Geometric Sequences: Find Nth Term Sums

Grade 12 · geometry · 75 practice problems · read aloud

🔊 Listen to this explanation

📐 Geometric Sequences

A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio (r). These sequences model exponential growth/decay in finance, biology, and physics.

🔢 Step-by-Step Guide

  1. Identify the first term (a₁) and common ratio (r)
  2. General formula: aₙ = a₁ × r⁽ⁿ⁻¹⁾
  3. Sum of n terms: Sₙ = a₁(1 - rⁿ)/(1 - r) when r ≠ 1
  4. Infinite sum: S∞ = a₁/(1 - r) only if |r| < 1

📊 Worked Examples

Example 1: Find the 8th term

Sequence: 3, 6, 12, 24, ...

Step 1: a₁ = 3, r = 6 ÷ 3 = 2

Step 2: a₈ = 3 × 2⁽⁸⁻¹⁾ = 3 × 2⁷ = 3 × 128 = 384

Example 2: Infinite Sum

Sequence: 16, 4, 1, 0.25, ...

Step 1: a₁ = 16, r = 4 ÷ 16 = 0.25

Step 2: Since |r| < 1, S∞ = 16 ÷ (1 - 0.25) = 16 ÷ 0.75 = 64/3

🚨 Common Mistakes

  • Using rⁿ instead of r⁽ⁿ⁻¹⁾ in the general term formula
  • Applying infinite sum formula when |r| ≥ 1 (it diverges!)
  • Mixing up arithmetic and geometric sequence formulas
  • Forgetting that r can be negative or fractional

💡 Tips & Tricks

  • Memory aid: "First times ratio to the (n minus one)"
  • Check your ratio by dividing any term by the previous term
  • For negative r, terms alternate signs
  • Graph geometric sequences to see exponential patterns

🎯 Practice Suggestions

  • Create your own sequences with different r values
  • Practice word problems involving population growth and compound interest
  • Mix arithmetic and geometric sequences in practice sets
  • Use graphing calculators to visualize sequences

Practice problems

6 of the 75, worked through step by step — try them before opening the answer.

1 ∑(n=1 to 5) 3 × 2^(n-1) = ?

Hint: This involves finding the sum of terms in a geometric sequence. Remember the formula for the sum of the first n terms of a geometric series.

Show the answer

Answer: 93

  1. Write out each term in the sum** For n = 1: 3 × 2^(1-1) = 3 × 2^0 = 3 × 1 = 3 For n = 2: 3 × 2^(2-1) = 3 × 2^1 = 3 × 2 = 6 For n = 3: 3 × 2^(3-1) = 3 × 2^2 = 3 × 4 = 12 For n = 4: 3 × 2^(4-1) = 3 × 2^3 = 3 × 8 = 24 For n = 5: 3 × 2^(5-1) = 3 × 2^4 = 3 × 16 = 48 --- **
  2. List all terms** The terms are: 3, 6, 12, 24, 48 --- **
  3. Add them step by step** Start with 3 3 + 6 = 9 9 + 12 = 21 21 + 24 = 45 45 + 48 = 93 --- **
  4. Final answer** The sum is 93.

Let's break this down step by step. We are given the sum: Sum from n = 1 to 5 of 3 × 2^(n-1) --- **

2 ∑(n=1 to 5) 2 × 3^(n-1) = ?

Hint: This involves finding the sum of terms in a geometric sequence. Remember the formula for the sum of the first n terms of a geometric series, where you need to identify the first term and common ratio.

Show the answer

Answer: 242

  1. Write out each term** For n = 1: 2 × 3^(1-1) = 2 × 3^0 = 2 × 1 = 2 For n = 2: 2 × 3^(2-1) = 2 × 3^1 = 2 × 3 = 6 For n = 3: 2 × 3^(3-1) = 2 × 3^2 = 2 × 9 = 18 For n = 4: 2 × 3^(4-1) = 2 × 3^3 = 2 × 27 = 54 For n = 5: 2 × 3^(5-1) = 2 × 3^4 = 2 × 81 = 162 --- **
  2. Add the terms** 2 + 6 = 8 8 + 18 = 26 26 + 54 = 80 80 + 162 = 242 --- **
  3. Conclusion** The sum is 242. --- **Answer:** 242

Let's solve the sum step by step. We are given: Sum from n = 1 to 5 of 2 × 3^(n-1) --- **

3 ∑(n=1 to 4) 5 × (-2)^(n-1) = ?

Hint: This is a geometric series where each term is found by multiplying the previous term by a constant ratio. Identify the first term and the common ratio to calculate the sum.

Show the answer

Answer: -25

  1. Identify the first term a₁ = 5 and common ratio r = -2
  2. The sum of a geometric series is S_n = a₁(1 - r^n)/(1 - r)
  3. For n = 4 terms: S₄ = 5(1 - (-2)⁴)/(1 - (-2))
  4. Calculate (-2)⁴ = 16
  5. S₄ = 5(1 - 16)/(1 + 2) = 5(-15)/3 = -75/3 = -25
  6. The sum is -25

4 ∑(n=1 to 6) 4 × (-2)^(n-1) = ?

Hint: This is a geometric series where each term is found by multiplying the previous term by a constant ratio. Identify the first term and the common ratio, then apply the formula for the sum of a finite geometric series.

Show the answer

Answer: -84

  1. Identify the first term (a₁) and common ratio (r) The first term is a₁ = 4 The common ratio is r = -2
  2. Use the formula for the sum of a finite geometric series Sₙ = a₁ × (1 - rⁿ)/(1 - r) For n = 6 terms: S₆ = 4 × (1 - (-2)⁶)/(1 - (-2))
  3. Calculate (-2)⁶ (-2)⁶ = (-2) × (-2) × (-2) × (-2) × (-2) × (-2) = 64
  4. Substitute into the formula S₆ = 4 × (1 - 64)/(1 + 2) S₆ = 4 × (-63)/3
  5. Simplify the expression S₆ = 4 × (-21) S₆ = -84

The answer is -84.

5 ∑(n=1 to 4) 3 × (-3)^(n-1) = ?

Hint: Consider the pattern of signs and how each term changes from the previous one in a geometric series.

Show the answer

Answer: -60

  1. Identify the first term a = 3 and common ratio r = -3
  2. Write out the terms: n=1: 3 × (-3)^0 = 3 × 1 = 3 n=2: 3 × (-3)^1 = 3 × (-3) = -9 n=3: 3 × (-3)^2 = 3 × 9 = 27 n=4: 3 × (-3)^3 = 3 × (-27) = -81
  3. Sum the terms: 3 + (-9) + 27 + (-81) = 3 - 9 + 27 - 81
  4. Calculate step by step: 3 - 9 = -6; -6 + 27 = 21; 21 - 81 = -60

The answer is -60.

6 ∑(n=1 to 4) 3 × (-2)^(n-1) = ?

Hint: This is a geometric series where each term is found by multiplying the previous term by a constant ratio. Identify the first term and the common ratio to apply the geometric series sum formula.

Show the answer

Answer: -15

  1. Identify the first term (a) and common ratio (r). The first term is 3, and the common ratio is -2.
  2. The number of terms (n) is 4.
  3. Use the geometric series sum formula: S_n = a * (1 - r^n) / (1 - r)
  4. Substitute the values: S_4 = 3 * (1 - (-2)^4) / (1 - (-2))
  5. Calculate (-2)^4 = 16
  6. Calculate 1 - 16 = -15
  7. Calculate 1 - (-2) = 3
  8. S_4 = 3 * (-15) / 3 = -15

The answer is -15.

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