01 · The rule
Find uₙ and u₅ for 7, 21, 63, …
Hint
There are four multiplications from term 1 to term 5.
Worked solution
a = 7, r = 3. Thus uₙ = 7 × 3^(n − 1), and u₅ = 567.
Understand · explore · practise
Geometric sequences
Find the nth term and common ratio of geometric sequences, handle negative and zero ratios, and solve membership and logarithmic threshold questions.
Before you startIndices, logarithms and inequalities
01 / A constant multiplier
Geometric means multiply by the same number each time.
A geometric sequence satisfies uₙ₊₁ = ruₙ for a fixed real number r, called the common ratio. When a term is nonzero, divide the following term by it to find r.
6, 18, 54, 162, … has r = 3
16, −8, 4, −2, … has r = −½
Check every available adjacent pair against the same multiplier. As with any finite list, assume it is geometric only when that rule is stated or justified; finitely many values cannot uniquely specify an infinite continuation.
02 / The nth term
From term 1 to term n there are n − 1 multiplications.
uₙ = ar^(n − 1), where u₁ = a
First term 6, common ratio 3Worked example
uₙ = 6 × 3^(n − 1)
At n = 1 the exponent is 0, so the first term is 6.
u₆ = 6 × 3⁵ = 1458
Six terms involve five multiplications after the initial value.
For r = 0, define the first term directly as a, then every later term is zero. This avoids needing to interpret 0⁰ in the formula. The recurrence definition remains valid.
Multiply at each stepExplore
The first term stays at 2. Dots are individual terms at integer indices.
u₄ = 2 × 2³ = 16.
Positive ratio above 1: these positive terms increase.
Watch repeated multiplication
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Negative and zero ratios
Keep the ratio inside brackets when raising it to a power.
uₙ = 16(−½)^(n − 1)Worked example
u₅ = 16(−½)⁴ = 1
An even exponent gives a positive multiplier.
u₆ = 16(−½)⁵ = −½
An odd exponent gives a negative multiplier.
With a positive first term and negative nonzero r, odd-indexed terms are positive and even-indexed terms are negative. Magnitudes grow if |r| > 1 and shrink if |r| < 1. The alternating signs mean the sequence is not simply increasing or decreasing.
If r = 1 the sequence is constant. If r = −1 and a ≠ 0, it repeats with period 2. If a = 0, every term is zero for every r: the terms do not determine a unique ratio. Never calculate 0/0.
A list such as 7, 0, 0, … is geometric with r = 0 under the recurrence definition. But 0, 0, 7 cannot be geometric: multiplying zero cannot produce 7.
04 / Find the ratio from two terms
An even index gap can leave two possible real ratios.
Given u₃ = 20 and u₅ = 80Worked example
u₅/u₃ = r² = 4
Division is valid because the known denominator is nonzero.
r = 2 or r = −2
An even power does not identify the sign.
a = u₃/r² = 5 in both cases
The possible sequences start 5, 10, 20,… or 5, −10, 20,…
u₂ = 10 or −10
A condition such as positive ratio would choose only one of these.
For q > p, u_q/u_p = r^(q − p), when u_p ≠ 0. Odd index gaps determine at most one real ratio. An even gap with a negative quotient has no real solution.
05 / Unknown parameters
Equal-ratio equations still need a final substitution check.
k − 1, k + 2, 2k + 4 are consecutive geometric termsWorked example
(k + 2)² = (k − 1)(2k + 4)
For three consecutive geometric terms A, B, C, we have B² = AC.
k² − 2k − 8 = (k − 4)(k + 2) = 0
The candidates are k = 4 and k = −2.
k = 4 gives 3, 6, 12 with r = 2
This is valid.
k = −2 gives −3, 0, 0 with r = 0
This is also valid under uₙ₊₁ = ruₙ.
B² = AC is necessary but is not sufficient when zero terms occur: 0, 0, 7 satisfies the equation but is not geometric. Check the original multiplication rule, and apply any restriction such as r ≠ 0 or positive terms.
06 / Does a value occur?
Both the exponent and the sign must fit an integer index.
Does uₙ = 5 × 3^(n − 1) contain 135 or 245?Worked example
135/5 = 27 = 3³
Thus n − 1 = 3, so 135 is term 4.
245/5 = 49
49 is between 3³ and 3⁴ and is not a power of 3, so 245 is not a term.
Does uₙ = 3(−2)^(n − 1) contain 96?Worked example
|96|/3 = 32 = 2⁵
The magnitude forces n = 6.
u₆ = 3(−2)⁵ = −96
The sign is wrong. Therefore +96 is not present, while −96 is term 6.
