Solved Exercises on Geometric Sequences Review in Algebra 2

Master geometric sequences: nth term, common ratio, recursive formulas, sum of terms, and applications through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Finding the nth Term
Exercise 1
Find the 8th term of the geometric sequence: 3, 6, 12, 24, ...
Definition:

Geometric sequence: A sequence where each term is obtained by multiplying the previous term by a constant value called the common ratio.

nth term method:
  1. Identify the first term (a₁)
  2. Find the common ratio (r)
  3. Apply the formula: aₙ = a₁ · r^(n-1)
  4. Substitute the given value of n
First term
a₁ = 3
Common ratio
r = 2
Step 1: Identify the first term

The first term of the sequence is \(a_1 = 3\)

Step 2: Find the common ratio

Calculate the ratio between consecutive terms:

\(r = \frac{a_2}{a_1} = \frac{6}{3} = 2\)

Verify: \(\frac{a_3}{a_2} = \frac{12}{6} = 2\), \(\frac{a_4}{a_3} = \frac{24}{12} = 2\)

So the common ratio is \(r = 2\)

Step 3: Apply the nth term formula

The formula for the nth term of a geometric sequence is:

\(a_n = a_1 \cdot r^{n-1}\)

Step 4: Substitute values for the 8th term

For \(n = 8\):

\(a_8 = 3 \cdot 2^{8-1} = 3 \cdot 2^7\)

\(a_8 = 3 \cdot 128\)

\(a_8 = 384\)

Step 5: Verify the result

We can verify by continuing the sequence: 3, 6, 12, 24, 48, 96, 192, 384

The 8th term is indeed 384

a₈ = 384
Final answer:

The 8th term of the geometric sequence is 384.

Applied rules:

• nth term formula: \(a_n = a_1 \cdot r^{n-1}\)

• Common ratio: \(r = \frac{a_{n+1}}{a_n}\) for any term

• Consistency: The common ratio is the same between any two consecutive terms

2 Finding Common Ratio
Exercise 2
Find the common ratio and write the explicit formula for the geometric sequence where a₄ = 48 and a₇ = 384.
Definition:

Explicit formula: A formula that allows you to find any term of a sequence directly using its position in the sequence.

Set up equations
a₄ = a₁ · r³ = 48, a₇ = a₁ · r⁶ = 384
Solve system
r = 2, a₁ = 6
Step 1: Write the general formula

The general formula for the nth term is: \(a_n = a_1 \cdot r^{n-1}\)

Step 2: Write equations using given information

For \(a_4 = 48\): \(a_1 \cdot r^{4-1} = 48\), so \(a_1 \cdot r^3 = 48\)

For \(a_7 = 384\): \(a_1 \cdot r^{7-1} = 384\), so \(a_1 \cdot r^6 = 384\)

Step 3: Solve the system of equations

Divide the second equation by the first:

\(\frac{a_1 \cdot r^6}{a_1 \cdot r^3} = \frac{384}{48}\)

\(r^3 = 8\)

\(r = 2\)

Step 4: Find the first term

Substitute \(r = 2\) into the first equation:

\(a_1 \cdot 2^3 = 48\)

\(a_1 \cdot 8 = 48\)

\(a_1 = 6\)

Step 5: Write the explicit formula

Substitute \(a_1 = 6\) and \(r = 2\) into the general formula:

\(a_n = 6 \cdot 2^{n-1}\)

Step 6: Verify the solution

Check: \(a_4 = 6 \cdot 2^{4-1} = 6 \cdot 2^3 = 6 \cdot 8 = 48\) ✓

Check: \(a_7 = 6 \cdot 2^{7-1} = 6 \cdot 2^6 = 6 \cdot 64 = 384\) ✓

r = 2, aₙ = 6 · 2^(n-1)
Final answer:

The common ratio is \(r = 2\).

The explicit formula is \(a_n = 6 \cdot 2^{n-1}\).

