WMA12/01 Jun 2023, Question 11: Sigma Notation
Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 8 marks
The question, in short
A recurrence relation defines terms using constants a and b with first term 3; derive early terms, use two given conditions to form a quadratic in a, then find the larger u₂.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (ai) · 1 mark
Work out the second term of the sequence using a and b.
- 1u₂ = b − a × u₁
- 2u₂ = b − 3a
Answeru₂ = b − 3a
Part (aii) · 1 mark
Work out the third term using a and b.
- 1u₃ = b − a × u₂
- 2u₃ = b − a(b − 3a) = b − ab + 3a²
Answeru₃ = b − a(b − 3a) = b − ab + 3a²
Part (b) · 3 marks
Using the sum of the first three terms equal to 153 and b = a + 9, derive the stated quadratic in a.
- 1u₁ + u₂ + u₃ = 3 + (b − 3a) + (b − ab + 3a²) = 153
- 2Substitute b = a + 9: 3 + (9 − 2a) + (a + 9 − a(a + 9) + 3a²) = 153
- 33 + 9 − 2a + a + 9 − a² − 9a + 3a² = 153
- 42a² − 10a + 21 = 153, so 2a² − 10a − 132 = 0
- 5a² − 5a − 66 = 0
AnswerShown: 3 + (b − 3a) + (b − ab + 3a²) = 153 with b = a + 9 gives 2a² − 10a − 132 = 0, so a² − 5a − 66 = 0
Part (c) · 3 marks
Using your quadratic, determine the greater of the two possible values for the second term.
- 1(a − 11)(a + 6) = 0, so a = 11 or a = −6
- 2u₂ = b − 3a = (a + 9) − 3a = 9 − 2a
- 3a = 11 gives u₂ = −13; a = −6 gives u₂ = 21
- 4Larger value: u₂ = 21
Answeru₂ = 21
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