Use these Sequences practice questions to review arithmetic sequences, geometric sequences, Fibonacci sequences, consecutive even numbers, sequence patterns, and nth-term formulas. After answering each question, open the explanation to see the step-by-step solution.
Sequences Topics Covered
- Finding the next term in a sequence
- Recognizing arithmetic patterns
- Recognizing geometric patterns
- Working with consecutive even numbers
- Using the Fibonacci sequence
- Finding nth terms
- Writing sequence formulas
Sample Questions
5, 13, 21, 29, ___
- A. 37
- B. 38
- C. 39
- D. 42
- E. 43
Show Answer
Look at the difference between each pair of consecutive terms.
13 − 5 = 8
21 − 13 = 8
29 − 21 = 8
Each term is 8 more than the previous term, so add 8 to 29.
29 + 8 = 37
The next number is 37.
- A. 6, 8, 10, 12, 14
- B. 8, 10, 12, 14, 16
- C. 9, 10, 11, 12, 13
- D. 10, 11, 12, 13, 14
- E. 10, 12, 14, 16, 18
Show Answer
Five consecutive even numbers increase by 2 each time.
Let x represent the first number. Then the five numbers are:
x, x + 2, x + 4, x + 6, x + 8
Their sum is 60:
x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 60
5x + 20 = 60
5x = 40
x = 8
So the sequence is 8, 10, 12, 14, 16.
6, 3, 32, 34, ___
- A. 14
- B. 38
- C. 37
- D. 12
- E. 32
Show Answer
Each number is half of the previous number.
6 ÷ 2 = 3
3 ÷ 2 = 32
32 ÷ 2 = 34
Now multiply the last term by one-half:
34 × 12 = 38
The next number is 38.
- A. 0, 1, 1, 2, 2, 4
- B. 0, 1, 1, 2, 3, 4
- C. 0, 1, 1, 2, 3, 5
- D. 0, 1, 2, 3, 5, 8
- E. 0, 1, 2, 3, 6, 9
Show Answer
The first two numbers are 0 and 1.
Add them to find the third number:
0 + 1 = 1
Continue adding consecutive terms:
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
The first six numbers are 0, 1, 1, 2, 3, 5.
- A. 2, 4, 8, 16, 32, 64
- B. 2, 4, 12, 32, 96, 288
- C. 2, 6, 12, 24, 72, 216
- D. 2, 6, 12, 36, 108, 324
- E. 2, 6, 18, 54, 162, 486
Show Answer
The first number is 2. Each number after that is three times the previous number.
2 × 3 = 6
6 × 3 = 18
18 × 3 = 54
54 × 3 = 162
162 × 3 = 486
The first six numbers are 2, 6, 18, 54, 162, 486.
- A. −4, −2, 0, 2, 4
- B. −3, −1, 1, 3, 5
- C. −2, 0, 2, 4, 6
- D. −2, −1, 0, 1, 2
- E. 0, 2, 4, 6, 8
Show Answer
Five consecutive even numbers increase by 2 each time.
The sequence must add to 0, so the positive and negative terms need to balance each other.
−4 + (−2) + 0 + 2 + 4 = 0
Therefore, the correct sequence is −4, −2, 0, 2, 4.
9, 15, 21, 27, 33, …
- A. 135
- B. 136
- C. 138
- D. 141
- E. 142
Show Answer
The sequence increases by 6 each time, so it is an arithmetic sequence.
The first term is 9, and the common difference is 6.
A formula for the nth term is:
f(n) = 6n + 3
Substitute 23 for n:
f(23) = 6(23) + 3
f(23) = 138 + 3
f(23) = 141
The 23rd term is 141.
32, 48, 72, 108, …
- A. 124, 140, 156, 172
- B. 4843, 1354, 4058, 12,07281
- C. 162, 243, 7292, 21874
- D. 162, 324, 648, 1296
- E. 172, 258, 387, 581
Show Answer
Each number is multiplied by 32 to get the next term.
108 × 32 = 162
162 × 32 = 243
243 × 32 = 7292
7292 × 32 = 21874
The next four numbers are 162, 243, 7292, and 21874.
0, 1, 3, 6, 10, ___
- A. 13
- B. 14
- C. 15
- D. 16
- E. 17
Show Answer
Look at the differences between consecutive terms:
1 − 0 = 1
3 − 1 = 2
6 − 3 = 3
10 − 6 = 4
The differences are increasing by 1 each time. The next difference should be 5.
10 + 5 = 15
The next number is 15.
69, 60, 51, 42, 33, …
- A. −279
- B. −201
- C. −189
- D. −188
- E. −31
Show Answer
The sequence decreases by 9 each time, so it is an arithmetic sequence.
A formula for the nth term is:
f(n) = −9n + 78
Substitute 31 for n:
f(31) = −9(31) + 78
f(31) = −279 + 78
f(31) = −201
The 31st term is −201.
How to Use These Sequences Practice Questions
Start by answering each question before opening the explanation. Then compare your work to the step-by-step solution. If you miss a question, review the pattern or formula used in the solution before moving on.
For extra review, focus on the sequence types you miss most often. Some questions require recognizing a constant difference, while others require recognizing a constant multiplier, adding previous terms, or using a formula for the nth term.