Applied rules:

• System of equations: Use given terms to create equations in \(a_1\) and \(r\)

• Division method: Divide equations to eliminate \(a_1\) and solve for \(r\)

• Substitution: Use the value of \(r\) to find \(a_1\)

3 Recursive Formula
Exercise 3
Write the recursive formula for the geometric sequence: 16, -8, 4, -2, ... Then find the 6th term using the recursive formula.
Definition:

Recursive formula: A formula that defines each term of a sequence using the preceding term(s).

First term
a₁ = 16
Common ratio
r = -1/2
Step 1: Identify the first term

The first term is \(a_1 = 16\)

Step 2: Find the common ratio

\(r = \frac{a_2}{a_1} = \frac{-8}{16} = -\frac{1}{2}\)

Verify: \(\frac{a_3}{a_2} = \frac{4}{-8} = -\frac{1}{2}\), \(\frac{a_4}{a_3} = \frac{-2}{4} = -\frac{1}{2}\)

So the common ratio is \(r = -\frac{1}{2}\)

Step 3: Write the recursive formula

For a geometric sequence: \(a_1 = \text{first term}\) and \(a_n = a_{n-1} \cdot r\)

Therefore: \(a_1 = 16\) and \(a_n = a_{n-1} \cdot \left(-\frac{1}{2}\right)\)

Step 4: Find the 6th term using recursion

We already know: \(a_1 = 16\)

\(a_2 = a_1 \cdot \left(-\frac{1}{2}\right) = 16 \cdot \left(-\frac{1}{2}\right) = -8\)

\(a_3 = a_2 \cdot \left(-\frac{1}{2}\right) = -8 \cdot \left(-\frac{1}{2}\right) = 4\)

\(a_4 = a_3 \cdot \left(-\frac{1}{2}\right) = 4 \cdot \left(-\frac{1}{2}\right) = -2\)

\(a_5 = a_4 \cdot \left(-\frac{1}{2}\right) = -2 \cdot \left(-\frac{1}{2}\right) = 1\)

\(a_6 = a_5 \cdot \left(-\frac{1}{2}\right) = 1 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2}\)

Step 5: Verify using explicit formula

Explicit formula: \(a_n = 16 \cdot \left(-\frac{1}{2}\right)^{n-1}\)

Check: \(a_6 = 16 \cdot \left(-\frac{1}{2}\right)^{6-1} = 16 \cdot \left(-\frac{1}{2}\right)^5 = 16 \cdot \left(-\frac{1}{32}\right) = -\frac{1}{2}\) ✓

a₁ = 16, aₙ = aₙ₋₁ · (-1/2); a₆ = -1/2
Final answer:

The recursive formula is: \(a_1 = 16\) and \(a_n = a_{n-1} \cdot \left(-\frac{1}{2}\right)\).

The 6th term is \(a_6 = -\frac{1}{2}\).

Applied rules:

• Recursive definition: \(a_1 = \text{initial value}\), \(a_n = a_{n-1} \cdot r\)

• Geometric sequence: Each term is the previous term multiplied by the common ratio

• Verification: Use explicit formula to verify recursive results

Key Concepts: Definitions, Rules, and Methods
\(a_n = a_1 \cdot r^{n-1}\)
Explicit Formula for Geometric Sequences
Key definitions:

Geometric Sequence: A sequence where the ratio between consecutive terms is constant.

Common Ratio (r): The constant value multiplied by each term to get the next term.

Explicit Formula: A formula that calculates any term directly using its position in the sequence.

Recursive Formula: A formula that defines each term using the previous term.

Term: Each individual number in the sequence.

Geometric sequence methodologies:
  1. Finding nth term: Use \(a_n = a_1 \cdot r^{n-1}\)
  2. Finding common ratio: \(r = \frac{a_{n+1}}{a_n}\)
  3. Finding first term: Use known term and common ratio
  4. Writing explicit formula: Identify \(a_1\) and \(r\), then substitute into formula
  5. Writing recursive formula: State \(a_1\) and \(a_n = a_{n-1} \cdot r\)
Tip 1: The common ratio is always the same between any two consecutive terms.
Tip 2: For finding the nth term, use the explicit formula instead of counting manually.
Tip 3: When given two terms, set up a system of equations to find \(a_1\) and \(r\).
Tip 4: The graph of a geometric sequence shows exponential growth or decay.
Common errors: Incorrect common ratio, wrong substitution in formulas, arithmetic mistakes, confusing with arithmetic sequences.
Exam preparation: Practice finding patterns, memorize formulas, verify solutions.
Essential rules to remember:

