Geometric Series Applications

This lesson covers advanced applications of geometric series, including proof of the sum formula, convergence analysis, and real-world modelling. These skills are frequently examined in Edexcel A-Level Mathematics (9MA0).

Proof of the Sum Formula

Prove that S(n) = a(1 - r^n) / (1 - r) for r ≠ 1.

Write the sum: S(n) = a + ar + ar² + ar³ + ... + ar^(n-1)

Multiply both sides by r: r x S(n) = ar + ar² + ar³ + ... + ar^(n-1) + ar^n

Subtract the second from the first: S(n) - r x S(n) = a - ar^n

S(n)(1 - r) = a(1 - r^n)

Since r ≠ 1: S(n) = a(1 - r^n) / (1 - r)

Proof of the Sum to Infinity Formula

Prove that for |r| < 1, S(infinity) = a / (1 - r).

Starting from S(n) = a(1 - r^n) / (1 - r):

When |r| < 1, as n tends to infinity, r^n tends to 0.

Therefore: S(n) tends to a(1 - 0) / (1 - r) = a / (1 - r)

Note: When |r| >= 1, r^n does not approach 0, so the series diverges.

Convergence and Divergence

Condition	Behaviour
	r
	r
r = 1	All terms equal a; S(n) = na tends to infinity
r = -1	Terms alternate a, -a, a, -a; no limit

Worked Example 1

Determine whether each geometric series converges or diverges. If it converges, find S(infinity).

(a) 10 + 5 + 2.5 + 1.25 + ...

r = 1/2, |r| < 1 so it converges. S(infinity) = 10 / (1 - 1/2) = 20

(b) 3 - 6 + 12 - 24 + ...

r = -2, |r| = 2 > 1 so it diverges.

r = 0.8, |r| < 1 so it converges. S(infinity) = 100 / (1 - 0.8) = 100 / 0.2 = 500

Finding r and a from S(infinity)

Worked Example 2

A convergent geometric series has S(infinity) = 36 and the second term is 8. Find the possible values of a and r.

S(infinity) = a/(1-r) = 36, so a = 36(1-r) ... (1)

Second term: ar = 8 ... (2)

Substitute (1) into (2): 36(1-r) x r = 8

36r - 36r² = 8

36r² - 36r + 8 = 0

9r² - 9r + 2 = 0

(3r - 1)(3r - 2) = 0

r = 1/3 or r = 2/3

If r = 1/3: a = 36(1 - 1/3) = 24 If r = 2/3: a = 36(1 - 2/3) = 12

Answer: a = 24, r = 1/3 or a = 12, r = 2/3

Depreciation and Growth Models

Worked Example 3

A machine costs £50,000 and depreciates by 20% each year. Its maintenance costs £3000 in the first year, increasing by £500 each year.

(a) Find the machine's value after 5 years.

Value = 50000 x 0.8^5 = 50000 x 0.32768 = £16,384

(b) Find the total maintenance cost over 5 years.

Maintenance costs form an AP: 3000, 3500, 4000, 4500, 5000

S(5) = 5/2 x (3000 + 5000) = 5/2 x 8000 = £20,000

Total = 50000 + 20000 - 16384 = £53,616

Worked Example 4

Fish are introduced into a lake. Each year 800 fish are added and 90% of the existing fish survive.

After n years, the total population is:

P(n) = 800 + 800(0.9) + 800(0.9²) + ... + 800(0.9^(n-1))

This is a geometric series with a = 800, r = 0.9.

P(n) = 800(1 - 0.9^n) / (1 - 0.9) = 8000(1 - 0.9^n)

As n tends to infinity: P(infinity) = 8000(1 - 0) = 8000 fish (long-term stable population)

Financial Models

Worked Example 5

£200 is deposited at the start of each year into an account earning 5% compound interest per year. Find the total value after 10 years (just after the 10th deposit).

The 1st deposit has earned interest for 9 years (it was deposited at the start): 200 x 1.05^9 ...but immediately after the 10th deposit, the 10th deposit has earned nothing yet: 200

Counting from the most recent deposit: Total = 200(1 + 1.05 + 1.05² + ... + 1.05^9)

This has 10 terms, a geometric series with a = 200, r = 1.05:

= 200 x (1.05^10 - 1) / (1.05 - 1)

= 200 x (1.62889 - 1) / 0.05

= 200 x 12.5779

= £2515.58

Recurring Decimals as Geometric Series

Worked Example 6

Show that 0.777... = 7/9 using a geometric series.

0.777... = 7/10 + 7/100 + 7/1000 + ...

This is a GP with a = 7/10, r = 1/10.

S(infinity) = (7/10) / (1 - 1/10) = (7/10) / (9/10) = 7/9

Worked Example 7

Express 0.454545... as a fraction.

0.454545... = 45/100 + 45/10000 + 45/1000000 + ...

