AQA A-Level Mathematics Paper 1, June 2025: Question 9

10 marks · Medium difficulty · Multi-step Problem

Work with geometric series to find the sum of five terms, express the sum to infinity in terms of the common ratio, and determine its range of values.

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Question

Question 9 consists of parts (a) and (b). Part (a) asks for the exact sum of the first five terms of a geometric series S having second term 60 and common ratio 0.2, worth 4 marks. Part (b) introduces a series T with second term 60 and positive common ratio r. Subpart (b)(i) asks to show that the sum to infinity is 60 divided by (r minus r squared) for 2 marks. Subpart (b)(ii) asks to find the maximum value of r minus r squared for 2 marks. Subpart (b)(iii) asks to hence find the range of possible values of the sum to infinity with full justification, worth 2 marks.
Question text

9 (a) A geometric series, S, has second term 60

The common ratio of S is 0.2

Find the exact value of the sum of the first five terms of S

[4 marks]

9 (b) A different geometric series, T, has second term 60 and positive common ratio r

The sum to infinity of T is T∞

9 (b) (i) Show that

T∞ = 2

r – r

[2 marks]

… 9

(08)

9 (b) (ii) Find the maximum value of r – r2

[2 marks]

9 (b) (iii) Hence find the range of possible values of T∞

Fully justify your answer.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 9 showing: 9(a) sets 0.2a = 60 to find a = 300, uses S_5 formula to obtain 374.88 or 9372/25 (4 marks). 9(b)(i) uses ar = 60 giving a = 60/r, substitutes into sum to infinity formula a/(1-r) to derive 60/(r - r^2) (2 marks). 9(b)(ii) finds the maximum occurs at r = 1/2 giving maximum value 1/4 (2 marks). 9(b)(iii) computes minimum T_infinity = 60/(1/4) = 240, deducing the range T_infinity >= 240 (2 marks). Total: 10 marks.

Q Marking instructions AO Marks Typical solution

9(a) Uses 0.2a = 60 0.2a = 60

1.1a M1

PI by a = 300 a =

Deduces a = 300

60 300 1(− 0.25 )

2.2a R1

PI by the use of S =

0.2 5 −

10.2

Uses a complete method to find =

374.88

the sum of the five terms.

For example

a (1− rn )

Uses Sn = with

1− r

their a 60 , r = 0.2 and n = 5

Or

a (1− rn )

3.1a M1

Uses Sn = with

1− r

a = 60, r = 0.2 and n = 4 and adds

their first term

Or

Uses their a = 300 and r = 0.2 to

calculate all five terms and finds

their sum

300 + 60 + 12 + 2.4 + 0.48

9372

Obtains

25 1.1b– A-LEVELA1 MATHEMATICS – –

OE

Subtotal 4

9(b)(i) a 60

States (T =) ar = 60 a =

1− r r

and a

60 T =

ar = 60 or a = 1− r

r 3.1a M1 60

Or =

60 r (1− r)

(T =) r

States 60

1− r =

r − r2

Completes a reasoned argument

to show T = 2

r − r

a

Must start from (T =) and 2.1 R1

1− r

ar = 60 or a =

r

AG

Subtotal 2

Q Marking instructions AO Marks Typical solution 9

9(b)(ii) 1 r − r2 is at its maximum when

Deduces (r =) 2.2a M1

r =

Deduces maximum value of 2

21 1.1b R1 2 1

r − r = Maximum value of r − r =

Subtotal 2

9(b)(iii) 60 60

Obtains 2 provided Min T = = 240

their r − r 1

2.2a M1

their r − r2 0 4

PI by 240

Deduces T 240 2.2a R1 Hence T 240

Subtotal 2

Question 9 Total 10

How to answer it

A-Level Mathematics: Geometric Series & Quadratic Optimization

WHAT THIS QUESTION TESTS

This question assesses your command of geometric sequences and series (terms, finite sum formula, and sum to infinity), along with algebraic proof, quadratic optimization (via completing the square or differentiation), and inequality reasoning when dealing with reciprocals.

PART 9 (a) • 4 MARKS

Finding the Exact Sum of the First Five Terms

Given: Second term u₂ = 60, Common ratio r = 0.2

📐 Step-by-Step Calculation

  1. Find the first term, a:
    The general term is un = a rn-1, so the second term is:
    u₂ = a r = 60
    a(0.2) = 60 ⇒ a = 60 / 0.2 = 300
    M1 Form equation 0.2a = 60  |  R1 Deduce a = 300
  2. Apply the geometric sum formula for n = 5:
    S₅ = a(1 - r⁵) / (1 - r)
    Substitute a = 300 and r = 0.2 :
    S₅ = 300(1 - 0.2⁵) / (1 - 0.2) = 300(1 - 0.00032) / 0.8 = 300(0.99968) / 0.8
    M1 Correct substitution into sum formula (or calculating and adding all 5 terms: 300 + 60 + 12 + 2.4 + 0.48)
  3. Evaluate to an exact value:
    S₅ = 374.88  (or exact fraction 9372 / 25 )
    A1 Correct exact value

✅ Correct Answer

374.88  or  9372/25

❌ Common Errors

  • Assuming the first term is 60 (misreading "second term").
  • Rounding the final answer: the question explicitly states exact value, so giving a rounded figure like 375 loses the final mark.
PART 9 (b)(i) • 2 MARKS

Proof: Expressing Sum to Infinity in Terms of r

Show that T∞ = 60 / (r - r²)

📐 Algebraic Derivation

  1. Express a in terms of r:
    Since the second term is 60: a r = 60 ⇒ a = 60 / r
  2. Substitute into the sum to infinity formula:
    T∞ = a / (1 - r)
    M1 Stating both standard formulas: T∞ = a / (1 - r) and ar = 60 (or a = 60/r )
  3. Eliminate the nested fraction clearly:
    T∞ = (60 / r) / (1 - r) = 60 / [r(1 - r)] = 60 / (r - r²)
    R1 Fully reasoned argument with no omitted steps leading to given result (AG)

🧠 Exam Technique: "Show That" Proofs

Never skip steps when the answer is given on the page. Clearly state the starting formula T∞ = a / (1 - r) before making substitutions. Multiplying numerator and denominator by r must be explicitly visible.

❌ Common Errors

  • Writing down 60 / (r - r²) immediately without showing r(1 - r) in the denominator.
  • Failing to explicitly define or quote ar = 60 .
PART 9 (b)(ii) • 2 MARKS

Maximizing the Quadratic Expression r - r²

Find the maximum value of r - r²

Method 1: Completing the Square

  1. Factor out -1: -(r² - r)
  2. Complete the square:
    -[ (r - 1/2)² - 1/4 ] = 1/4 - (r - 1/2)²
  3. Since (r - 1/2)² ≥ 0 , the expression is maximized when r = 1/2 .
  4. Maximum value = 1/4 (or 0.25 ).

Method 2: Calculus (Differentiation)

  1. Let f(r) = r - r²
  2. Differentiate with respect to r: f'(r) = 1 - 2r
  3. Set to zero for stationary point:
    1 - 2r = 0 ⇒ r = 1/2
  4. Substitute back:
    f(1/2) = (1/2) - (1/2)² = 1/4

✅ Mark Scheme Breakdown

  • M1 Deduces r = 1/2 (or achieves turning point form).
  • R1 Obtains maximum value of 1/4 (or 0.25 ).

❌ Common Errors

  • Stopping at r = 1/2 ! The question asks for the maximum value of the expression, not just the value of r where it occurs.
  • Sign errors when completing the square with a negative coefficient of r² .
PART 9 (b)(iii) • 2 MARKS

Range of Possible Values of T∞

Hence find the range of possible values of T∞ with full justification

📐 Step-by-Step Justification

  1. Consider the denominator of T∞:
    From part (b)(ii), the denominator is D = r - r² , which has a maximum value of 1/4 .
  2. Consider the constraints on r:
    A geometric series has a sum to infinity if and only if |r| < 1 . We are given that r is positive, so 0 < r < 1 .
    Within this interval, r - r² = r(1 - r) > 0 .
    Therefore: 0 < r - r² ≤ 1/4 .
  3. Deduce the behavior of the reciprocal:
    Because the denominator is strictly positive and has an upper bound of 1/4 , dividing 60 by this denominator gives a minimum value:
    Minimum T∞ = 60 / (1/4) = 240
    M1 Computes 60 / (their maximum) = 60 / (1/4) = 240
  4. State the range:
    As r → 0 or r → 1 , the denominator (r - r²) → 0⁺ , which means T∞ → ∞ .
    Hence, T∞ ≥ 240 .
    R1 Deduces T∞ ≥ 240 with full justification.

✅ Correct Answer

T∞ ≥ 240

💡 The "Reciprocal Inversion" Principle

When a fraction has a positive, constant numerator and a positive variable denominator:

Maximizing the denominator MINIMIZES the overall fraction.

Since the maximum of the denominator is 1/4 , the minimum value of 60 / (r - r²) is 240 .

❌ Common Errors

  • Writing T∞ ≤ 240 : Forgetting that a maximum denominator produces a minimum fraction!
  • Writing T∞ > 240 with a strict inequality: Since r = 0.5 is valid ( |0.5| < 1 ), T∞ can equal 240 exactly.
  • Giving only the single number 240 instead of an inequality representing the full range.

🧠 Examiner Insight

Top-performing students explicitly linked part (b)(ii) using the word "Hence". They clearly justified that because the series converges, 0 < r < 1 , confirming the denominator is strictly positive, which guarantees that T∞ is positive and unbounded above as the denominator approaches 0.

Topics

Pure Mathematics · B: Algebra and functions · D: Sequences and series

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.