A geometric sequence multiplies by a fixed common ratio r at every step. Its nth term is uₙ = arⁿ⁻¹, where a is the first term and r is any term divided by the one before. At A-Level you find a and r from two terms, show that r satisfies a quadratic, use the fact that the middle of three terms squared equals the product of the outer two, and use logarithms to find when a term passes a value.
Geometric sequences
- nth term: uₙ = arⁿ⁻¹
- Common ratio: r = uₙ₊₁ / uₙ
- Dividing two terms cancels a: u₅/u₃ = r²
- Three consecutive terms x, y, z: y² = xz
The nth term arⁿ⁻¹
Each term is the previous term multiplied by r, so uₙ = arⁿ⁻¹. The power is n − 1, not n, because the first term has not been multiplied at all. If r is negative, the terms alternate in sign; keep r in brackets when raising it to a power. Find r by dividing any term by the term before it.
Finding a and r from two terms
Write each given term as arᵏ, then divide one equation by the other to cancel a and leave a power of r. If the power is even, there are two possible values, positive and negative, so look for a condition such as “r > 0” or “all terms are positive” to choose. Then substitute back to find a.
Conditions that give a quadratic in r
A condition such as “the sum of the second and third terms is 30” gives an equation like ar + ar² = 30. With a known, this is a quadratic in r, so there can be two possible ratios. Divide through by any common factor before factorising.
Three terms in geometric progression
For three consecutive terms of a geometric sequence, the ratio from the first to the second equals the ratio from the second to the third. So the middle term squared equals the product of the outer two: y² = xz. If the terms are expressions in an unknown, this gives a quadratic to solve.
Growth, depreciation and the smallest n
A quantity that changes by a fixed percentage each period is geometric: a 15% loss means multiplying by 0.85; a 4% gain means multiplying by 1.04. To find when a term first passes a value, form an inequality and take logarithms to bring the power down. If r < 1, log r is negative, so dividing by it reverses the inequality.
Worked examples
Example 1. Find the 7th term of 3, 6, 12, 24, …
- a = 3 and r = 2.
- u₇ = 3 × 2⁶ = 3 × 64.
Answer: 192.
Marks: M1 for 3 × 2⁶, A1 for 192.
Example 2. Given u₃ = 20, u₅ = 12.8 and r > 0, show that r = 0.8 and find a.
- u₅/u₃ = r² = 12.8/20 = 0.64, so r = 0.8 since r > 0.
- u₃ = ar² = 0.64a = 20.
Answer: r = 0.8 and a = 31.25.
Marks: M1 for r² = 0.64, A1 for r, A1 for a.
Example 3. A geometric sequence has first term 8, and u₂ + u₃ = 30. Show that 4r² + 4r − 15 = 0 and find the possible values of r.
- 8r + 8r² = 30, so 8r² + 8r − 30 = 0, which divides by 2 to give 4r² + 4r − 15 = 0.
- (2r − 3)(2r + 5) = 0.
Answer: r = 3/2 or r = −5/2.
Marks: M1 for forming the equation, M1 for factorising, A1 for both values.
Example 4. k, k + 4 and 4k are consecutive terms of a geometric sequence. Find the possible values of k.
- The middle term squared equals the product of the others: (k + 4)² = 4k².
- k² + 8k + 16 = 4k², so 3k² − 8k − 16 = 0.
- (3k + 4)(k − 4) = 0.
Answer: k = 4 or k = −4/3.
Marks: M1 for (k + 4)² = 4k², M1 for the quadratic, A1 for both values.
Example 5. A £20 000 car loses 15% of its value each year. Find its value after 5 years, and the first year in which its value is below £8000.
- Value after n years: 20 000 × 0.85ⁿ.
- After 5 years: 20 000 × 0.85⁵ = 8874.1…
- Below 8000: 0.85ⁿ < 0.4. Taking logs, n log 0.85 < log 0.4. Dividing by the negative log 0.85 reverses the inequality: n > 5.64.
Answer: About £8874 after 5 years; it first falls below £8000 during year 6.
Marks: M1 for 20 000 × 0.85⁵, A1 for 8874, M1 for the log inequality, A1 for year 6.
Common mistakes
- Using rⁿ instead of rⁿ⁻¹. The nth term is arⁿ⁻¹.
- Dropping the negative ratio. r² = 0.64 gives r = ±0.8; use the given condition to choose.
- Losing brackets with negative r. (−2)⁴ = 16 but −2⁴ = −16.
- Not reversing the inequality. Dividing by log r when r < 1 means dividing by a negative number.
- Using the percentage as the ratio. A 15% decrease means r = 0.85, not 0.15.
How geometric sequences are examined
Geometric sequences are part of Pure 2 (WMA12) in Edexcel International A-Level, within sequences and series; Edexcel A-Level (9MA0) covers them in Year 2. Typical tasks:
- find a and r from two given terms, choosing between ± values
- show that r satisfies a quadratic, then solve it
- find an unknown when three expressions are consecutive terms
- use logs to find the smallest n for a term above or below a value
Mark-schemed questions on all of these are in the Pure 2 sequences and series exam question set. For more practice, try the sequences and series worksheet or generate questions in the revision engine for sequences and series.
Next, geometric series add the terms up. The log step uses solving equations with logarithms.
Frequently asked questions
What is the nth term of a geometric sequence?
uₙ = arⁿ⁻¹, where a is the first term and r is the common ratio.
How do you find the common ratio?
Divide any term by the term before it. From two terms, divide one by the other to get a power of r.
How do you show three numbers are in geometric progression?
Show that the middle number squared equals the product of the other two.