• Explicit formula: \(a_n = a_1 \cdot r^{n-1}\)

• Recursive formula: \(a_1 = \text{first term}\), \(a_n = a_{n-1} \cdot r\)

• Common ratio: \(r = \frac{a_{n+1}}{a_n}\)

• Constant ratio: The value of \(r\) is the same throughout the sequence

• Exponential relationship: Geometric sequences represent exponential functions with domain restricted to positive integers

r = \frac{aₙ₊₁}{aₙ}
Common Ratio Formula
aₙ = aₘ · r^(n-m)
General Term Formula
Sₙ = \frac{a₁(1-r^n)}{1-r} \text{ (when } r ≠ 1\text{)}
Sum of Geometric Series
Solution: Exercises 4 to 5
4 Application Problem
Exercise 4
A bacterial culture starts with 500 bacteria and triples every hour. How many bacteria will there be after 6 hours? How many hours will it take for the population to exceed 100,000 bacteria?
Definition:

Application problems: Real-world scenarios that can be modeled using geometric sequences.

Sequence parameters
a₁ = 500, r = 3
After 6 hours
a₇ = 364,500
Step 1: Identify the sequence parameters

This is a geometric sequence where:

Initial amount: \(a_1 = 500\) (bacteria at hour 0)

Common ratio: \(r = 3\) (triples each hour)

Time: Hours correspond to sequence positions (hour 1 = a₂, hour 2 = a₃, etc.)

Step 2: Find the population after 6 hours

After 6 hours means we need \(a_7\) (since \(a_1\) is hour 0):

Use the explicit formula: \(a_n = a_1 \cdot r^{n-1}\)

\(a_7 = 500 \cdot 3^{7-1} = 500 \cdot 3^6\)

\(a_7 = 500 \cdot 729\)

\(a_7 = 364,500\)

Step 3: Find when population exceeds 100,000

We need to find the smallest \(n\) such that \(a_n > 100,000\):

\(500 \cdot 3^{n-1} > 100,000\)

\(3^{n-1} > \frac{100,000}{500}\)

\(3^{n-1} > 200\)

Step 4: Solve the inequality

Take the natural logarithm of both sides:

\((n-1) \ln(3) > \ln(200)\)

\(n-1 > \frac{\ln(200)}{\ln(3)}\)

\(n-1 > \frac{5.298}{1.099} \approx 4.82\)

\(n > 5.82\)

Since \(n\) must be an integer, \(n \geq 6\)

Step 5: Interpret the result

When \(n = 6\): \(a_6 = 500 \cdot 3^5 = 500 \cdot 243 = 121,500\)

This corresponds to hour 5 (since \(a_1\) is hour 0)

So the population will exceed 100,000 after 5 hours.

After 6 hours: 364,500 bacteria; Exceeds 100,000 after 5 hours
Final answer:

After 6 hours, there will be 364,500 bacteria.

The population will exceed 100,000 bacteria after 5 hours.

Applied rules:

• Explicit formula: \(a_n = a_1 \cdot r^{n-1}\) for finding specific terms

• Exponential growth: Real-world phenomena often follow geometric patterns

• Logarithms: Used to solve exponential inequalities

5 Mixed Problem
Exercise 5
In a geometric sequence, the sum of the first 4 terms is 120, and the 4th term is 81. Find the first term and the common ratio.
Definition:

Mixed problems: Problems that require using multiple formulas simultaneously to solve for unknowns.

