Geometric Sequences

A geometric sequence (also called a geometric progression, or GP) is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio. Geometric sequences model exponential growth and decay and are widely tested on the Edexcel A-Level (9MA0).

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Key Definitions

Term	Meaning
Geometric sequence	A sequence with a constant ratio between consecutive terms
Common ratio (r)	The multiplier: r = u(n+1) / u(n)
First term (a)	The starting value, u(1)
Convergent GP	A GP where

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The nth Term Formula

For a geometric sequence with first term a and common ratio r:

u(n) = a x r^(n-1)

Worked Example 1

Find the 8th term of the GP 3, 6, 12, 24, ...

• a = 3, r = 6/3 = 2
• u(8) = 3 x 2^7 = 3 x 128 = 384

Worked Example 2

The 2nd term of a GP is 12 and the 5th term is 324. Find a and r.

• u(2) = ar = 12 ... (1)
• u(5) = ar^4 = 324 ... (2)

Divide (2) by (1): r^3 = 324/12 = 27, so r = 3

Substitute back: a(3) = 12, so a = 4

Answer: a = 4, r = 3

Worked Example 3

A GP has first term 500 and common ratio 0.8. Find which term is the first to be less than 1.

We need ar^(n-1) < 1:

500 x 0.8^(n-1) < 1

0.8^(n-1) < 1/500 = 0.002

Taking logarithms: (n-1) x log(0.8) < log(0.002)

Since log(0.8) is negative, dividing reverses the inequality:

n - 1 > log(0.002) / log(0.8) = -2.699 / -0.09691 ≈ 27.85

So n - 1 > 27.85, meaning n > 28.85

The 29th term is the first less than 1.

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Sum of n Terms

The sum of the first n terms of a geometric series is:

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

or equivalently:

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

The first form is generally more convenient when |r| < 1, and the second when r > 1.

Worked Example 4

Find the sum of the first 10 terms of 2 + 6 + 18 + 54 + ...

• a = 2, r = 3, n = 10
• S(10) = 2(3^10 - 1) / (3 - 1)
• S(10) = 2(59049 - 1) / 2
• S(10) = 59048

Worked Example 5

Find the sum of the first 8 terms of 100 + 50 + 25 + 12.5 + ...

• a = 100, r = 0.5, n = 8
• S(8) = 100(1 - 0.5^8) / (1 - 0.5)
• S(8) = 100(1 - 0.00390625) / 0.5
• S(8) = 100 x 0.99609375 / 0.5
• S(8) = 199.22 (to 2 d.p.)

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Sum to Infinity

When |r| < 1, the terms get closer and closer to zero, and the sum converges to a finite limit:

S(infinity) = a / (1 - r) provided |r| < 1

This is an extremely important result that appears frequently in A-Level exams.

Worked Example 6

Find the sum to infinity of 8 + 4 + 2 + 1 + ...

• a = 8, r = 1/2
• Since |r| = 1/2 < 1, the sum converges
• S(infinity) = 8 / (1 - 1/2) = 8 / (1/2) = 16

Worked Example 7

A GP has sum to infinity 20 and first term 12. Find the common ratio.

S(infinity) = a / (1 - r)

20 = 12 / (1 - r)

1 - r = 12/20 = 3/5

r = 1 - 3/5 = 2/5

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Compound Interest and Depreciation

Geometric sequences naturally model compound interest and depreciation.

Worked Example 8

£5000 is invested at 3% compound interest per annum. Find the value after 10 years.

Each year the amount is multiplied by 1.03:

Value after n years = 5000 x 1.03^n

After 10 years: 5000 x 1.03^10 = 5000 x 1.34392... = £6719.58 (to nearest penny)

Worked Example 9

A car worth £18,000 depreciates by 15% each year. After how many complete years will it first be worth less than £5000?

Value after n years = 18000 x 0.85^n

We need 18000 x 0.85^n < 5000:

0.85^n < 5000/18000 = 5/18

n x log(0.85) < log(5/18)

n > log(5/18) / log(0.85) = -0.5563 / -0.07058 ≈ 7.88

After 8 complete years.

