C. Discrete Functions · Sequences and SeriesGeometric Sequences and the General Term
C2.1 C2.2 C1.4About 75 minutes
Learning goals- I can identify a geometric sequence and find its common ratio.
- I can explain where the formula tₙ = arⁿ⁻¹ comes from.
- I can use the formula to find any term, the number of terms, or a and r from two terms.
- I can connect geometric sequences to exponential functions.
Words to know
geometric sequence
A sequence in which each term is found by multiplying the previous term by the same non-zero number.
common ratio (r)
The constant multiplier: r = tₙ ÷ tₙ₋₁. It can be a whole number, a fraction or negative (but not 0).
general term of a geometric sequence
tₙ = arⁿ⁻¹, where a is the first term.
alternating sequence
A sequence whose terms switch between positive and negative, as happens when r is negative.
Tell three friends
A school club starts a phone tree to cancel practice on a snow day. The coach calls 3 students. Each of them calls 3 more, and each of those calls 3 more. The numbers of calls in each round are 3, 9, 27, 81, …. The jumps are not the same size: 6, 18, 54, …. Instead, each term is the previous term multiplied by the same number. That makes it a geometric sequence, and geometric sequences grow (or shrink) much faster than arithmetic ones.
Definition
A sequence is geometric if the ratio of consecutive terms is constant. That constant is the common ratio r:
r = t₂ ÷ t₁ = t₃ ÷ t₂ = … = tₙ ÷ tₙ₋₁
- 3, 9, 27, 81, … is geometric with r = 3.
- 80, 40, 20, 10, … is geometric with r = 1/2.
- 2, −6, 18, −54, … is geometric with r = −3 (the signs alternate).
- 1, 4, 9, 16, … is not geometric: 4 ÷ 1 = 4 but 9 ÷ 4 = 2.25.
Its recursion formula is t₁ = a, tₙ = r·tₙ₋₁.
Developing the general term
Connect a number example with the algebra. Use 3, 6, 12, 24, … (a = 3, r = 2):
- t₁ = 3 ↔ t₁ = a
- t₂ = 3 × 2 ↔ t₂ = ar
- t₃ = 3 × 2 × 2 = 3 × 2² ↔ t₃ = ar²
- t₄ = 3 × 2³ ↔ t₄ = ar³
The exponent on r is one less than the position. So
tₙ = arⁿ⁻¹
For 3, 6, 12, 24, …: tₙ = 3(2)ⁿ⁻¹, and t₁₀ = 3(2)⁹ = 3 × 512 = 1536.
Each arrow multiplies by r = 2. To reach the 5th term you multiply the first term by 2 four times, so t₅ = 3 × 2⁴ = 48.Geometric sequences are exponential
In the unit Exponential Functions you studied functions like f(x) = 2ˣ. A geometric sequence is a discrete exponential function: tₙ = arⁿ⁻¹ can be rewritten as tₙ = (a/r) × rⁿ, an exponential function of n with base r. Its graph is a set of points on an exponential curve.
- If r > 1 (and a > 0), the terms grow faster and faster.
- If 0 < r < 1, the terms decay towards 0.
- If r < 0, the terms alternate in sign.
Three geometric sequences: 2.5, 5, 10, 20 (r = 2) grows; 32, 16, 8, 4, 2 (r = 1/2) decays towards 0; 1, −2, 4, −8 (r = −2) alternates above and below the axis.Worked exampleExample 1: Finding a term
For 64, −32, 16, −8, …, find the general term and t₈.
- r = −32 ÷ 64 = −1/2 (check: 16 ÷ (−32) = −1/2 ✓). a = 64.
- tₙ = 64(−1/2)ⁿ⁻¹
- t₈ = 64(−1/2)⁷ = 64 × (−1/128) = −1/2. (An odd power of a negative number is negative.)
tₙ = 64(−1/2)ⁿ⁻¹; t₈ = −0.5
Worked exampleExample 2: How many terms?
How many terms are in the geometric sequence 2, 6, 18, …, 4374?
- a = 2, r = 3, last term 4374.
- 4374 = 2(3)ⁿ⁻¹ → 3ⁿ⁻¹ = 2187.
- Write 2187 as a power of 3: 3⁵ = 243, 3⁶ = 729, 3⁷ = 2187. So 3ⁿ⁻¹ = 3⁷.