Set up equations
S₄ = 120, a₄ = 81
Solve system
a₁ = 3, r = 3
Step 1: Write the known information

We know: \(S_4 = 120\) and \(a_4 = 81\)

Step 2: Write the formulas with known values

Sum formula: \(S_n = \frac{a_1(1-r^n)}{1-r}\) (when \(r \neq 1\))

So: \(S_4 = \frac{a_1(1-r^4)}{1-r} = 120\)

Explicit formula: \(a_n = a_1 \cdot r^{n-1}\)

So: \(a_4 = a_1 \cdot r^{4-1} = a_1 \cdot r^3 = 81\)

Step 3: Express a₁ in terms of r

From the second equation: \(a_1 = \frac{81}{r^3}\)

Step 4: Substitute into the sum equation

\(\frac{\frac{81}{r^3}(1-r^4)}{1-r} = 120\)

\(\frac{81(1-r^4)}{r^3(1-r)} = 120\)

\(\frac{81(1-r^4)}{r^3(1-r)} = 120\)

Step 5: Simplify using factorization

Notice that \(1-r^4 = (1-r^2)(1+r^2) = (1-r)(1+r)(1+r^2)\)

So: \(\frac{81(1-r)(1+r)(1+r^2)}{r^3(1-r)} = 120\)

\(\frac{81(1+r)(1+r^2)}{r^3} = 120\)

Step 6: Solve for r

\(81(1+r)(1+r^2) = 120r^3\)

Let's try integer values for r. Testing \(r = 3\):

\(81(1+3)(1+9) = 81 \cdot 4 \cdot 10 = 3240\)

\(120 \cdot 3^3 = 120 \cdot 27 = 3240\) ✓

So \(r = 3\)

Step 7: Find a₁

Substitute \(r = 3\) into \(a_1 = \frac{81}{r^3}\):

\(a_1 = \frac{81}{3^3} = \frac{81}{27} = 3\)

Step 8: Verify the solution

Sequence: 3, 9, 27, 81

Sum: \(3 + 9 + 27 + 81 = 120\) ✓

Fourth term: \(a_4 = 81\) ✓

a₁ = 3, r = 3
Final answer:

The first term is \(a_1 = 3\).

The common ratio is \(r = 3\).

Applied rules:

• System of equations: Use multiple formulas to create equations with unknowns

• Substitution method: Solve one equation and substitute into another

• Factorization: Simplify expressions using algebraic identities

• Verification: Check both conditions with calculated values

Geometric Sequences: Complete Review Guide
\(a_n = a_1 \cdot r^{n-1}\)
Explicit Formula
Key definitions:

Geometric Sequence: A sequence where each term differs from the previous term by a constant ratio.

Common Ratio (r): The constant value multiplied by each term to get the next term.

Explicit Formula: A formula that calculates any term directly using its position in the sequence.

Recursive Formula: A formula that defines each term using the previous term.

Geometric Series: The sum of the terms in a geometric sequence.

Complete methodology:
  1. Identify sequence type: Verify that the ratio between consecutive terms is constant
  2. Find parameters: Determine \(a_1\) and \(r\) from given information
  3. Choose appropriate formula: Explicit for single terms, recursive for sequential calculations
  4. Solve for unknowns: Use algebraic techniques to find missing values
  5. Verify solutions: Check answers using alternative methods or by substituting back
  6. Interpret results: Ensure answers make sense in the context of the problem
Tip 1: Always check that the common ratio is consistent throughout the sequence.
Tip 2: For finding distant terms, use the explicit formula rather than recursion.

Tip 3: When given two terms, set up a system of equations to find both \(a_1\) and \(r\).

Tip 4: The graph of a geometric sequence shows exponential growth when |r| > 1 or decay when |r| < 1.

Common errors: Incorrect common ratio, wrong formula substitution, arithmetic mistakes, confusing with arithmetic sequences, forgetting to verify solutions.
Exam preparation: Practice identifying sequence types, memorize all formulas, work with various problem formats.
Essential formulas and rules:

• Explicit formula: \(a_n = a_1 \cdot r^{n-1}\)

• Recursive formula: \(a_1 = \text{first term}\), \(a_n = a_{n-1} \cdot r\)

• Common ratio: \(r = \frac{a_{n+1}}{a_n}\)

• Alternative explicit formula: \(a_n = a_m \cdot r^{n-m}\)

• Sum of geometric series: \(S_n = \frac{a_1(1-r^n)}{1-r}\) when \(r \neq 1\)

• Infinite geometric series: \(S = \frac{a_1}{1-r}\) when \(|r| < 1\)

r = \frac{aₙ₊₁}{aₙ}
Common Ratio Formula
aₙ = aₘ · r^(n-m)
General Term Formula
Sₙ = \frac{a₁(1-r^n)}{1-r}
Sum of Geometric Series (r ≠ 1)
Applications and contexts:

Exponential growth: Population growth, compound interest

Exponential decay: Radioactive decay, depreciation

Finance: Investment returns, loan payments

Biology: Bacterial growth, viral spread

Problem-solving strategies:
  1. Read carefully: Identify what is given and what is asked
  2. Model: Recognize geometric sequence pattern
  3. Plan: Select appropriate formula based on given information
  4. Solve: Execute calculations systematically
  5. Check: Verify answer makes sense in context
Key note: Geometric sequences represent exponential functions with domain restricted to positive integers.
Key note: The common ratio determines if the sequence grows (>1) or shrinks (<1).

Questions & Answers

Question: I'm confused about when to use the explicit formula versus the recursive formula. How do I decide?

Answer: The choice depends on what you're trying to find:

  • Use explicit formula (\(a_n = a_1 \cdot r^{n-1}\)) when:
    • You need to find a specific term far along in the sequence (like the 100th term)
    • You want to find any term directly without knowing the previous term
    • You're writing a general formula for the sequence
  • Use recursive formula (\(a_n = a_{n-1} \cdot r\)) when:
    • You need to generate several consecutive terms
    • You're programming a sequence generator
    • You want to emphasize the relationship between consecutive terms

For example, if you need the 50th term, use explicit: \(a_{50} = a_1 \cdot r^{49}\).

If you need terms 5, 6, and 7 and you know \(a_4\), use recursive: \(a_5 = a_4 \cdot r\), \(a_6 = a_5 \cdot r\), etc.

Question: How do I find the common ratio if I'm only given two terms that aren't consecutive? For example, if I know \(a_3 = 20\) and \(a_6 = 160\), how do I find \(r\)?

Answer: You can use the general term formula to find the common ratio when given non-consecutive terms.

Using the formula \(a_n = a_1 \cdot r^{n-1}\):

  • For \(a_3 = 20\): \(a_1 \cdot r^{3-1} = 20\), so \(a_1 \cdot r^2 = 20\)
  • For \(a_6 = 160\): \(a_1 \cdot r^{6-1} = 160\), so \(a_1 \cdot r^5 = 160\)

Divide the second equation by the first:

\(\frac{a_1 \cdot r^5}{a_1 \cdot r^2} = \frac{160}{20}\)

\(r^3 = 8\), so \(r = 2\)

Alternatively, you can use the formula: \(r^{n-m} = \frac{a_n}{a_m}\)

In this case: \(r^{6-3} = \frac{a_6}{a_3} = \frac{160}{20} = 8\), so \(r^3 = 8\) and \(r = 2\)

This method works for any two terms in a geometric sequence.

Question: What's the relationship between geometric sequences and exponential functions? They seem similar but I'm not sure how they connect.

Answer: Geometric sequences are essentially exponential functions with a restricted domain.

Compare the geometric sequence formula with the exponential function formula:

  • Geometric sequence: \(a_n = a_1 \cdot r^{n-1}\)
  • Exponential function: \(y = a \cdot b^x\)

If we let \(n\) be the input (instead of \(x\)) and \(a_n\) be the output (instead of \(y\)): \(a_n = a_1 \cdot r^{n-1}\)

So the first term \(a_1\) corresponds to the initial value \(a\), and the common ratio \(r\) corresponds to the base \(b\).

The key difference is that geometric sequences have a domain restricted to positive integers (1, 2, 3, ...), while exponential functions typically have a domain of all real numbers.

This is why the graph of a geometric sequence appears as discrete points along an exponential curve.