Arithmetic and Geometric Sequences

Understanding patterns in sequences and finding explicit and recursive formulas

Arithmetic and Geometric Sequences

What is a Sequence?

A sequence is an ordered list of numbers. Each number in the sequence is called a term.

Notation: a1,a2,a3,a4,…,an,…a_1, a_2, a_3, a_4, \ldots, a_n, \ldots

  • a1a_1 is the first term
  • ana_n is the nnth term
  • nn is the term number (position)

Arithmetic Sequences

An arithmetic sequence has a constant difference between consecutive terms.

Common difference: d=an+1−and = a_{n+1} - a_n

Formulas

Explicit (Direct) Formula: an=a1+(n−1)da_n = a_1 + (n - 1)d

where:

  • ana_n = nnth term
  • a1a_1 = first term
  • dd = common difference
  • nn = term number

Recursive Formula: an=an−1+d,a1=(given)a_n = a_{n-1} + d, \quad a_1 = \text{(given)}

Example: 3,7,11,15,19,…3, 7, 11, 15, 19, \ldots

  • First term: a1=3a_1 = 3
  • Common difference: d=4d = 4
  • Explicit: an=3+(n−1)(4)=3+4n−4=4n−1a_n = 3 + (n-1)(4) = 3 + 4n - 4 = 4n - 1
  • Recursive: an=an−1+4,a1=3a_n = a_{n-1} + 4, \quad a_1 = 3
  • 10th term: a10=4(10)−1=39a_{10} = 4(10) - 1 = 39

Geometric Sequences

A geometric sequence has a constant ratio between consecutive terms.

Common ratio: r=an+1anr = \frac{a_{n+1}}{a_n}

Formulas

Explicit (Direct) Formula: an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}

where:

  • ana_n = nnth term
  • a1a_1 = first term
  • rr = common ratio
  • nn = term number

Recursive Formula: an=an−1⋅r,a1=(given)a_n = a_{n-1} \cdot r, \quad a_1 = \text{(given)}

Example: 2,6,18,54,162,…2, 6, 18, 54, 162, \ldots

  • First term: a1=2a_1 = 2
  • Common ratio: r=3r = 3
  • Explicit: an=2⋅3n−1a_n = 2 \cdot 3^{n-1}
  • Recursive: an=3an−1,a1=2a_n = 3a_{n-1}, \quad a_1 = 2
  • 6th term: a6=2⋅35=2⋅243=486a_6 = 2 \cdot 3^5 = 2 \cdot 243 = 486

Identifying Sequence Type

Arithmetic: Check if differences are constant

  • 7,11,15,19,…7, 11, 15, 19, \ldots → differences: 4,4,44, 4, 4 ✓

Geometric: Check if ratios are constant

  • 3,12,48,192,…3, 12, 48, 192, \ldots → ratios: 4,4,44, 4, 4 ✓

Neither: If differences and ratios both vary

  • 1,4,9,16,…1, 4, 9, 16, \ldots (perfect squares) → neither

Sum of Arithmetic Sequence (Finite)

Sum of first nn terms: Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}

or

Sn=n[2a1+(n−1)d]2S_n = \frac{n[2a_1 + (n-1)d]}{2}

Example

Sum of first 10 terms of 3,7,11,15,…3, 7, 11, 15, \ldots: S10=10(3+39)2=10(42)2=210S_{10} = \frac{10(3 + 39)}{2} = \frac{10(42)}{2} = 210

Sum of Geometric Sequence (Finite)

Sum of first nn terms: Sn=a1⋅1−rn1−r,r≠1S_n = a_1 \cdot \frac{1 - r^n}{1 - r}, \quad r \neq 1

Example

Sum of first 5 terms of 2,6,18,54,…2, 6, 18, 54, \ldots: S5=2⋅1−351−3=2⋅1−243−2=2⋅−242−2=242S_5 = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot \frac{-242}{-2} = 242

Applications

  • Arithmetic: Linear growth, evenly spaced values

    • Saving $50 per month
    • Theater seating rows

  • Geometric: Exponential growth/decay

    • Population growth
    • Radioactive decay
    • Compound interest

📚 Practice Problems

1Problem 1easy

❓ Question:

For the arithmetic sequence 5,9,13,17,…5, 9, 13, 17, \ldots, find the 20th term and write both explicit and recursive formulas.

💡 Show Solution

Identify the sequence:

  • First term: a1=5a_1 = 5
  • Common difference: d=9−5=4d = 9 - 5 = 4

Explicit formula: an=a1+(n−1)da_n = a_1 + (n-1)d an=5+(n−1)(4)a_n = 5 + (n-1)(4) an=5+4n−4a_n = 5 + 4n - 4 an=4n+1a_n = 4n + 1

Recursive formula: an=an−1+4,a1=5a_n = a_{n-1} + 4, \quad a_1 = 5

Find the 20th term: a20=4(20)+1=80+1=81a_{20} = 4(20) + 1 = 80 + 1 = 81

Verify: We can check by adding dd nineteen times to a1a_1: 5+19(4)=5+76=815 + 19(4) = 5 + 76 = 81 ✓

Answers:

  • Explicit: an=4n+1a_n = 4n + 1
  • Recursive: an=an−1+4,a1=5a_n = a_{n-1} + 4, a_1 = 5
  • 20th term: 8181

2Problem 2easy

❓ Question:

An arithmetic sequence has first term a1=5a_1 = 5 and common difference d=3d = 3.

a) Write the first five terms. b) Find the 20th term. c) Find the sum of the first 20 terms.

