Solved Exercises on Arithmetic Sequences Review in Algebra 2

Master arithmetic sequences: nth term, common difference, 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 15th term of the arithmetic sequence: 7, 12, 17, 22, ...
Definition:

Arithmetic sequence: A sequence where each term differs from the previous term by a constant value called the common difference.

nth term method:
  1. Identify the first term (a₁)
  2. Find the common difference (d)
  3. Apply the formula: aₙ = a₁ + (n-1)d
  4. Substitute the given value of n
First term
a₁ = 7
Common difference
d = 5
Step 1: Identify the first term

The first term of the sequence is \(a_1 = 7\)

Step 2: Find the common difference

Calculate the difference between consecutive terms:

\(d = a_2 - a_1 = 12 - 7 = 5\)

Verify: \(a_3 - a_2 = 17 - 12 = 5\), \(a_4 - a_3 = 22 - 17 = 5\)

So the common difference is \(d = 5\)

Step 3: Apply the nth term formula

The formula for the nth term of an arithmetic sequence is:

\(a_n = a_1 + (n-1)d\)

Step 4: Substitute values for the 15th term

For \(n = 15\):

\(a_{15} = 7 + (15-1) \cdot 5\)

\(a_{15} = 7 + 14 \cdot 5\)

\(a_{15} = 7 + 70\)

\(a_{15} = 77\)

Step 5: Verify the result

We can verify by continuing the sequence: 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77

The 15th term is indeed 77

a₁₅ = 77
Final answer:

The 15th term of the arithmetic sequence is 77.

Applied rules:

• nth term formula: \(a_n = a_1 + (n-1)d\)

• Common difference: \(d = a_{n+1} - a_n\) for any term

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

2 Finding Common Difference
Exercise 2
Find the common difference and write the explicit formula for the arithmetic sequence where a₃ = 14 and a₇ = 34.
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
a₃ = a₁ + 2d = 14, a₇ = a₁ + 6d = 34
Solve system
d = 5, a₁ = 4
Step 1: Write the general formula

The general formula for the nth term is: \(a_n = a_1 + (n-1)d\)

Step 2: Write equations using given information

For \(a_3 = 14\): \(a_1 + (3-1)d = 14\), so \(a_1 + 2d = 14\)

For \(a_7 = 34\): \(a_1 + (7-1)d = 34\), so \(a_1 + 6d = 34\)

Step 3: Solve the system of equations

Subtract the first equation from the second:

\((a_1 + 6d) - (a_1 + 2d) = 34 - 14\)

\(a_1 + 6d - a_1 - 2d = 20\)

\(4d = 20\)

\(d = 5\)

Step 4: Find the first term

Substitute \(d = 5\) into the first equation:

\(a_1 + 2(5) = 14\)

\(a_1 + 10 = 14\)

\(a_1 = 4\)

Step 5: Write the explicit formula

Substitute \(a_1 = 4\) and \(d = 5\) into the general formula:

\(a_n = 4 + (n-1) \cdot 5\)

\(a_n = 4 + 5n - 5\)

\(a_n = 5n - 1\)

Step 6: Verify the solution

Check: \(a_3 = 5(3) - 1 = 15 - 1 = 14\) ✓

Check: \(a_7 = 5(7) - 1 = 35 - 1 = 34\) ✓

d = 5, aₙ = 5n - 1
Final answer:

The common difference is \(d = 5\).

The explicit formula is \(a_n = 5n - 1\).

Applied rules:

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

• Elimination method: Subtract equations to eliminate \(a_1\) and solve for \(d\)

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

3 Recursive Formula
Exercise 3
Write the recursive formula for the arithmetic sequence: 10, 7, 4, 1, ... Then find the 8th 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 = 10
Common difference
d = -3
Step 1: Identify the first term

The first term is \(a_1 = 10\)

Step 2: Find the common difference

\(d = a_2 - a_1 = 7 - 10 = -3\)

Verify: \(a_3 - a_2 = 4 - 7 = -3\), \(a_4 - a_3 = 1 - 4 = -3\)

So the common difference is \(d = -3\)

Step 3: Write the recursive formula

For an arithmetic sequence: \(a_1 = \text{first term}\) and \(a_n = a_{n-1} + d\)

Therefore: \(a_1 = 10\) and \(a_n = a_{n-1} + (-3) = a_{n-1} - 3\)

Step 4: Find the 8th term using recursion

We already know: \(a_1 = 10\)

\(a_2 = a_1 - 3 = 10 - 3 = 7\)

\(a_3 = a_2 - 3 = 7 - 3 = 4\)

\(a_4 = a_3 - 3 = 4 - 3 = 1\)

\(a_5 = a_4 - 3 = 1 - 3 = -2\)

\(a_6 = a_5 - 3 = -2 - 3 = -5\)

\(a_7 = a_6 - 3 = -5 - 3 = -8\)

\(a_8 = a_7 - 3 = -8 - 3 = -11\)

Step 5: Verify using explicit formula

Explicit formula: \(a_n = 10 + (n-1)(-3) = 10 - 3n + 3 = 13 - 3n\)

Check: \(a_8 = 13 - 3(8) = 13 - 24 = -11\) ✓

a₁ = 10, aₙ = aₙ₋₁ - 3; a₈ = -11
Final answer:

The recursive formula is: \(a_1 = 10\) and \(a_n = a_{n-1} - 3\).

The 8th term is \(a_8 = -11\).

Applied rules:

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

• Arithmetic sequence: Each term is the previous term plus the common difference

• Verification: Use explicit formula to verify recursive results

Key Concepts: Definitions, Rules, and Methods
\(a_n = a_1 + (n-1)d\)
Explicit Formula for Arithmetic Sequences
Key definitions:

Arithmetic Sequence: A sequence where the difference between consecutive terms is constant.

Common Difference (d): The constant value added to 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.

Arithmetic sequence methodologies:
  1. Finding nth term: Use \(a_n = a_1 + (n-1)d\)
  2. Finding common difference: \(d = a_{n+1} - a_n\)
  3. Finding first term: Use known term and common difference
  4. Writing explicit formula: Identify \(a_1\) and \(d\), then substitute into formula
  5. Writing recursive formula: State \(a_1\) and \(a_n = a_{n-1} + d\)
Tip 1: The common difference 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 \(d\).
Tip 4: The graph of an arithmetic sequence is always a straight line.
Common errors: Incorrect common difference, wrong substitution in formulas, arithmetic mistakes.
Exam preparation: Practice finding patterns, memorize formulas, verify solutions.
Essential rules to remember:

• Explicit formula: \(a_n = a_1 + (n-1)d\)

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

• Common difference: \(d = a_{n+1} - a_n\)

• Constant difference: The value of \(d\) is the same throughout the sequence

• Linear relationship: Arithmetic sequences represent linear functions with domain restricted to positive integers

d = aₙ₊₁ - aₙ
Common Difference Formula
aₙ = aₘ + (n-m)d
General Term Formula
Sₙ = \frac{n}{2}(a₁ + aₙ)
Sum of Arithmetic Series
Solution: Exercises 4 to 5
4 Application Problem
Exercise 4
A theater has 25 rows of seats. The first row has 12 seats, and each subsequent row has 3 more seats than the previous row. How many seats are in the last row? How many total seats are there?
Definition:

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

Sequence parameters
a₁ = 12, d = 3, n = 25
Last row seats
a₂₅ = 84
Step 1: Identify the sequence parameters

This is an arithmetic sequence where:

First term: \(a_1 = 12\) (seats in first row)

Common difference: \(d = 3\) (each row has 3 more seats than previous)

Number of terms: \(n = 25\) (total rows)

Step 2: Find the number of seats in the last row

Use the explicit formula: \(a_n = a_1 + (n-1)d\)

For the 25th row: \(a_{25} = 12 + (25-1) \cdot 3\)

\(a_{25} = 12 + 24 \cdot 3\)

\(a_{25} = 12 + 72\)

\(a_{25} = 84\)

Step 3: Find the total number of seats

Use the formula for the sum of an arithmetic series: \(S_n = \frac{n}{2}(a_1 + a_n)\)

Where \(n = 25\), \(a_1 = 12\), and \(a_{25} = 84\)

\(S_{25} = \frac{25}{2}(12 + 84)\)

\(S_{25} = \frac{25}{2} \cdot 96\)

\(S_{25} = 12.5 \cdot 96\)

\(S_{25} = 1200\)

Step 4: Verify the result

We can also use the alternative sum formula: \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\)

\(S_{25} = \frac{25}{2}[2(12) + (25-1)(3)]\)

\(S_{25} = \frac{25}{2}[24 + 72]\)

\(S_{25} = \frac{25}{2} \cdot 96 = 1200\) ✓

Last row: 84 seats, Total: 1200 seats
Final answer:

The last row has 84 seats.

The total number of seats in the theater is 1200.

