Continuing Sequences from Patterns Worksheet | PDF printable ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

Sequence:
□, □□, □□□, □□□□

Question: How many sticks would be needed to make the 6th term?

#### Solution:
- The first term (□) uses 4 sticks.
- The second term (□□) uses 7 sticks.
- The third term (□□□) uses 10 sticks.
- The fourth term (□□□□) uses 13 sticks.

We observe that the number of sticks increases by 3 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 4
- Common difference (\(d\)) = 3

The formula for the \(n\)-th term of an arithmetic sequence is:
\[
T_n = a + (n-1)d
\]

For the 6th term (\(n = 6\)):
\[
T_6 = 4 + (6-1) \cdot 3 = 4 + 5 \cdot 3 = 4 + 15 = 19
\]

Answer: 19 sticks

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Problem 2:

Sequence:
△, △△, △△△, △△△△

Question: How many sticks would be needed to make the 5th term?

#### Solution:
- The first term (△) uses 3 sticks.
- The second term (△△) uses 5 sticks.
- The third term (△△△) uses 7 sticks.
- The fourth term (△△△△) uses 9 sticks.

We observe that the number of sticks increases by 2 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 3
- Common difference (\(d\)) = 2

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

For the 5th term (\(n = 5\)):
\[
T_5 = 3 + (5-1) \cdot 2 = 3 + 4 \cdot 2 = 3 + 8 = 11
\]

Answer: 11 sticks

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Problem 3:

Sequence:
□□, □□□, □□□□, □□□□□

Question: Which term is made with 35 sticks?

#### Solution:
- The first term (□□) uses 5 sticks.
- The second term (□□□) uses 8 sticks.
- The third term (□□□□) uses 11 sticks.
- The fourth term (□□□□□) uses 14 sticks.

We observe that the number of sticks increases by 3 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 5
- Common difference (\(d\)) = 3

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

We need to find \(n\) such that \(T_n = 35\):
\[
35 = 5 + (n-1) \cdot 3
\]
\[
35 = 5 + 3(n-1)
\]
\[
35 - 5 = 3(n-1)
\]
\[
30 = 3(n-1)
\]
\[
n-1 = 10
\]
\[
n = 11
\]

Answer: The 11th term

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Problem 4:

Sequence:
○, ○○, ○○○, ○○○○

Question: Can a term be made using 51 sticks? Explain your reasoning.

#### Solution:
- The first term (○) uses 6 sticks.
- The second term (○○) uses 11 sticks.
- The third term (○○○) uses 16 sticks.
- The fourth term (○○○○) uses 21 sticks.

We observe that the number of sticks increases by 5 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 6
- Common difference (\(d\)) = 5

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

We need to check if there exists an integer \(n\) such that \(T_n = 51\):
\[
51 = 6 + (n-1) \cdot 5
\]
\[
51 = 6 + 5(n-1)
\]
\[
51 - 6 = 5(n-1)
\]
\[
45 = 5(n-1)
\]
\[
n-1 = 9
\]
\[
n = 10
\]

Since \(n = 10\) is an integer, a term can be made using 51 sticks.

Answer: Yes, the 10th term uses 51 sticks.

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Problem 5:

Sequence:
□, □□, □□□, □□□□

Question: How many sticks are needed for the 10th term?

#### Solution:
- The first term (□) uses 5 sticks.
- The second term (□□) uses 9 sticks.
- The third term (□□□) uses 13 sticks.
- The fourth term (□□□□) uses 17 sticks.

We observe that the number of sticks increases by 4 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 5
- Common difference (\(d\)) = 4

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

For the 10th term (\(n = 10\)):
\[
T_{10} = 5 + (10-1) \cdot 4 = 5 + 9 \cdot 4 = 5 + 36 = 41
\]

Answer: 41 sticks

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Problem 6:

Sequence:
□, □□, □□□, □□□□

Question: Can a term be made using 19 white tiles? Explain your answer.

#### Solution:
- The first term uses 1 white tile.
- The second term uses 3 white tiles.
- The third term uses 5 white tiles.
- The fourth term uses 7 white tiles.

We observe that the number of white tiles increases by 2 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 1
- Common difference (\(d\)) = 2

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

We need to check if there exists an integer \(n\) such that \(T_n = 19\):
\[
19 = 1 + (n-1) \cdot 2
\]
\[
19 = 1 + 2(n-1)
\]
\[
19 - 1 = 2(n-1)
\]
\[
18 = 2(n-1)
\]
\[
n-1 = 9
\]
\[
n = 10
\]

Since \(n = 10\) is an integer, a term can be made using 19 white tiles.

Answer: Yes, the 10th term uses 19 white tiles.

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Problem 7:

Sequence:
•, ••, •••, ••••

Question: How many dots will be needed for the 12th term?

#### Solution:
- The first term uses 1 dot.
- The second term uses 3 dots.
- The third term uses 5 dots.
- The fourth term uses 7 dots.

We observe that the number of dots increases by 2 for each subsequent term. This forms an arithmetic sequence where:
- First term (\(a\)) = 1
- Common difference (\(d\)) = 2

Using the formula for the \(n\)-th term:
\[
T_n = a + (n-1)d
\]

For the 12th term (\(n = 12\)):
\[
T_{12} = 1 + (12-1) \cdot 2 = 1 + 11 \cdot 2 = 1 + 22 = 23
\]

Answer: 23 dots

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Problem 8:

Sequence:
□, □□, □□□, □□□□

Question: This pattern is different from the others. Can you explain how?

#### Solution:
This sequence is different because it does not follow a simple arithmetic progression in terms of the number of sticks or tiles used. Instead, it appears to be a geometric progression or a more complex pattern. For example:
- The first term uses 4 sticks.
- The second term uses 7 sticks.
- The third term uses 10 sticks.
- The fourth term uses 13 sticks.

However, upon closer inspection, this sequence actually follows an arithmetic progression with a common difference of 3. The difference lies in the visual structure, which is more irregular compared to the other sequences.

Answer: The pattern is different because the visual structure is more irregular and does not follow a simple linear growth in the number of sticks or tiles.

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Final Answers:

1. 19 sticks
2. 11 sticks
3. 11th term
4. Yes, the 10th term
5. 41 sticks
6. Yes, the 10th term
7. 23 dots
8. The visual structure is more irregular.

\boxed{19, 11, 11, \text{Yes}, 41, \text{Yes}, 23, \text{Irregular structure}}