoting S∞ without checking |r| < 1.
• Proofs that start from the answer, or skip the line showing why terms cancel.
• Declaring a sequence increasing or decreasing from the first few terms only.

Next steps

• Re-read the study guide for any question you missed.
• Use the revision notes for formulae and the proofs.
• See the Edexcel A Level Mathematics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Unit P2, section 4 (Pearson Education Limited).