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Geometric Progressions (GP)

CSEC Additional Mathematics Essential Knowledge: Geometric Progressions (GP) are sequences where each term is obtained by multiplying the previous term by a constant factor. Understanding GP is crucial for modeling exponential growth and decay, calculating compound interest, and solving problems involving repeated multiplication patterns. Mastery of GP formulas is essential for success in sequences, series, and financial mathematics.

Key Concept: A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the “common ratio” (denoted by \(r\)). If \(a\) is the first term, then the sequence is: \(a, ar, ar^2, ar^3, \ldots\)

Part 1: Recognizing Geometric Progressions

🔍

Identifying GP Sequences

📏
The Common Ratio Test

For a sequence to be a GP, the ratio between consecutive terms must be constant:

\(r = \frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} = \cdots\)

Where \(r\) is the common ratio, and \(T_n\) represents the \(n\)th term.

\(a\)
→ ×\(r\)
\(ar\)
→ ×\(r\)
\(ar^2\)
→ ×\(r\)
\(ar^3\)
→ ×\(r\)
⋯

Example: GP Sequence

Sequence: 2, 6, 18, 54, 162, …

\(r = \frac{6}{2} = 3\)

\(r = \frac{18}{6} = 3\) ✓

\(r = \frac{54}{18} = 3\) ✓

Constant ratio: This is a GP with \(r = 3\)

Example: NOT a GP

Sequence: 3, 6, 10, 15, 21, …

\(\frac{6}{3} = 2\)

\(\frac{10}{6} = 1.67\) ✗

\(\frac{15}{10} = 1.5\) ✗

Different ratios: This is NOT a GP

📝 Example 1: Identifying GP and Finding Common Ratio

Determine if each sequence is a GP. If yes, find the common ratio \(r\).

(a)
5, 15, 45, 135, …
\(\frac{15}{5} = 3\), \(\frac{45}{15} = 3\), \(\frac{135}{45} = 3\) ✓
GP with \(r = 3\)
(b)
16, 8, 4, 2, 1, …
\(\frac{8}{16} = 0.5\), \(\frac{4}{8} = 0.5\), \(\frac{2}{4} = 0.5\) ✓
GP with \(r = 0.5\)
(c)
-2, 4, -8, 16, -32, …
\(\frac{4}{-2} = -2\), \(\frac{-8}{4} = -2\), \(\frac{16}{-8} = -2\) ✓
GP with \(r = -2\)
(d)
\(x, x^2, x^3, x^4, …\)
\(\frac{x^2}{x} = x\), \(\frac{x^3}{x^2} = x\) ✓
GP with \(r = x\)

Key Insight: The common ratio \(r\) can be positive (all terms same sign), negative (alternating signs), greater than 1 (growth), between 0 and 1 (decay), or negative with absolute value less than 1 (alternating decay). All are valid geometric progressions!

Part 2: The nth Term Formula

📐

Finding Any Term in a GP

🎯
The General Term Formula
\(T_n = ar^{n-1}\)

Where:

  • \(T_n\) = the nth term
  • \(a\) = first term (\(T_1\))
  • \(r\) = common ratio
  • \(n\) = term number
Term 1
\(a\)
× \(r^0\)
Term 2
\(ar\)
× \(r^1\)
Term 3
\(ar^2\)
× \(r^2\)
Term \(n\)
\(ar^{n-1}\)
× \(r^{n-1}\)
📝 Example 2: Finding Specific Terms

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

1
Identify \(a = 3\) (first term)
2
Find \(r = \frac{6}{3} = 2\)
3
Use formula: \(T_n = ar^{n-1}\)
4
For \(n = 8\): \(T_8 = 3 × 2^{8-1} = 3 × 2^7\)
5
Calculate: \(T_8 = 3 × 128 = 384\)
6
Answer: The 8th term is 384
📝 Example 3: Finding Term Number

Which term of the GP 2, 6, 18, 54, … is 4374?