- Equal bases, so the exponents are equal: n − 1 = 7, n = 8.
Worked exampleExample 3: Two terms given
In a geometric sequence, t₂ = 12 and t₅ = 324. Find a, r and the general term.
- Going from t₂ to t₅ is 3 multiplications by r: r³ = 324 ÷ 12 = 27, so r = 3.
- t₂ = ar: 12 = 3a, so a = 4.
- tₙ = 4(3)ⁿ⁻¹. Check t₅: 4(81) = 324 ✓.
a = 4, r = 3, tₙ = 4(3)ⁿ⁻¹
Worked exampleExample 4: Two possible sequences
In a geometric sequence, t₃ = 20 and t₅ = 80. Find the possible sequences.
- From t₃ to t₅ is 2 multiplications: r² = 80 ÷ 20 = 4, so r = 2 or r = −2.
- t₃ = ar² = 4a = 20, so a = 5 in both cases.
- r = 2: 5, 10, 20, 40, 80, … and r = −2: 5, −10, 20, −40, 80, …
tₙ = 5(2)ⁿ⁻¹ or tₙ = 5(−2)ⁿ⁻¹
Worked exampleExample 5: The first term above a target
Which is the first term of 5, 10, 20, 40, … that is greater than 1000?
- tₙ = 5(2)ⁿ⁻¹ > 1000 means 2ⁿ⁻¹ > 200.
- Powers of 2: 2⁷ = 128 (too small), 2⁸ = 256 (big enough). So n − 1 = 8, n = 9.
- t₉ = 5(256) = 1280. (Check: t₈ = 5(128) = 640 < 1000.)
- Systematic trial like this, or a table on a calculator or spreadsheet, works when the numbers are not exact powers. (In Grade 12, Advanced Functions (MHF4U) adds logarithms for solving these equations exactly.)
Just for funGeometric sequences are terrible at keeping secrets.
Tell one of them something, and soon 2 know, then 4, then 8, then 16…
Hands-on activityTear, stack, repeat
You need: one sheet of scrap paper and a ruler.
- Tear the sheet in half and stack the two pieces. Tear the stack in half and stack again. Repeat as many times as you can (usually 5 or 6 times before the stack is too thick to tear neatly).
- Record the number of pieces after each tear. Write the general term.
- A sheet of printer paper is about 0.1 mm thick. Write a general term for the thickness of the stack after n tears.
- Predict the number of pieces and the thickness after 20 tears (if you could keep going).
What to look for: pieces: 2, 4, 8, 16, 32, … so tₙ = 2(2)ⁿ⁻¹ = 2ⁿ. Thickness: 0.2, 0.4, 0.8, … mm, so Tₙ = 0.2(2)ⁿ⁻¹ = 0.1(2)ⁿ mm. After 20 tears: 2²⁰ = 1 048 576 pieces, and the stack would be 0.1 × 1 048 576 ≈ 104 858 mm ≈ 105 m tall, roughly the height of a 30-storey building. That is how fast r = 2 grows.
Your turn
1Is 5, 15, 45, 135, … geometric? If so, give a and r.
2Find the general term of 5, 15, 45, … and find t₇.
tₙ = 5(3)ⁿ⁻¹; t₇ = 5(3)⁶ = 5 × 729 = 3645.
3Find t₆ for 1000, 100, 10, ….
r = 0.1: t₆ = 1000(0.1)⁵ = 0.01.
4Find t₉ for 1, −2, 4, −8, ….
r = −2: t₉ = 1(−2)⁸ = 256.
5How many terms are in 3, 6, 12, …, 768?
768 = 3(2)ⁿ⁻¹ → 2ⁿ⁻¹ = 256 = 2⁸ → n = 9.
6In a geometric sequence t₁ = 7 and t₄ = 189. Find r and t₆.
r³ = 189 ÷ 7 = 27 → r = 3; t₆ = 7(3)⁵ = 7 × 243 = 1701.
7Which is the first term of 3, 9, 27, … greater than 10 000?
3ⁿ > 10 000 (since tₙ = 3(3)ⁿ⁻¹ = 3ⁿ). 3⁸ = 6561, 3⁹ = 19 683. So t₉ = 19 683.
More practice: warm-up
1What is the common ratio of 4, 12, 36, …?
2What is the common ratio of 50, 10, 2, …?
3Write the first four terms of the geometric sequence with a = 2 and r = 5.