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Exam Tips

Tip	Detail
State convergence condition	When using S(infinity), always state that
Ratio from two terms	Divide one term equation by another to eliminate a
Negative r	If r < 0, terms alternate in sign — still a valid GP
Percentage problems	5% increase means r = 1.05; 12% decrease means r = 0.88
Exact vs decimal	Give exact answers unless told otherwise

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Practice Problems

1. Find the 6th term of the GP 2, 10, 50, ... [Answer: 6250]
2. A GP has first term 81 and 4th term 3. Find r. [Answer: r = 1/3]
3. Find S(infinity) for 12 + 6 + 3 + 1.5 + ... [Answer: 24]
4. A GP has S(infinity) = 40 and r = 3/4. Find a. [Answer: 10]
5. £2000 is invested at 4.5% compound interest. Find the value after 6 years. [Answer: £2601.59]

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Summary

• A geometric sequence has constant common ratio r.
• nth term: u(n) = ar^(n-1)
• Sum of n terms: S(n) = a(1 - r^n) / (1 - r)
• Sum to infinity: S(infinity) = a / (1 - r), valid only when |r| < 1
• Compound interest and depreciation are modelled by geometric sequences.

A-Level Deep Dive: Geometric Sequences

Spec mapping

Edexcel 9MA0 specification — Sequences and series (geometric) covers geometric sequences and series; the formula for the $n$nth term and the formula for the sum of a finite geometric series; the formula for the sum to infinity of a convergent geometric series, including the use of $|r| < 1$∣r∣<1; modulus notation (refer to the official specification document for exact wording). This sub-strand sits in the Pure content of 9MA0-01 (Paper 1) and is also examinable on 9MA0-02 (Paper 2). The geometric sum formulae $S_n = \dfrac{a(1 - r^n)}{1 - r}$Sn​=1−ra(1−rn)​ and $S_\infty = \dfrac{a}{1 - r}$S∞​=1−ra​ appear in section 5 of the Edexcel formula booklet, but the $n$nth term formula $u_n = ar^{n-1}$un​=arn−1 does not — that must be memorised. Synoptically the topic threads into arithmetic sequences (the parallel structure of $u_n$un​ and $S_n$Sn​ formulae), binomial expansion (the special case $(1 + x)^{-1} = 1 + x + x^2 + \cdots$(1+x)−1=1+x+x2+⋯ for $|x| < 1$∣x∣<1 is exactly a GP with first term $1$1 and ratio $x$x, and convergence is the same condition), compound interest and depreciation modelling (every annuity, mortgage, or savings calculation is a GP sum), the convergence test $|r| < 1$∣r∣<1 (which becomes the ratio test in Year 2 series work and undergraduate analysis), and recurring decimals, which Edexcel A-Level reframes as geometric series rather than as a long-division curiosity.

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Worked example with full mark scheme

Question (8 marks):

A geometric sequence has third term $u_3 = 12$u3​=12 and sixth term $u_6 = 96$u6​=96.

(a) Find the first term $a$a and the common ratio $r$r. (3)

(b) Find the sum of the first 10 terms, giving your answer as an exact integer. (3)

(c) Determine, with reasoning, whether the sum to infinity exists. If it does, evaluate it; if it does not, explain why. (2)

Solution with mark scheme:

(a) Step 1 — set up the ratio of consecutive given terms.

Using $u_n = ar^{n-1}$un​=arn−1:

$\dfrac{u_6}{u_3} = \dfrac{ar^5}{ar^2} = r^3 = \dfrac{96}{12} = 8$u3​u6​​=ar2ar5​=r3=1296​=8

M1 — forming the quotient $u_6 / u_3$u6​/u3​ to eliminate $a$a and isolate a power of $r$r. Common error: subtracting $u_6 - u_3$u6​−u3​ as if the sequence were arithmetic. That misuses the structure entirely and earns nothing.