1
Identify \(a = 2\), \(r = \frac{6}{2} = 3\)
2
We want \(T_n = 4374\)
3
Use formula: \(4374 = 2 × 3^{n-1}\)
4
Divide by 2: \(2187 = 3^{n-1}\)
5
Express 2187 as power of 3: \(2187 = 3^7\)
6
So: \(3^{n-1} = 3^7\)
7
Equate exponents: \(n-1 = 7\)
8
\(n = 8\)
9
Answer: 4374 is the 8th term

Three Variables Formula

\(T_n = ar^{n-1}\)

Given any three, you can find the fourth

Relationship Between Terms

\(\frac{T_n}{T_m} = r^{n-m}\)

Ratio between terms m and n

Middle Term Property

\(T_k^2 = T_{k-1} × T_{k+1}\)

Square of middle term equals product of neighbors

Part 3: Sum of n Terms of a GP

∑

Calculating Finite Sums

💰
The Sum Formulas

When \(r \neq 1\)

\(S_n = \frac{a(1 – r^n)}{1 – r}\)

Alternative form: \(S_n = \frac{a(r^n – 1)}{r – 1}\)

Use this for all cases except \(r = 1\)

When \(r = 1\)

\(S_n = na\)

Reason: All terms are equal to \(a\)

Example: 5, 5, 5, 5, … has sum \(5n\)

Memory Aid: For \(r \neq 1\), use \(S_n = \frac{a(1 – r^n)}{1 – r}\) when \(|r| < 1\) (usually gives positive denominator), and \(S_n = \frac{a(r^n - 1)}{r - 1}\) when \(|r| > 1\) (avoids negative denominator).

📝 Example 4: Finding Sum of Terms

Find the sum of the first 6 terms of the GP: 2, 6, 18, 54, …

1
Identify: \(a = 2\), \(r = \frac{6}{2} = 3\), \(n = 6\)
2
Since \(r \neq 1\), use: \(S_n = \frac{a(r^n – 1)}{r – 1}\)
3
Substitute: \(S_6 = \frac{2(3^6 – 1)}{3 – 1}\)
4
Calculate: \(3^6 = 729\)
5
Simplify: \(S_6 = \frac{2(729 – 1)}{2} = \frac{2 × 728}{2}\)
6
Final: \(S_6 = 728\)
7
Answer: Sum of first 6 terms is 728
📝 Example 5: Sum with Fractional Ratio

Find the sum of the first 5 terms of the GP: 64, 32, 16, 8, …

1
Identify: \(a = 64\), \(r = \frac{32}{64} = 0.5\), \(n = 5\)
2
Use formula: \(S_n = \frac{a(1 – r^n)}{1 – r}\)
3
Substitute: \(S_5 = \frac{64(1 – 0.5^5)}{1 – 0.5}\)
4
Calculate: \(0.5^5 = 0.03125\)
5
Simplify: \(S_5 = \frac{64(1 – 0.03125)}{0.5} = \frac{64 × 0.96875}{0.5}\)
6
Calculate: \(64 × 0.96875 = 62\)
7
Divide: \(S_5 = \frac{62}{0.5} = 124\)
8
Answer: Sum of first 5 terms is 124

Part 4: Sum to Infinity

∞

Infinite Geometric Series

For a GP with \(|r| < 1\), the sum to infinity converges to a finite value:

\(S_\infty = \frac{a}{1 – r}\) where \(|r| < 1\)

If \(|r| ≥ 1\), the sum to infinity does not exist (diverges)

🎯
Why This Works

When \(|r| < 1\), as \(n → ∞\), \(r^n → 0\). So:

\(S_n = \frac{a(1 – r^n)}{1 – r} → \frac{a(1 – 0)}{1 – r} = \frac{a}{1 – r}\)

Visual Example: Adding \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots\) gets closer and closer to 1, but never exceeds it.

📝 Example 6: Sum to Infinity

Find the sum to infinity of the GP: 16, 4, 1, 0.25, …

1
Identify: \(a = 16\), \(r = \frac{4}{16} = 0.25\)
2
Check condition: \(|r| = 0.25 < 1\) ✓ So sum exists
3
Use formula: \(S_\infty = \frac{a}{1 – r}\)
4
Substitute: \(S_\infty = \frac{16}{1 – 0.25} = \frac{16}{0.75}\)
5
Calculate: \(S_\infty = \frac{16}{0.75} = \frac{16}{3/4} = 16 × \frac{4}{3} = \frac{64}{3}\)
6
Answer: \(S_\infty = \frac{64}{3} ≈ 21.333…\)
📝 Example 7: Expressing Recurring Decimal as Fraction

Express 0.777… (0.\overline{7}) as a fraction using GP.

1
Write as sum: \(0.777… = 0.7 + 0.07 + 0.007 + 0.0007 + \cdots\)
2
Identify GP: \(a = 0.7\), \(r = \frac{0.07}{0.7} = 0.1\)
3
Check: \(|r| = 0.1 < 1\) ✓ Sum exists
4
Use formula: \(S_\infty = \frac{a}{1 – r} = \frac{0.7}{1 – 0.1} = \frac{0.7}{0.9}\)
5
Simplify: \(\frac{0.7}{0.9} = \frac{7/10}{9/10} = \frac{7}{9}\)
6
Answer: \(0.\overline{7} = \frac{7}{9}\)

Critical Check: The sum to infinity formula \(S_\infty = \frac{a}{1 – r}\) ONLY applies when \(|r| < 1\). If \(|r| ≥ 1\), the series diverges (sum to infinity doesn't exist). Always verify this condition before using the formula!

Part 5: Real-World Applications

🌍

GP in Practical Situations

Common Applications of Geometric Progressions:

Compound interest: \(A = P(1 + r)^n\)
Population growth (with constant percentage increase)
Radioactive decay (half-life calculations)
Depreciation of assets
Multilevel marketing/commission structures
Bacteria growth in biology
Fractal geometry (self-similar patterns)
💰
Compound Interest Example

Problem: $1000 is invested at 5% per annum compound interest. Find the amount after 10 years.

1
Each year multiplies by \(1 + 0.05 = 1.05\)
2
This forms a GP: \(a = 1000\), \(r = 1.05\), \(n = 11\) (Year 0 to Year 10)
3
Amount after 10 years: \(T_{11} = 1000 × 1.05^{10}\)
4
Calculate: \(1.05^{10} ≈ 1.62889\)
5
Amount: \(1000 × 1.62889 ≈ $1628.89\)
6
Answer: Approximately $1628.89
📈
Population Growth Example

Problem: A bacteria culture doubles every hour. If initially there are 100 bacteria, how many will there be after 6 hours?

1
Doubling means \(r = 2\)
2
GP: \(a = 100\), \(r = 2\), \(n = 7\) (Hour 0 to Hour 6)
3
After 6 hours: \(T_7 = 100 × 2^{6} = 100 × 64 = 6400\)
4
Answer: 6400 bacteria

Part 6: CSEC Past Paper Questions

📚

Exam-Style Questions

CSEC May/June 2019 Paper 2

Question: The third term of a geometric progression is 36 and the sixth term is 972.

(a) Find the common ratio.

(b) Find the first term.

(c) Calculate the sum of the first 8 terms.

(a)

\(T_3 = ar^2 = 36\)

\(T_6 = ar^5 = 972\)

Divide: \(\frac{T_6}{T_3} = \frac{ar^5}{ar^2} = r^3 = \frac{972}{36} = 27\)

So \(r^3 = 27\) ⇒ \(r = 3\) (since \(3^3 = 27\))

(b)

Using \(T_3 = ar^2 = 36\):

\(a × 3^2 = 36\) ⇒ \(9a = 36\) ⇒ \(a = 4\)

(c)

\(a = 4\), \(r = 3\), \(n = 8\)

\(S_8 = \frac{a(r^n – 1)}{r – 1} = \frac{4(3^8 – 1)}{3 – 1}\)

\(3^8 = 6561\)

\(S_8 = \frac{4(6561 – 1)}{2} = \frac{4 × 6560}{2} = 2 × 6560 = 13120\)

CSEC January 2018 Paper 2

Question: The sum to infinity of a geometric progression is 27. The second term is 6.

(a) Find the common ratio.

(b) Find the first term.

(c) Calculate the sum of the first 5 terms.

(a)

\(S_\infty = \frac{a}{1 – r} = 27\) … (1)

\(T_2 = ar = 6\) … (2)

From (2): \(a = \frac{6}{r}\)

Substitute into (1): \(\frac{6/r}{1 – r} = 27\)

\(\frac{6}{r(1 – r)} = 27\)

\(6 = 27r(1 – r)\)

Divide by 3: \(2 = 9r(1 – r)\)

\(2 = 9r – 9r^2\)

\(9r^2 – 9r + 2 = 0\)

Solve: \((3r – 1)(3r – 2) = 0\)

\(r = \frac{1}{3}\) or \(r = \frac{2}{3}\)

Both satisfy \(|r| < 1\)

(b)

If \(r = \frac{1}{3}\): \(a = \frac{6}{1/3} = 18\)

If \(r = \frac{2}{3}\): \(a = \frac{6}{2/3} = 9\)

Two possible GPs: (18, 6, 2, …) or (9, 6, 4, …)

(c)

Case 1: \(a = 18\), \(r = \frac{1}{3}\)

\(S_5 = \frac{18(1 – (1/3)^5)}{1 – 1/3} = \frac{18(1 – 1/243)}{2/3} = \frac{18 × 242/243}{2/3} = \frac{18 × 242 × 3}{243 × 2} = \frac{13068}{486} = 26.888…\)

Case 2: \(a = 9\), \(r = \frac{2}{3}\)

\(S_5 = \frac{9(1 – (2/3)^5)}{1 – 2/3} = \frac{9(1 – 32/243)}{1/3} = \frac{9 × 211/243}{1/3} = \frac{9 × 211 × 3}{243} = \frac{5697}{243} = 23.444…\)

Quiz: Test Your Understanding

Geometric Progressions Quiz
Question 1: Find the 7th term of the GP: 2, 6, 18, 54, …
Answer:
\(a = 2\), \(r = 3\), \(n = 7\)
\(T_7 = 2 × 3^{6} = 2 × 729 = 1458\)
Final answer: 1458
Question 2: Find the sum of the first 6 terms of the GP: 5, 10, 20, 40, …
Answer:
\(a = 5\), \(r = 2\), \(n = 6\)
\(S_6 = \frac{5(2^6 – 1)}{2 – 1} = \frac{5(64 – 1)}{1} = 5 × 63 = 315\)
Final answer: 315
Question 3: Find the sum to infinity of the GP: 12, 4, \(\frac{4}{3}\), \(\frac{4}{9}\), …
Answer:
\(a = 12\), \(r = \frac{4}{12} = \frac{1}{3}\)
\(|r| = \frac{1}{3} < 1\) ✓
\(S_\infty = \frac{12}{1 – 1/3} = \frac{12}{2/3} = 12 × \frac{3}{2} = 18\)
Final answer: 18
Question 4: The first term of a GP is 3 and the common ratio is 2. Which term is equal to 768?
Answer:
\(a = 3\), \(r = 2\), \(T_n = 768\)
\(768 = 3 × 2^{n-1}\)
\(256 = 2^{n-1}\)
\(2^8 = 256\), so \(n-1 = 8\)
\(n = 9\)
Final answer: 9th term
Question 5: Express the recurring decimal 0.454545… as a fraction using GP.
Answer:
\(0.454545… = 0.45 + 0.0045 + 0.000045 + \cdots\)
GP: \(a = 0.45\), \(r = 0.01\)
\(|r| = 0.01 < 1\) ✓
\(S_\infty = \frac{0.45}{1 – 0.01} = \frac{0.45}{0.99} = \frac{45}{99} = \frac{5}{11}\)
Final answer: \(\frac{5}{11}\)

🎯 Key Concepts Summary

  • Definition: GP = sequence with constant ratio between terms
  • Common Ratio: \(r = \frac{T_n}{T_{n-1}}\) (constant for all \(n\))
  • nth Term Formula: \(T_n = ar^{n-1}\)
  • Sum of n Terms:
    • When \(r \neq 1\): \(S_n = \frac{a(1 – r^n)}{1 – r}\) or \(S_n = \frac{a(r^n – 1)}{r – 1}\)
    • When \(r = 1\): \(S_n = na\)
  • Sum to Infinity: \(S_\infty = \frac{a}{1 – r}\) where \(|r| < 1\)
  • Important Properties:
    • Three terms \(p, q, r\) in GP: \(q^2 = pr\)
    • \(\frac{T_n}{T_m} = r^{n-m}\)
    • GP for compound interest: \(A = P(1 + r)^n\)
  • Common CSEC Question Types:
    • Find specific term using \(T_n\) formula
    • Find sum of finite number of terms
    • Find sum to infinity (when \(|r| < 1\))
    • Express recurring decimals as fractions
    • Word problems involving growth/decay
  • Exam Strategy:
    • Always identify \(a\) and \(r\) first
    • For sum to infinity, check \(|r| < 1\)
    • For recurring decimals, write as infinite GP
    • Show all steps for full method marks
    • Check if answer makes sense in context

CSEC Exam Strategy: When solving GP problems: (1) Write down what you know: \(a = ?\), \(r = ?\), \(n = ?\), \(T_n = ?\), \(S_n = ?\), \(S_\infty = ?\). (2) For sum problems, choose the correct formula based on whether \(r = 1\), \(|r| < 1\), or \(|r| > 1\). (3) For sum to infinity, ALWAYS verify \(|r| < 1\) first. (4) For "three terms in GP" problems, use the property \(q^2 = pr\). (5) Recurring decimals can always be expressed as infinite GP sums.

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