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
Sequences Practice Questions
5, 13, 21, 29, __
- 37
- 38
- 39
- 42
- 43
Show Answer
Answer: A. 37
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.
- 6, 8, 10, 12, 14
- 8, 10, 12, 14, 16
- 9, 10, 11, 12, 13
- 10, 11, 12, 13, 14
- 10, 12, 14, 16, 18
Show Answer
Answer: B. 8, 10, 12, 14, 16
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,
__
- 14
- 38
- 37
- 12
- 32
Show Answer
Answer: B. 38
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.
- 0, 1, 1, 2, 2, 4
- 0, 1, 1, 2, 3, 4
- 0, 1, 1, 2, 3, 5
- 0, 1, 2, 3, 5, 8
- 0, 1, 2, 3, 6, 9
Show Answer
Answer: C. 0, 1, 1, 2, 3, 5
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.
- 2, 4, 8, 16, 32, 64
- 2, 4, 12, 32, 96, 288
- 2, 6, 12, 24, 72, 216
- 2, 6, 12, 36, 108, 324
- 2, 6, 18, 54, 162, 486
Show Answer
Answer: E. 2, 6, 18, 54, 162, 486
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.
- −4, −2, 0, 2, 4
- −3, −1, 1, 3, 5
- −2, 0, 2, 4, 6
- −2, −1, 0, 1, 2
- 0, 2, 4, 6, 8
Show Answer
Answer: A. −4, −2, 0, 2, 4
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, …
- 135
- 136
- 138
- 141
- 142
Show Answer
Answer: D. 141
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, …
- 124, 140, 156, 172
- 4843,
1354,
4058,
12,07281 - 162, 243,
7292,
21874 - 162, 324, 648, 1296
- 172, 258, 387, 581
Show Answer
Answer: C. 162, 243, 7292, 21874
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, __
- 13
- 14
- 15
- 16
- 17
Show Answer
Answer: C. 15
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, …
- −279
- −201
- −189
- −188
- −31
Show Answer
Answer: B. −201
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.