Step 2 — solve for $r$r.

$r^3 = 8 \implies r = 2$r3=8⟹r=2.

A1 — correct $r = 2$r=2. (No need to consider $r = -2$r=−2 here, because cube roots over the reals are unique. If the question had given $u_6 / u_2 = r^4$u6​/u2​=r4, both signs would need consideration.)

Step 3 — back-substitute to find $a$a.

$u_3 = ar^2 = a \cdot 4 = 12 \implies a = 3$u3​=ar2=a⋅4=12⟹a=3.

A1 — correct $a = 3$a=3.

(b) Step 1 — apply the sum-to-$n$n-terms formula.

With $a = 3$a=3, $r = 2$r=2, $n = 10$n=10:

$S_{10} = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{3(1 - 2^{10})}{1 - 2}$S10​=1−ra(1−rn)​=1−23(1−210)​

M1 — substituting into the correct formula. Note the question gives an exact answer, so calculator decimals are not enough; $2^{10} = 1024$210=1024 must be handled exactly.

Step 2 — evaluate.

$2^{10} = 1024$210=1024, so $1 - 1024 = -1023$1−1024=−1023 and $1 - 2 = -1$1−2=−1:

$S_{10} = \dfrac{3 \cdot (-1023)}{-1} = 3 \cdot 1023 = 3069$S10​=−13⋅(−1023)​=3⋅1023=3069

M1 — correct arithmetic with the negative signs. Sign-handling is the most common slip: candidates write $\dfrac{3(1 - 1024)}{-1} = -3069$−13(1−1024)​=−3069, dropping the cancellation of negatives.

A1 — exact integer answer $S_{10} = 3069$S10​=3069.

(c) Since $r = 2$r=2 and $|r| = 2 \not< 1$∣r∣=2<1, the geometric series diverges: the partial sums grow without bound. The sum to infinity does not exist.

B1 — explicit reference to the convergence condition $|r| < 1$∣r∣<1.

B1 — correct conclusion (divergence) with brief justification.

Total: 8 marks (M3 A3 B2, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A different geometric sequence $\{v_n\}${vn​} has first term $v_1 = 16$v1​=16 and third term $v_3 = 4$v3​=4.

(a) Find the two possible values of the common ratio $r$r. (2)

(b) For each value of $r$r, determine whether the sum to infinity exists, and if so evaluate it. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using $v_3 = v_1 r^2 = 16 r^2 = 4 \implies r^2 = \tfrac{1}{4}$v3​=v1​r2=16r2=4⟹r2=41​.
• A1 (AO1.1b) — both roots stated: $r = \tfrac{1}{2}$r=21​ and $r = -\tfrac{1}{2}$r=−21​.

(b)

• B1 (AO2.4) — for each $r$r, check $|r| < 1$∣r∣<1: both $|\tfrac{1}{2}| = \tfrac{1}{2} < 1$∣21​∣=21​<1 and $|-\tfrac{1}{2}| = \tfrac{1}{2} < 1$∣−21​∣=21​<1, so both series converge and the sum to infinity exists in both cases.
• M1 (AO1.1b) — substituting into $S_\infty = \dfrac{a}{1 - r}$S∞​=1−ra​ for at least one value of $r$r.
• A1 (AO1.1b) — for $r = \tfrac{1}{2}$r=21​: $S_\infty = \dfrac{16}{1 - 1/2} = \dfrac{16}{1/2} = 32$S∞​=1−1/216​=1/216​=32.
• A1 (AO1.1b) — for $r = -\tfrac{1}{2}$r=−21​: $S_\infty = \dfrac{16}{1 - (-1/2)} = \dfrac{16}{3/2} = \dfrac{32}{3}$S∞​=1−(−1/2)16​=3/216​=332​.

Total: 6 marks split AO1 = 5, AO2 = 1. The AO2 mark is the explicit convergence check — Edexcel rewards stating "since $|r| < 1$∣r∣<1 the sum exists" before substituting. Candidates who jump straight to the formula without that check forfeit the reasoning mark even when the arithmetic is correct.