💡 Show Solution

Solution:

Part (a): For arithmetic sequence: an=a1+(n−1)da_n = a_1 + (n-1)d

a1=5a_1 = 5 a2=5+3=8a_2 = 5 + 3 = 8 a3=8+3=11a_3 = 8 + 3 = 11 a4=11+3=14a_4 = 11 + 3 = 14 a5=14+3=17a_5 = 14 + 3 = 17

First five terms: 5, 8, 11, 14, 17

Part (b): a20=a1+(20−1)d=5+19(3)=5+57=62a_{20} = a_1 + (20-1)d = 5 + 19(3) = 5 + 57 = 62

Part (c): Sum formula: Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}

S20=20(5+62)2=20(67)2=10(67)=670S_{20} = \frac{20(5 + 62)}{2} = \frac{20(67)}{2} = 10(67) = 670

3Problem 3medium

❓ Question:

A geometric sequence has first term a1=2a_1 = 2 and common ratio r=3r = 3.

a) Find the 6th term. b) Find the sum of the first 8 terms.

💡 Show Solution

Solution:

Part (a): For geometric sequence: an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}

a6=2⋅36−1=2⋅35=2⋅243=486a_6 = 2 \cdot 3^{6-1} = 2 \cdot 3^5 = 2 \cdot 243 = 486

Part (b): Sum formula: Sn=a1⋅rn−1r−1S_n = a_1 \cdot \frac{r^n - 1}{r - 1} (for r≠1r \neq 1)

S8=2⋅38−13−1S_8 = 2 \cdot \frac{3^8 - 1}{3 - 1}

=2⋅6561−12= 2 \cdot \frac{6561 - 1}{2}

=2⋅65602= 2 \cdot \frac{6560}{2}

=2⋅3280= 2 \cdot 3280

=6560= 6560

4Problem 4medium

❓ Question:

A geometric sequence has first term a1=3a_1 = 3 and common ratio r=2r = 2. Find the 8th term and the sum of the first 8 terms.

💡 Show Solution

Given information:

  • a1=3a_1 = 3
  • r=2r = 2

Find the 8th term using explicit formula: an=a1⋅rn−1a_n = a_1 \cdot r^{n-1} a8=3⋅28−1a_8 = 3 \cdot 2^{8-1} a8=3⋅27a_8 = 3 \cdot 2^7 a8=3⋅128=384a_8 = 3 \cdot 128 = 384

Find the sum of first 8 terms: Sn=a1⋅1−rn1−rS_n = a_1 \cdot \frac{1 - r^n}{1 - r} S8=3⋅1−281−2S_8 = 3 \cdot \frac{1 - 2^8}{1 - 2} S8=3⋅1−256−1S_8 = 3 \cdot \frac{1 - 256}{-1} S8=3⋅−255−1S_8 = 3 \cdot \frac{-255}{-1} S8=3⋅255=765S_8 = 3 \cdot 255 = 765

Verify the sequence: 3,6,12,24,48,96,192,3843, 6, 12, 24, 48, 96, 192, 384 Sum: 3+6+12+24+48+96+192+384=7653 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765 ✓

Answers:

  • 8th term: 384384
  • Sum of first 8 terms: 765765

5Problem 5hard

❓ Question:

The 3rd term of an arithmetic sequence is 14 and the 7th term is 30. Find the first term, common difference, and the explicit formula.

💡 Show Solution

Set up equations using an=a1+(n−1)da_n = a_1 + (n-1)d:

For the 3rd term: a3=a1+2d=14a_3 = a_1 + 2d = 14

For the 7th term: a7=a1+6d=30a_7 = a_1 + 6d = 30

Solve the system by elimination:

Subtract first equation from second: (a1+6d)−(a1+2d)=30−14(a_1 + 6d) - (a_1 + 2d) = 30 - 14 4d=164d = 16 d=4d = 4

Find a1a_1: Substitute d=4d = 4 into first equation: a1+2(4)=14a_1 + 2(4) = 14 a1+8=14a_1 + 8 = 14 a1=6a_1 = 6

Write the explicit formula: an=a1+(n−1)da_n = a_1 + (n-1)d an=6+(n−1)(4)a_n = 6 + (n-1)(4) an=6+4n−4a_n = 6 + 4n - 4 an=4n+2a_n = 4n + 2

Verify:

  • a3=4(3)+2=14a_3 = 4(3) + 2 = 14 ✓
  • a7=4(7)+2=30a_7 = 4(7) + 2 = 30 ✓

Answers:

  • First term: a1=6a_1 = 6
  • Common difference: d=4d = 4
  • Explicit formula: an=4n+2a_n = 4n + 2
🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics