How to Solve: Geometric Progression (GP)

How to Solve: Geometric Progression (GP)

Introduction

"Imagine you invest $100, and every month it grows by 10%. How much will you have in 5 years? That’s a geometric progression—master this, and you’ll ace finance questions, growth problems, and even tricky exam word problems!

What You Need To Know First

Before diving into geometric progressions (GP), ensure you understand: 1. Exponents – How to calculate powers (e.g., (2^3 = 8)). 2. Sequences – What a sequence is (a list of numbers following a pattern). 3. Basic Algebra – Solving for unknowns (e.g., (2x = 10 \implies x = 5)).

Key Vocabulary

Formulas To Know

Step-by-Step Method

Step 1: Identify the First Term ((a)) and Common Ratio ((r))

• First term ((a)): The first number in the GP.
• Common ratio ((r)): Divide the second term by the first term (or any term by its previous term).
• Example: In (3, 6, 12, 24, \dots), (r = \frac{6}{3} = 2).

Step 2: Decide What You Need to Find

• Find the (n^{th}) term? → Use (T_n = a \cdot r^{n-1}).
• Find the sum of the first (n) terms? → Use (S_n = \frac{a(r^n - 1)}{r - 1}) (or the alternative formula).
• Find the sum of an infinite GP? → Use (S_\infty = \frac{a}{1 - r}) (only if (|r| < 1)).

Step 3: Plug Values into the Correct Formula

• Substitute (a), (r), and (n) into the formula.
• Double-check: Did you use the right formula? Is (r) correct?

Step 4: Simplify and Solve

• Calculate exponents first (e.g., (r^n)).
• Then multiply/divide as needed.
• For infinite GP: Ensure (|r| < 1) (otherwise, the sum doesn’t exist).

Step 5: Verify Your Answer

• For (n^{th}) term: Check if the term fits the pattern.
• For sum: Add the first few terms manually to see if it matches.
• For infinite GP: Does the sum make sense? (e.g., (1 + \frac{1}{2} + \frac{1}{4} + \dots = 2)).

Worked Example Using the Steps

Problem: Find the 6th term and the sum of the first 6 terms of the GP: (5, 10, 20, 40, \dots).

Step 1: Identify (a) and (r) - First term ((a)) = 5 - Common ratio ((r)) = (\frac{10}{5} = 2)

Step 2: Find the 6th term ((T_6)) - Formula: (T_n = a \cdot r^{n-1}) - (T_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160)

Step 3: Find the sum of the first 6 terms ((S_6)) - Formula: (S_n = \frac{a(r^n - 1)}{r - 1}) - (S_6 = \frac{5(2^6 - 1)}{2 - 1} = \frac{5(64 - 1)}{1} = 5 \times 63 = 315)

Step 4: Verify - Terms: (5, 10, 20, 40, 80, 160) → 6th term = 160 ✔️ - Sum: (5 + 10 + 20 + 40 + 80 + 160 = 315) ✔️

Final Answer: - 6th term = 160 - Sum of first 6 terms = 315

Worked Examples

Example 1 – Basic (Find the (n^{th}) Term)

Problem: Find the 7th term of the GP: (3, 6, 12, 24, \dots).

Solution: 1. (a = 3), (r = \frac{6}{3} = 2) 2. (T_7 = 3 \cdot 2^{7-1} = 3 \cdot 2^6 = 3 \cdot 64 = 192)

What we did and why: - We identified (a) and (r) first. - Used the (n^{th}) term formula because we needed a specific term.

Example 2 – Medium (Sum of GP)

Problem: Find the sum of the first 5 terms of the GP: (16, 8, 4, 2, \dots).

Solution: 1. (a = 16), (r = \frac{8}{16} = \frac{1}{2}) 2. (S_5 = \frac{16 \left( \left( \frac{1}{2} \right)^5 - 1 \right)}{\frac{1}{2} - 1} = \frac{16 \left( \frac{1}{32} - 1 \right)}{- \frac{1}{2}} = \frac{16 \left( -\frac{31}{32} \right)}{- \frac{1}{2}} = \frac{-15.5}{-0.5} = 31)

What we did and why: - We used the sum formula because we needed the total of the first 5 terms. - Simplified carefully to avoid sign errors.

Example 3 – Exam Style (Disguised GP)

Problem: A ball is dropped from a height of 20 m. After each bounce, it reaches half its previous height. What is the total distance traveled by the ball when it hits the ground for the 5th time?

Solution: 1. First drop: 20 m (down). 2. First bounce up: 10 m, then down 10 m. 3. Second bounce up: 5 m, then down 5 m. 4. Pattern: Total distance = (20 + 2(10 + 5 + 2.5 + 1.25))
- The (2) accounts for up and down after the first drop. 5. GP part: (10 + 5 + 2.5 + 1.25) is a GP with (a = 10), (r = \frac{1}{2}), (n = 4). 6. Sum of GP: (S_4 = \frac{10 \left( 1 - \left( \frac{1}{2} \right)^4 \right)}{1 - \frac{1}{2}} = \frac{10 \left( 1 - \frac{1}{16} \right)}{\frac{1}{2}} = 20 \times \frac{15}{16} = 18.75) 7. Total distance: (20 + 2(18.75) = 20 + 37.5 = 57.5) m

What we did and why: - Recognized the GP hidden in the bouncing pattern. - Used the sum formula for the repeated up-down distances. - Added the initial drop separately.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in—tonight, before your exam, here’s what you need to remember about geometric progressions:

1. First term ((a)) and common ratio ((r)) are everything. Find them first by dividing any term by its previous term.
2. For the (n^{th}) term: Use (T_n = a \cdot r^{n-1}). Not (r^n)—that’s a common trap!
3. For the sum of the first (n) terms: Use (S_n = \frac{a(r^n - 1)}{r - 1}). If (r = 1), the sum is just (n \times a).
4. For infinite GP: Only works if (|r| < 1). Use (S_\infty = \frac{a}{1 - r}).
5. Watch out for disguised GPs in word problems. If something is growing or shrinking by a factor, it’s probably a GP.

Now, go practice 3 problems—one for each formula. You’ve got this!