GP with a = 45/100, r = 1/100.

S(infinity) = (45/100) / (1 - 1/100) = (45/100) / (99/100) = 45/99 = 5/11

Exam Tips

Tip	Detail
Proof questions	Be ready to prove both the S(n) and S(infinity) formulae
State convergence	Always state
Financial problems	Draw a timeline showing when each deposit/payment earns interest
Show direction of inequality	When using logs with a negative log base, remember to flip the inequality
Recurring decimals	Write the decimal as a GP, then use S(infinity)

Practice Problems

Prove the formula S(n) = a(r^n - 1)/(r - 1) starting from S(n) = a + ar + ... + ar^(n-1).
A GP has S(infinity) = 60 and first term 15. Find r. [Answer: 3/4]
A machine depreciates by 12% per year from an initial value of £80,000. Find its value after 4 years. [Answer: £47,946.25]
Express 0.363636... as a fraction. [Answer: 4/11]
£500 is invested annually at 3% interest. Find the total after 8 years. [Answer: £4468.35]

Summary

S(n) = a(1 - r^n)/(1 - r) is proved by multiplying by r and subtracting.
S(infinity) = a/(1 - r) follows from r^n tending to 0 when |r| < 1.
Depreciation, population growth, and compound interest are modelled by geometric sequences.
Recurring decimals can be expressed as fractions using the sum to infinity formula.

A-Level Deep Dive: Geometric Series Applications

Spec mapping

Edexcel 9MA0 specification section 4 — Sequences and series, sub-strands 4.2 and 4.3 covers sigma notation for sums of series; understand and work with geometric sequences and series including the formulae for the $n$ th term and the sum of a finite geometric series; understand and use the sum to infinity for a convergent geometric series, including the use of $|r| < 1$ ; use sequences and series in modelling (refer to the official specification document for exact wording). Application questions are explicitly tested in 9MA0-01 (Pure Mathematics 1), often as the longest single question on Paper 1, and reappear synoptically in 9MA0-02 (Pure 2) when integrated with logarithms (solving $ar^{n-1} > T$ requires $\log$ ). Section 6 (Exponentials and logarithms) and section 14 (Numerical methods, in Year 2) both lean on geometric reasoning. The Edexcel formula booklet does list $S_n = \dfrac{a(1 - r^n)}{1 - r}$ and $S_\infty = \dfrac{a}{1 - r}$ — but not the compound-interest formula $A = P(1 + r/100)^n$ , which must be derived or memorised.

Worked example with full mark scheme

Question (8 marks):

A sum of $\pounds 5000$ is invested in an account paying $4\%$ compound interest per annum. Interest is added at the end of each year.

(a) Show that the balance at the end of year $n$ is $5000(1.04)^n$ . (2)

(b) Find the balance at the end of year 10, giving your answer to the nearest penny. (2)

Solution with mark scheme:

(a) Step 1 — set up the geometric progression.

After year 1: $5000 \times 1.04$ . After year 2: $5000 \times 1.04 \times 1.04 = 5000(1.04)^2$ . By induction, the balance at the end of year $n$ is $5000(1.04)^n$ , the $n$ th term of a GP with multiplier $r = 1.04$ applied $n$ times to the initial $5000$ .

M1 — recognising the multiplicative structure (multiply by $1.04$ each year), not additive.

A1 — stating the closed form $5000(1.04)^n$ with correct base.

(b) Step 1 — substitute $n = 10$ .

$5000(1.04)^{10} = 5000 \times 1.48024428\ldots = 7401.2214\ldots$

M1 — substitution into the correct expression.

A1 — $\pounds 7401.22$ to the nearest penny. Common error: rounding $1.04^{10}$ prematurely to $1.48$ gives $\pounds 7400$ , losing the A1 for accuracy.

$5000(1.04)^n > 8000 \implies (1.04)^n > 1.6$

M1 — correct rearrangement.

Step 2 — take logs.

$n \ln(1.04) > \ln(1.6) \implies n > \dfrac{\ln 1.6}{\ln 1.04}$

M1 — applying $\ln$ to both sides correctly. Since $\ln(1.04) > 0$ , the inequality direction is preserved.

Step 3 — evaluate.

$n > \dfrac{0.4700\ldots}{0.03922\ldots} = 11.98\ldots$

A1 — correct numerical bound.

Step 4 — interpret.

Since $n$ must be an integer and the balance first exceeds $\pounds 8000$ at the end of a year, the smallest such $n$ is $n = 12$ .

A1 — correct integer answer with justification (the integer-ceiling step, not just rounding).

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A car is bought for $\pounds 18\,000$ . Each year its value depreciates by $15\%$ of the value at the start of that year.