For a positive ratio other than 1, logarithms can find a candidate n. It must be a positive integer, and substituting it into the original term rule confirms the value. Handle constant and zero-ratio sequences separately.
07 / Growth thresholds
Use logarithms, then check the integer boundary.
First term exceeding 100 when a = 10 and r = 1.4Worked example
10(1.4)^(n − 1) > 100 ⇒ (1.4)^(n − 1) > 10
Both sides are positive.
(n − 1)ln 1.4 > ln 10
Logarithms preserve order on positive numbers.
n > 1 + ln 10/ln 1.4 ≈ 7.843
Here ln 1.4 > 0, so division preserves the inequality.
u₇ ≈ 75.2954 and u₈ ≈ 105.4135
The terms increase, so the first crossing is n = 8.
Keep the unrounded logarithmic value until deciding the integer. If r is negative, ordinary real ln r is undefined: separate signs and parity before using magnitudes.
08 / Decay thresholds
Dividing by a negative logarithm reverses the inequality.
First term below 10 when a = 80 and r = ¾Worked example
80(¾)^(n − 1) < 10 ⇒ (¾)^(n − 1) < ⅛
The positive terms decrease.
(n − 1)ln(¾) < ln(⅛)
The logarithm itself is increasing.
n − 1 > ln(⅛)/ln(¾) ≈ 7.228
Division reverses the inequality because ln(¾) < 0.
u₈ ≈ 10.6787 and u₉ ≈ 8.0090
The first term below 10 is term 9.
Strictness matters at an exact boundary: 64(½)^(n − 1) is at most 4 first at n = 5, but is strictly below 4 first at n = 6.
09 / Your turn
Track ratio signs, exponent gaps and integer indices.
Use exact arithmetic where possible. All sequence rules are real.
02 · A negative ratio
Find u₆ for 48, −24, 12, …
Hint
Keep −½ in brackets.
Worked solution
u₆ = 48(−½)⁵ = −3/2.
03 · An odd index gap
u₂ = 12 and u₅ = 96. Find r and a.
Hint
Divide the later term by the earlier one.
Worked solution
r³ = 96/12 = 8, so r = 2. Since ar = 12, a = 6.
04 · An even gap
u₂ = 18 and u₄ = 162. Find all possible real a and r.
Hint
The ratio equation is quadratic.
Worked solution
r² = 9, so r = 3 with a = 6, or r = −3 with a = −6. Both pairs satisfy the supplied terms.
05 · A missing term
4, x, 36 are consecutive geometric terms. Find x and the next term.
Hint
Check both square roots.
Worked solution
x² = 144, so x = 12 with r = 3 and next term 108; or x = −12 with r = −3 and next term −108.
06 · Membership
Does 320 occur in uₙ = 5 × 2^(n − 1)? If so, where?
Hint
Divide by the first term.
Worked solution
320/5 = 64 = 2⁶, so n = 7.
07 · Check the sign
Does +160 occur in uₙ = 5(−2)^(n − 1)?
Hint
The magnitude alone is not enough.
Worked solution
160/5 = 32 = 2⁵ forces n = 6, but u₆ = −160. Therefore +160 is not a term.
08 · Growth
Find the first term above 1000 in uₙ = 4 × 3^(n − 1).
Hint
Find the adjacent powers of 3 surrounding 250.
Worked solution
u₆ = 4 × 243 = 972 and u₇ = 4 × 729 = 2916. The positive terms increase, so term 7 is the first above 1000.
09 · Decay and equality
For uₙ = 96(½)^(n − 1), find the first term at most 3 and the first term below 3.
Hint
Identify where equality occurs.
Worked solution
u₆ = 3 is the first at most 3; u₇ = 3/2 is the first strictly below 3.
10 · Zero terms
Using uₙ₊₁ = ruₙ, decide whether 9, 0, 0,… and 0, 0, 9,… can be geometric. Does an all-zero sequence determine r?
Hint
A fixed multiplier cannot turn zero into a nonzero value.
Worked solution
9, 0, 0,… is geometric with r = 0. The second list is not geometric. An all-zero sequence works for any r, so it cannot identify a unique ratio.
10 / Recap
Count multiplications and retain valid signs.
- Geometric means uₙ₊₁ = ruₙ with one fixed r.
- For nonzero r, uₙ = ar^(n − 1); treat r = 0 directly.
- An even index gap can give two real ratios.
- Zero terms require the multiplication rule, not division by zero.
- Membership needs an integer index and the correct sign.
- Decay thresholds reverse when dividing by ln r < 0.
Section 1 of 10 · A constant multiplier