4For tₙ = 6(2)ⁿ⁻¹, find t₅.
5Is 2, 4, 6, 8, … geometric? Explain.
No. 4 ÷ 2 = 2 but 6 ÷ 4 = 1.5: no common ratio. (It is arithmetic.)
More practice: on your own
1Find the general term of 96, 48, 24, … and find the first term that is less than 1.
tₙ = 96(1/2)ⁿ⁻¹. Terms: 96, 48, 24, 12, 6, 3, 1.5, 0.75. t₈ = 0.75 is the first below 1.
2How many terms are in 1/9, 1/3, 1, …, 729?
r = 3, a = 1/9: (1/9)(3)ⁿ⁻¹ = 729 → 3ⁿ⁻¹ = 6561 = 3⁸ → n = 9.
3In a geometric sequence t₄ = 24 and t₇ = −192. Find a and r.
r³ = −192 ÷ 24 = −8 → r = −2; a(−2)³ = 24 → −8a = 24 → a = −3. Sequence: −3, 6, −12, 24, …
4Write tₙ = 5(4)ⁿ⁻¹ as a recursion formula, and describe its graph.
t₁ = 5, tₙ = 4tₙ₋₁. Its graph is separate points on an increasing exponential curve (discrete exponential growth).
5Show that x, x³, x⁵, … (x ≠ 0) is geometric and find t₁₀.
x³ ÷ x = x², x⁵ ÷ x³ = x². r = x². t₁₀ = x(x²)⁹ = x¹⁹.
Challenge
1Find all values of k so that k − 2, k + 2, 3k + 2 are consecutive terms of a geometric sequence.
Equal ratios: (k + 2)/(k − 2) = (3k + 2)/(k + 2) → (k + 2)² = (k − 2)(3k + 2) → k² + 4k + 4 = 3k² − 4k − 4 → 0 = 2k² − 8k − 8 → k² − 4k − 4 = 0 → k = (4 ± √32)/2 = 2 ± 2√2. (Check k = 2 + 2√2 ≈ 4.83: terms ≈ 2.83, 6.83, 16.49, ratios ≈ 2.414 ✓.)
2Insert three numbers between 2 and 162 so that the five numbers form a geometric sequence. Give all possibilities.
162 = 2r⁴ → r⁴ = 81 → r = 3 or r = −3. 2, 6, 18, 54, 162 or 2, −6, 18, −54, 162.
3A geometric sequence has t₂ = 6 and t₄ = 54, and an arithmetic sequence has the same t₁ and t₂. Which sequence has the larger 6th term?
Geometric: r² = 9, r = ±3, a = ±2 (a = 6/r). If r = 3: 2, 6, 18, …, t₆ = 2(3)⁵ = 486. If r = −3: −2, 6, −18, 54, −162, 486, so t₆ = (−2)(−3)⁵ = 486 as well. Arithmetic with t₁ = 2, t₂ = 6: d = 4, t₆ = 22. (Or with t₁ = −2, t₂ = 6: d = 8, t₆ = 38.) The geometric sequence's 6th term (486) is larger in every case.
Check your understandingQuiz
1. What is the common ratio of 2, −8, 32, −128, …?
2. Which is the general term of 7, 14, 28, …?
3. For a = 81 and r = 1/3, what is t₅?
4. How many terms are in 4, 8, 16, …, 1024?
5. In a geometric sequence t₁ = 3 and t₃ = 75. What are the possible values of r?
6. The graph of a geometric sequence with a > 0 and r > 1 is:
For parents and teachersBig idea: Expectation C2.1 asks students to identify geometric sequences, and C2.2 asks them to develop tₙ = arⁿ⁻¹ by connecting the steps of a number example to the algebra, then apply it. Geometric sequences connect to the Exponential Functions unit, and they are the backbone of compound interest in the Financial Applications unit.
How to help: Compare "add the same" (arithmetic) with "multiply by the same" (geometric). Ask your teen to predict which grows faster over 20 steps, then check with a calculator.
Common difficulties:
- Using rⁿ instead of rⁿ⁻¹.
- Order of operations: in 3(2)⁹, raise 2 to the 9th power first, then multiply by 3 (not 6⁹).
- Negative ratios: brackets are essential on a calculator, (−2)⁸ = 256 but −2⁸ = −256.
- Forgetting the negative root when solving r² = 4 (r = ±2).