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Synoptic links

Connects to:

1. Arithmetic sequences (parallel structure): the AP formulae $u_n = a + (n-1)d$un​=a+(n−1)d and $S_n = \tfrac{n}{2}(2a + (n-1)d)$Sn​=2n​(2a+(n−1)d) run in lock-step with the GP formulae $u_n = ar^{n-1}$un​=arn−1 and $S_n = \dfrac{a(1 - r^n)}{1 - r}$Sn​=1−ra(1−rn)​. The single most diagnostic A-Level error on this topic is using one set of formulae for the other type of sequence; the cure is always to first identify whether successive terms differ by adding (AP) or multiplying (GP).

2. Compound interest, depreciation and annuity modelling (Section 13 — Numerical methods and modelling): an investment of $£P$£P at annual rate $r\%$r% has value $P(1 + r/100)^n$P(1+r/100)n after $n$n years — a GP with first term $P$P and ratio $1 + r/100$1+r/100. Annuity sums (regular yearly deposits) are partial GP sums. Every mortgage repayment schedule is a finite GP whose convergence under $|r| < 1$∣r∣<1 is the condition for the loan to ever be repaid.

3. Binomial expansion for non-integer index (Section 4 — Algebra and functions): the expansion $(1 - x)^{-1} = 1 + x + x^2 + x^3 + \cdots$(1−x)−1=1+x+x2+x3+⋯ for $|x| < 1$∣x∣<1 is literally the GP sum $\sum_{n=0}^{\infty} x^n = \dfrac{1}{1 - x}$∑n=0∞​xn=1−x1​. The convergence range "$|x| < 1$∣x∣<1" that Edexcel demands you state for binomial expansions is the geometric convergence condition wearing a different costume.

4. Calculus — Taylor series convergence (preview of Year 2 / undergraduate): every Taylor series has a radius of convergence established by comparison with a geometric series. The series $\sum x^n / n!$∑xn/n! for $e^x$ex converges everywhere because the ratio of successive terms $\to 0$→0, beating any GP. The series $\sum x^n / n$∑xn/n for $-\ln(1 - x)$−ln(1−x) converges only for $|x| < 1$∣x∣<1 — exactly the GP condition.

5. The ratio test (introduction): for a general series $\sum a_n$∑an​, computing $L = \lim_{n \to \infty} |a_{n+1} / a_n|$L=limn→∞​∣an+1​/an​∣ generalises the GP ratio. If $L < 1$L<1 the series converges, if $L > 1$L>1 it diverges, mirroring $|r| < 1$∣r∣<1 for GPs. The GP is the prototype every other convergence theorem reduces to.

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Mark-scheme literacy

Geometric-sequence questions on 9MA0 split AO marks across all three objectives, with a notable AO2 reasoning component for convergence:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Substituting into $u_n = ar^{n-1}$un​=arn−1, $S_n$Sn​, $S_\infty$S∞​; solving for $a$a and $r$r from given terms; arithmetic with powers and signs
AO2 (reasoning / interpretation)	20–30%	Stating and applying the $
AO3 (problem-solving / modelling)	5–15%	Compound-interest, savings or population-decay modelling questions where the candidate must identify the GP structure from a real-world description

Examiner-rewarded phrasing: "since $|r| < 1$∣r∣<1, the sum to infinity exists and equals…"; "the common ratio is $r = u_{n+1} / u_n$r=un+1​/un​, constant for all $n$n"; "the series diverges because $|r| \geq 1$∣r∣≥1, so partial sums grow without bound". Phrases that lose marks: "the sum to infinity is" with no convergence check (loses the AO2 mark even with correct arithmetic); "the series goes on forever" instead of "diverges" (vague, loses B-mark); using $\dfrac{a(r^n - 1)}{r - 1}$r−1a(rn−1)​ when $r < 1$r<1 — although algebraically equivalent to the standard form, mismatched signs in working often produce wrong numerical answers.