AQA A-Level Mathematics Paper 3, June 2025: Question 7

3 marks · Easy difficulty · Short Answer

Solve the equation involving sigma notation $\sum_{r=1}^{3} (ar + 5) = 57$ to find the value of the constant $a$.

Practise this question

Question

Question 7 asks to solve the equation: the sum from r equals 1 to 3 of (ar + 5) equals 57, to find the value of a. The question is worth 3 marks.

Mark scheme

Show the mark scheme Mark scheme for Question 7 shows: B1 for obtaining the three terms a + 5, 2a + 5, and 3a + 5; M1 for forming an equation summing these three terms (or using the arithmetic series sum formula) equated to 57; A1 for the final correct answer a = 7.

How to answer it

Solving Linear Equations Involving Sigma Notation

What this question tests

Understanding and expanding finite series expressed in sigma (∑) notation, constructing a linear algebraic equation from a given total, and solving for an unknown constant accurately.

Question 7 • 3 Marks

Full Solution & Step-by-Step Breakdown

Solve ∑r=13 (ar + 5) = 57 to find the value of a

📐 Calculations (Step-by-Step)

Step 1: Expand the summation term by term
Substitute r = 1, r = 2, and r = 3 into the expression (ar + 5):
• For r = 1: a(1) + 5 = a + 5
• For r = 2: a(2) + 5 = 2a + 5
• For r = 3: a(3) + 5 = 3a + 5
Step 2: Set up the linear equation
Sum all three terms and equate to 57:
(a + 5) + (2a + 5) + (3a + 5) = 57
Collect like terms:
(a + 2a + 3a) + (5 + 5 + 5) = 57
6a + 15 = 57
Step 3: Solve for a
Subtract 15 from both sides:
6a = 57 − 15
6a = 42
Divide by 6:
a = 42 ÷ 6 = 7

✅ Mark Scheme Breakdown

  • B1 (AO 2.5): Obtains a + 5 , 2a + 5 , and 3a + 5 .
    Note: Can be implied (PI) by seeing the simplified expression 6a + 15 or the final correct answer.
  • M1 (AO 3.1a): Forms an equation using the sum of three terms containing at least two correct terms equal to 57 (e.g. 6a + 15 = 57 ), OR correctly applies an arithmetic series sum formula Sn with n = 3.
  • A1 (AO 1.1b): Correct final value of a = 7.
Total: 3 / 3 Marks (Fully correct mathematical argument with precise evaluation).

💡 Key Knowledge

  • Sigma Definition: ∑r=1n f(r) means f(1) + f(2) + ... + f(n).
  • Constant Terms: A constant inside the summation is added for every value of r. For n terms, adding 5 gives 5 × 3 = 15.
  • Alternative AP Method: Since (ar + 5) is linear in r, it forms an arithmetic progression with first term A = a + 5, common difference d = a, and last term L = 3a + 5:
    S₃ = 3/2(first + last) = 3/2(4a + 10) = 6a + 15

🧠 Exam Technique

  • Keep it simple: When the upper limit is small (like n = 3), expanding term-by-term is much faster and less error-prone than quoting arithmetic series formulae.
  • Check via back-substitution: If a = 7, the terms are (7 + 5) = 12, (14 + 5) = 19, (21 + 5) = 26. Sum = 12 + 19 + 26 = 57. Verified!
  • Show intermediate steps: Always write down 6a + 15 = 57 to secure B1 and M1 marks even if a slips in mental arithmetic.

❌ Common Errors & Pitfalls

  • Adding the constant only once: Writing (a + 2a + 3a) + 5 = 57 leading to 6a + 5 = 57. The constant (+5) belongs to each individual term inside the brackets and must be added 3 times.
  • Index confusion: Substituting r = 0, 1, 2 instead of reading the limits carefully (r runs from 1 to 3, giving 3 terms total).
  • Arithmetic slips in division: Solving 6a = 42 incorrectly under time pressure.

Topics

Pure Mathematics · D: Sequences and series

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.