(a) Find an expression for the value of the car at the end of year $n$ . (2)

(b) The owner decides to sell when the value first falls below $\pounds 6000$ . Find, with justification, the year in which this happens. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO3.3) — recognising that depreciation by $15\%$ multiplies by $0.85$ each year (modelling).
A1 (AO1.1b) — value at end of year $n$ is $18000(0.85)^n$ .

(b)

M1 (AO1.1a) — set up $18000(0.85)^n < 6000$ , equivalently $(0.85)^n < \tfrac{1}{3}$ .
M1 (AO1.1b) — apply $\ln$ : $n \ln(0.85) < \ln(1/3)$ . Crucial: since $\ln(0.85) < 0$ , the inequality flips on division: $n > \dfrac{\ln(1/3)}{\ln(0.85)}$ .
A1 (AO1.1b) — $n > \dfrac{-1.0986}{-0.1625} = 6.76\ldots$ .
A1 (AO2.5) — interpreting: the value first falls below $\pounds 6000$ during year 7 (so the owner sells at the end of year 7).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a standard modelling question — the AO3 mark for setting up the multiplicative factor, the AO2 mark for handling the inequality flip when $\ln r < 0$ .

Synoptic links

Connects to:

Compound interest formula $A = P(1 + r/100)^n$ (KS4 / Year 1): this is just the GP $n$ th-term structure $ar^n$ with $a = P$ and ratio $1 + r/100$ . A-Level reframes the GCSE formula as a special case, exposing the underlying geometric structure.
Section 6 — Exponentials and logarithms: every "find $n$ " application question requires solving $ar^n = T$ for $n$ , which forces $\ln$ . The pair (geometric model, logarithmic solver) is the most heavily examined synoptic combination on 9MA0-01.
Section 7 — Exponential modelling (Year 2): the discrete GP $P(1 + r)^n$ becomes the continuous model $P e^{kt}$ in the limit as compounding intervals shrink. A-Level Year 2 derives $\lim_{n \to \infty}(1 + 1/n)^n = e$ , which is exactly the bridge between geometric and exponential growth.
Recurring decimals (KS3 / Year 1): $0.\dot{3} = \sum_{n=1}^{\infty} 3 \times 10^{-n} = \dfrac{3/10}{1 - 1/10} = \dfrac{1}{3}$ . Every recurring decimal is an infinite GP with $r = 10^{-k}$ where $k$ is the period length.
Zeno's paradox (history of mathematics): the dichotomy paradox — to traverse a distance, one must first traverse half, then half of the remainder, and so on — produces the GP $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = 1$ . The sum-to-infinity formula resolves the paradox, and is the historical motivation for studying convergent series.

Mark-scheme literacy

Application questions on 9MA0 split AO marks more evenly than pure manipulation questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	40–50%	Applying GP formulae, taking logs, evaluating with a calculator, stating exact answers
AO2 (reasoning / interpretation)	20–30%	Justifying inequality direction (flip when $\ln r < 0$ ), interpreting "first exceeds" as a ceiling, naming the GP and stating $a$ , $r$
AO3 (problem-solving / modelling)	25–35%	Translating real-world percentages into a multiplicative ratio ( $15\%$ depreciation $\to r = 0.85$ ), recognising the model is geometric not arithmetic, validating the model assumption

Examiner-rewarded phrasing: "the balance forms a geometric progression with $a = 5000$ and $r = 1.04$ "; "since $|r| = 0.85 < 1$ , the value tends to zero"; "since $n$ must be a positive integer and $n > 11.98$ , the smallest valid value is $n = 12$ ". Phrases that lose marks: "round up to 12" (without stating why — the integer constraint and the strictness of "first exceeds" must be named); "the value will be zero eventually" (untrue — depreciation is asymptotic); " $15\%$ off, so $r = 0.15$ " (this is the percentage removed, not the ratio).

A specific Edexcel pattern to watch: questions about compound interest paid in advance (annuities-due) versus in arrears (ordinary annuities) shift the index by one. Read the timing carefully — "interest paid at the start of the year" gives $P(1+r)^{n+1}$ over $n$ years, not $P(1+r)^n$ .

Grade-band model answers

3-mark question

Question: A population of bacteria doubles every hour. The initial population is 200. Find the population after 8 hours.

Grade C response (~210 words):

The population is a GP with $a = 200$ and ratio $r = 2$ . After 8 hours, the population is $200 \times 2^8 = 200 \times 256 = 51200$ .

So the population after 8 hours is 51200 bacteria.

Geometric Series Applications

Geometric Series Applications

Proof of the Sum Formula

Proof of the Sum to Infinity Formula

Convergence and Divergence

Worked Example 1

Finding r and a from S(infinity)

Worked Example 2

Depreciation and Growth Models

Worked Example 3

Worked Example 4

Financial Models

Worked Example 5

Recurring Decimals as Geometric Series

Worked Example 6

Worked Example 7

Exam Tips

Practice Problems

Summary

A-Level Deep Dive: Geometric Series Applications

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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