Arithmetic and Geometric Sequences Worksheet Answer Key – Algebra 1

Problem 1 Common difference

13 − 7 = 6, 19 − 13 = 6, 25 − 19 = 6, so d = 6.

Next term: 25 + 6 = 31.

d = 6; next term 31

Problem 2 Common ratio

30 ÷ 6 = 5, 150 ÷ 30 = 5, 750 ÷ 150 = 5, so r = 5.

Next term: 750 × 5 = 3,750.

r = 5; next term 3,750

Problem 3 List an arithmetic sequence

12, then 12 − 4 = 8, 8 − 4 = 4, 4 − 4 = 0, 0 − 4 = −4.

12, 8, 4, 0, −4

Problem 4 List a geometric sequence

2, 2 × 5 = 10, 10 × 5 = 50, 50 × 5 = 250.

2, 10, 50, 250

Problem 5 Classify sequences

(a) Differences 5, 7, 9 and ratios 2.25, 1.78, 1.56: neither.

(b) Ratios 12/24 = 1/2, 6/12 = 1/2, 3/6 = 1/2: geometric.

(c) Differences −3, −3, −3: arithmetic.

(a) neither (b) geometric (c) arithmetic

Problem 6 Formula from a graph

The terms are 5, 8, 11, 14, so d = 3 and a₁ = 5.

a_n = 5 + (n − 1)(3) = 3n + 2. Then a₂₀ = 3(20) + 2 = 62.

a_n = 3n + 2; a₂₀ = 62

Problem 7 Explicit arithmetic formula

d = −4 and a₁ = 15, so aₙ = 15 + (n − 1)(−4) = 19 − 4n.

a₁₂ = 19 − 48 = −29.

a_n = 19 − 4n; a₁₂ = −29

Problem 8 Explicit geometric formula

r = 3 and a₁ = 6, so aₙ = 6 · 3^{n − 1}.

a₆ = 6 · 3⁵ = 6 · 243 = 1,458.

a_n = 6 · 3^{n − 1}; a₆ = 1,458

Problem 9 Recursive to explicit

9, 14, 19, 24, 29.

aₙ = 9 + (n − 1)(5) = 5n + 4.

Check: a₅ = 5(5) + 4 = 29.

a_n = 5n + 4

Problem 10 Recursive geometric

Each term is half the previous one: a₁ = 80, aₙ = 0.5·aₙ₋₁.

a₈ = 80 · (1/2)⁷ = 80 ÷ 128 = 0.625.

a₁ = 80, a_n = 0.5a_{n − 1}; a₈ = 0.625

Problem 11 Which term equals 128?

d = 6, so aₙ = 8 + 6(n − 1) = 6n + 2.

6n + 2 = 128 gives 6n = 126 and n = 21.

Check: 6(21) + 2 = 128.

The 21st term

Problem 12 Two terms of an arithmetic sequence

d = (37 − 17) ÷ (7 − 3) = 20 ÷ 4 = 5.

a₁ = 17 − 2(5) = 7, so aₙ = 7 + 5(n − 1) = 5n + 2.

Check: a₃ = 17, a₇ = 37.

d = 5; a₁ = 7; a_n = 5n + 2

Problem 13 Two terms of a geometric sequence

From term 2 to term 5 the ratio is applied 3 times: r³ = 324 ÷ 12 = 27, so r = 3.

a₁ = 12 ÷ 3 = 4.

Check: 4, 12, 36, 108, 324.

r = 3; a₁ = 4

Problem 14 Sequence as a function

The terms are 2, 6, 18, 54, so r = 6 ÷ 2 = 3.

f(n) = 2 · 3^{n − 1}. Then f(5) = 2 · 3⁴ = 2 · 81 = 162.

The domain is the positive integers 1, 2, 3, ….

f(n) = 2 · 3^{n − 1}; f(5) = 162

Problem 15 Weekly savings

a₁ = 15 and d = 8, so aₙ = 15 + 8(n − 1) = 8n + 7.

8n + 7 > 150 gives 8n > 143, so n > 17.875.

Week 18 gives 8(18) + 7 = 151; week 17 gives 143.

Week 18

Problem 16 Stadium rows

aₙ = 22 + 3(n − 1) = 3n + 19.

Row 30: 22 + 3(29) = 109 seats.

3n + 19 = 70 gives 3n = 51 and n = 17.

109 seats in row 30; row 17 has 70 seats

Problem 17 Viral video views

aₙ = 3 · 3^{n − 1} = 3ⁿ.

Day 7: 3⁷ = 2,187.

3⁸ = 6,561 is too small and 3⁹ = 19,683 exceeds 10,000.

aₙ = 3ⁿ; 2,187 views on day 7; day 9

Problem 18 Two depreciation models

Model A is geometric with r = 0.8: a₄ = 18,000(0.8)³ = 18,000(0.512) = 9,216.

Model B is arithmetic: 18,000 − 3(2,500) = 10,500.

Recursive formula for A: a₁ = 18,000, aₙ = 0.8·aₙ₋₁.

A: $9,216; B: $10,500; Model A is lower

Problem 19 Toothpick squares

1 square needs 4 toothpicks and each additional square adds 3, so aₙ = 4 + 3(n − 1) = 3n + 1.

25 squares: 3(25) + 1 = 76. For 100 toothpicks: 3n + 1 = 100 gives n = 33.

aₙ = 3n + 1; 76 toothpicks; 33 squares

Problem 20 Bouncing ball

h(n) = 160(0.75)ⁿ.

Bounce 5: 160(0.75)⁵ = 160(0.2373046875) ≈ 37.97 inches.

Bounce 7: 160(0.75)⁷ ≈ 21.36, still above 20. Bounce 8: 160(0.75)⁸ ≈ 16.02, below 20.

h(n) = 160(0.75)ⁿ; 37.97 inches; bounce 8