Applied rules:

• Explicit formula: \(a_n = a_1 + (n-1)d\) for finding specific terms

• Sum formula: \(S_n = \frac{n}{2}(a_1 + a_n)\) for total of arithmetic series

• Real-world modeling: Identify sequence parameters from context

5 Mixed Problem
Exercise 5
In an arithmetic sequence, the sum of the first 10 terms is 145, and the 10th term is 28. Find the first term and the common difference.
Definition:

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

Set up equations
S₁₀ = 145, a₁₀ = 28
Solve system
a₁ = 1, d = 3
Step 1: Write the known information

We know: \(S_{10} = 145\) and \(a_{10} = 28\)

Step 2: Write the formulas with known values

Sum formula: \(S_n = \frac{n}{2}(a_1 + a_n)\)

So: \(S_{10} = \frac{10}{2}(a_1 + a_{10}) = 5(a_1 + 28) = 145\)

Explicit formula: \(a_n = a_1 + (n-1)d\)

So: \(a_{10} = a_1 + (10-1)d = a_1 + 9d = 28\)

Step 3: Solve for the first term

From the sum equation: \(5(a_1 + 28) = 145\)

\(a_1 + 28 = \frac{145}{5} = 29\)

\(a_1 = 29 - 28 = 1\)

Step 4: Solve for the common difference

Substitute \(a_1 = 1\) into the second equation:

\(a_1 + 9d = 28\)

\(1 + 9d = 28\)

\(9d = 27\)

\(d = 3\)

Step 5: Verify the solution

Check: \(a_{10} = 1 + 9(3) = 1 + 27 = 28\) ✓

Check: \(S_{10} = \frac{10}{2}(1 + 28) = 5(29) = 145\) ✓

a₁ = 1, d = 3
Final answer:

The first term is \(a_1 = 1\).

The common difference is \(d = 3\).

Applied rules:

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

• Substitution method: Solve one equation and substitute into another

• Verification: Check both conditions with calculated values

Arithmetic Sequences: Complete Review Guide
\(a_n = a_1 + (n-1)d\)
Explicit Formula
Key definitions:

Arithmetic Sequence: A sequence where each term differs from the previous term by a constant value (common difference).

Common Difference (d): The constant value added to 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.

Arithmetic Series: The sum of the terms in an arithmetic sequence.

Complete methodology:
  1. Identify sequence type: Verify that the difference between consecutive terms is constant
  2. Find parameters: Determine \(a_1\) and \(d\) 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 difference 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 \(d\).
Tip 4: The graph of an arithmetic sequence is always a straight line when plotted against term number.
Common errors: Incorrect common difference, wrong formula substitution, arithmetic mistakes, forgetting to verify solutions.
Exam preparation: Practice identifying sequence types, memorize all formulas, work with various problem formats.
Essential formulas and rules:

• Explicit formula: \(a_n = a_1 + (n-1)d\)

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

• Common difference: \(d = a_{n+1} - a_n\)

• Alternative explicit formula: \(a_n = a_m + (n-m)d\)

• Sum of arithmetic series: \(S_n = \frac{n}{2}(a_1 + a_n)\) or \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\)

d = aₙ₊₁ - aₙ
Common Difference Formula
aₙ = aₘ + (n-m)d
General Term Formula
Sₙ = \frac{n}{2}(a₁ + aₙ)
Sum of Arithmetic Series
Applications and contexts:

Linear growth: Situations with constant rate of change

Financial planning: Regular deposits, salary increases

Geometry: Number of objects in patterns

Physics: Uniform motion with constant acceleration

Problem-solving strategies:
  1. Read carefully: Identify what is given and what is asked
  2. Model: Recognize arithmetic 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: Arithmetic sequences represent linear functions with domain restricted to positive integers.
Key note: The sum formula can be remembered as "number of terms × average of first and last terms".

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 (\(a_n = a_1 + (n-1)d\)) 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 (\(a_n = a_{n-1} + d\)) 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: \(a_{50} = a_1 + 49d\).

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

Question: How do I find the common difference if I'm only given two terms that aren't consecutive? For example, if I know \(a_3 = 10\) and \(a_7 = 22\), how do I find \(d\)?

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

Using the formula \(a_n = a_1 + (n-1)d\):

  • For \(a_3 = 10\): \(a_1 + 2d = 10\)
  • For \(a_7 = 22\): \(a_1 + 6d = 22\)

Subtract the first equation from the second:

\((a_1 + 6d) - (a_1 + 2d) = 22 - 10\)

\(4d = 12\), so \(d = 3\)

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

In this case: \(d = \frac{a_7 - a_3}{7 - 3} = \frac{22 - 10}{4} = \frac{12}{4} = 3\)

This method works for any two terms in an arithmetic sequence.

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

Answer: Arithmetic sequences are essentially linear functions with a restricted domain.

Compare the arithmetic sequence formula with the linear function formula:

  • Arithmetic sequence: \(a_n = a_1 + (n-1)d\)
  • Linear function: \(y = mx + b\)

If we let \(n\) be the input (instead of \(x\)) and \(a_n\) be the output (instead of \(y\)): \(a_n = dn + (a_1 - d)\)

So the common difference \(d\) corresponds to the slope \(m\), and the expression \((a_1 - d)\) corresponds to the y-intercept \(b\).

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

This is why the graph of an arithmetic sequence appears as discrete points along a straight line.