Arithmetic Sequences: Find Nth Term Sums

Grade 12 · algebra · 100 practice problems · read aloud

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Arithmetic Sequences: The Linear Pattern

An arithmetic sequence is an ordered list of numbers where the difference between consecutive terms is constant. This operation is foundational for modeling linear growth and decay in real-world scenarios like finance (simple interest, depreciation) and physics (constant motion).

🔍 Key Formula & Components

The nth term of an arithmetic sequence is found using:

aₙ = a₁ + (n - 1)d

  • aₙ = the nth term you're solving for
  • a₁ = the first term in the sequence
  • n = the term number (position)
  • d = the common difference (a₂ - a₁)

📝 Step-by-Step Problem Solving

  1. Identify the given values: a₁, n, d, and/or aₙ.
  2. Determine the common difference (d) if not given.
  3. Substitute the known values into the formula aₙ = a₁ + (n - 1)d.
  4. Solve for the unknown variable using algebraic manipulation.

✨ Worked Examples

Example 1: Finding a Specific Term

Find the 15th term of the sequence: 2, 5, 8, 11, ...

Step 1: a₁ = 2, n = 15, d = 5 - 2 = 3

Step 2: a₁₅ = 2 + (15 - 1) * 3

Step 3: a₁₅ = 2 + (14) * 3 = 2 + 42 = 44

Example 2: Finding the Common Difference

The 4th term is 10 and the 9th term is 25. Find the common difference.

Step 1: Set up two equations: a₄ = a₁ + 3d = 10 and a₉ = a₁ + 8d = 25

Step 2: Subtract the first equation from the second: (a₁ + 8d) - (a₁ + 3d) = 25 - 10

Step 3: 5d = 15 → d = 3

⚠️ Common Mistakes to Avoid

  • Incorrect "n" value: Remember the formula uses (n - 1), so for the 10th term, you multiply d by 9, not 10.
  • Misidentifying d: Always subtract a term from the one that follows it (a₂ - a₁), not the other way around.
  • Formula confusion: Don't mix up the arithmetic sequence formula (aₙ = a₁ + (n-1)d) with the arithmetic series sum formula.

💡 Tips & Tricks

  • Memory Aid: "First term, plus d, times one less than n."
  • Quick Check: The graph of an arithmetic sequence's terms forms a straight line. If your terms don't increase linearly, you've made an error.
  • Shortcut: To find d from any two terms aₘ and aₙ, use d = (aₘ - aₙ) / (m - n).

🎯 Practice Suggestions

  • Start with basic problems finding aₙ given a₁ and d.
  • Progress to problems where you need to find a₁ or d given other terms.
  • Create your own sequences and verify the pattern holds.
  • Connect to real-world applications: calculate monthly savings, car depreciation, or simple interest.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 ∑(n=1 to 10) (3n - 2) = ?

Hint: This is an arithmetic series. Find the first term, last term, and number of terms to use the sum formula.

Show the answer

Answer: 145

  1. Identify the arithmetic sequence: a_n = 3n - 2
  2. Find the first term: a_1 = 3(1) - 2 = 1
  3. Find the last term: a_10 = 3(10) - 2 = 28
  4. Use the arithmetic series sum formula: S_n = n/2 × (a_1 + a_n)
  5. Substitute values: S_10 = 10/2 × (1 + 28) = 5 × 29
  6. Calculate: 5 × 29 = 145

The answer is 145.

2 ∫(3x² - 4x + 2) dx from 0 to 2 = ?

Hint: Find the antiderivative of each term, then evaluate at the upper and lower limits of integration.

Show the answer

Answer: 4

  1. Find the antiderivative of 3x² - 4x + 2 Antiderivative of 3x² is (3/3)x³ = x³ Antiderivative of -4x is (-4/2)x² = -2x² Antiderivative of 2 is 2x So the antiderivative is x³ - 2x² + 2x
  2. Evaluate from 0 to 2 At x = 2: (2)³ - 2(2)² + 2(2) = 8 - 8 + 4 = 4 At x = 0: (0)³ - 2(0)² + 2(0) = 0 - 0 + 0 = 0
  3. Subtract: 4 - 0 = 4

The answer is 4.

3 ∫(3x² - 6x + 2) dx from 0 to 2 = ?

Hint: Find the antiderivative of the polynomial, then evaluate it at the upper and lower limits of integration.

Show the answer

Answer: 0

  1. Find the antiderivative of 3x² - 6x + 2 Antiderivative = (3x³/3) - (6x²/2) + 2x = x³ - 3x² + 2x
  2. Evaluate the antiderivative at the upper limit (x = 2) F(2) = (2)³ - 3(2)² + 2(2) = 8 - 12 + 4 = 0
  3. Evaluate the antiderivative at the lower limit (x = 0) F(0) = (0)³ - 3(0)² + 2(0) = 0 - 0 + 0 = 0
  4. Apply the Fundamental Theorem of Calculus ∫(3x² - 6x + 2) dx from 0 to 2 = F(2) - F(0) = 0 - 0 = 0

The answer is 0.

4 Sequence: 6, 14, 22, 30... Find a₁₂ and sum of first 12 terms

Hint: Identify the first term and common difference, then apply the formulas for nth term and sum of arithmetic sequences.

Show the answer

Answer: a₁₂ = 94, S₁₂ = 600

  1. Identify the first term a₁ = 6
  2. Find the common difference d = 14 - 6 = 8
  3. Calculate the 12th term using aₙ = a₁ + (n-1)d a₁₂ = 6 + (12-1)×8 = 6 + 11×8 = 6 + 88 = 94
  4. Calculate the sum of first 12 terms using Sₙ = n(a₁ + aₙ)/2 S₁₂ = 12(6 + 94)/2 = 12×100/2 = 1200/2 = 600 Therefore, a₁₂ = 94 and S₁₂ = 600

5 Sequence: 2, 7, 12, 17... Find a₁₂ and sum of first 12 terms.

Hint: Identify the first term and the common difference between consecutive terms. Then apply the formulas for the nth term and the sum of an arithmetic series.

Show the answer

Answer: a₁₂ = 57, S₁₂ = 354

  1. Identify the first term a₁ = 2.
  2. Find the common difference d = 7 - 2 = 5.
  3. Use the nth term formula aₙ = a₁ + (n-1)d. For n = 12: a₁₂ = 2 + (12-1)×5 = 2 + 11×5 = 2 + 55 = 57.
  4. Use the sum formula Sₙ = n(a₁ + aₙ)/2. For n = 12: S₁₂ = 12(2 + 57)/2 = 12×59/2 = 708/2 = 354. Therefore, a₁₂ = 57 and S₁₂ = 354.

6 Sequence: 1, 7, 13, 19... Find a₁₇ and sum of first 17 terms.

Hint: Identify the first term and the common difference between consecutive terms. Then apply the formulas for the nth term and the sum of an arithmetic series.

Show the answer

Answer: a₁₇ = 97, S₁₇ = 833

  1. Identify the first term a₁ = 1.
  2. Find the common difference d = 7 - 1 = 6.
  3. Use the nth term formula aₙ = a₁ + (n-1)d. For n = 17: a₁₇ = 1 + (17-1)×6 = 1 + 16×6 = 1 + 96 = 97.
  4. Use the sum formula Sₙ = n(a₁ + aₙ)/2. For n = 17: S₁₇ = 17(1 + 97)/2 = 17×98/2 = 1666/2 = 833. Therefore, a₁₇ = 97 and S₁₇ = 833.
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