Sequences review | KS3 maths worksheet | Teachit - Free Printable

Problem Analysis and Solution

The provided worksheet focuses on sequences, including finding terms, describing rules, identifying sequence types, and working with the \( n \)-th term formula. Below is a detailed solution for each section.

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Section 1: Sequences

#### a. \( 2, 6, 10, 14, 18, \ldots \)
- Next Term: The sequence increases by 4 each time. So, the next term is \( 18 + 4 = 22 \).
- Term to Term Rule: Add 4 to the previous term.
- Sequence Type: Linear (arithmetic sequence).

#### b. \( 2, 4, 8, 14, 22, \ldots \)
- Next Term: The differences between consecutive terms are increasing by 2 each time: \( 2, 4, 6, 8, \ldots \). So, the next difference is 10, and the next term is \( 22 + 10 = 32 \).
- Term to Term Rule: Add an increasing even number (starting from 2).
- Sequence Type: Quadratic.

#### c. \( 1, 2, 4, 8, 16, \ldots \)
- Next Term: Each term is multiplied by 2. So, the next term is \( 16 \times 2 = 32 \).
- Term to Term Rule: Multiply the previous term by 2.
- Sequence Type: Geometric.

#### d. \( 20, 17, 14, 11, 8, \ldots \)
- Next Term: The sequence decreases by 3 each time. So, the next term is \( 8 - 3 = 5 \).
- Term to Term Rule: Subtract 3 from the previous term.
- Sequence Type: Linear (arithmetic sequence).

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Section 2: What is meant by the nth term or \( T_n \)?

The \( n \)-th term (\( T_n \)) is a formula that allows you to find any term in the sequence directly by substituting the position \( n \). For example, if \( T_n = 3n + 2 \), then the 5th term is \( T_5 = 3(5) + 2 = 17 \).

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Section 3: Find the value of the term

#### a. \( 3n + 5 \)
- Term 7: Substitute \( n = 7 \): \( T_7 = 3(7) + 5 = 21 + 5 = 26 \).

#### b. \( 2n - 3 \)
- Term 2: Substitute \( n = 2 \): \( T_2 = 2(2) - 3 = 4 - 3 = 1 \).

#### c. \( 4n + 6 \)
- Term 20: Substitute \( n = 20 \): \( T_{20} = 4(20) + 6 = 80 + 6 = 86 \).

#### d. \( 3n - 5 \)
- Term 1: Substitute \( n = 1 \): \( T_1 = 3(1) - 5 = 3 - 5 = -2 \).

#### e. \( n^2 + 4 \)
- Term 6: Substitute \( n = 6 \): \( T_6 = 6^2 + 4 = 36 + 4 = 40 \).

#### f. \( 2n^2 \)
- Term 3: Substitute \( n = 3 \): \( T_3 = 2(3^2) = 2(9) = 18 \).

#### g. \( n(n - 1) \)
- Term 7: Substitute \( n = 7 \): \( T_7 = 7(7 - 1) = 7 \times 6 = 42 \).

#### h. \( 20 - 5n \)
- Term 6: Substitute \( n = 6 \): \( T_6 = 20 - 5(6) = 20 - 30 = -10 \).

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Section 4: Correct Laura's answers for the 5th term

#### a. \( 2n + 3 \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 2(5) + 3 = 10 + 3 = 13 \). Laura's answer (28) is incorrect.

#### b. \( 3n - 1 \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 3(5) - 1 = 15 - 1 = 14 \). Laura's answer (14) is correct.

#### c. \( 30 - 2n \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 30 - 2(5) = 30 - 10 = 20 \). Laura's answer (5) is incorrect.

#### d. \( 20 - 3n \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 20 - 3(5) = 20 - 15 = 5 \). Laura's answer (5) is correct.

#### e. \( n^2 - 3 \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 5^2 - 3 = 25 - 3 = 22 \). Laura's answer (22) is correct.

#### f. \( 2n^2 \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 2(5^2) = 2(25) = 50 \). Laura's answer (100) is incorrect.

#### g. \( n(n + 1) \)
- Term 5: Substitute \( n = 5 \): \( T_5 = 5(5 + 1) = 5 \times 6 = 30 \). Laura's answer (30) is correct.

#### h. \( (4n)^2 \)
- Term 5: Substitute \( n = 5 \): \( T_5 = (4 \times 5)^2 = 20^2 = 400 \). Laura's answer (200) is incorrect.

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Section 5: Write the first five terms for each sequence given by the nth term

#### a. \( T_n = 3n + 2 \)
- \( T_1 = 3(1) + 2 = 5 \)
- \( T_2 = 3(2) + 2 = 8 \)
- \( T_3 = 3(3) + 2 = 11 \)
- \( T_4 = 3(4) + 2 = 14 \)
- \( T_5 = 3(5) + 2 = 17 \)
- First five terms: \( 5, 8, 11, 14, 17 \)

#### b. \( T_n = 4n - 1 \)
- \( T_1 = 4(1) - 1 = 3 \)
- \( T_2 = 4(2) - 1 = 7 \)
- \( T_3 = 4(3) - 1 = 11 \)
- \( T_4 = 4(4) - 1 = 15 \)
- \( T_5 = 4(5) - 1 = 19 \)
- First five terms: \( 3, 7, 11, 15, 19 \)

#### c. \( T_n = 5n - 3 \)
- \( T_1 = 5(1) - 3 = 2 \)
- \( T_2 = 5(2) - 3 = 7 \)
- \( T_3 = 5(3) - 3 = 12 \)
- \( T_4 = 5(4) - 3 = 17 \)
- \( T_5 = 5(5) - 3 = 22 \)
- First five terms: \( 2, 7, 12, 17, 22 \)

#### d. \( T_n = 3n - 5 \)
- \( T_1 = 3(1) - 5 = -2 \)
- \( T_2 = 3(2) - 5 = 1 \)
- \( T_3 = 3(3) - 5 = 4 \)
- \( T_4 = 3(4) - 5 = 7 \)
- \( T_5 = 3(5) - 5 = 10 \)
- First five terms: \( -2, 1, 4, 7, 10 \)

#### e. \( T_n = (n - 1)(n + 1) \)
- \( T_1 = (1 - 1)(1 + 1) = 0 \)
- \( T_2 = (2 - 1)(2 + 1) = 3 \)
- \( T_3 = (3 - 1)(3 + 1) = 8 \)
- \( T_4 = (4 - 1)(4 + 1) = 15 \)
- \( T_5 = (5 - 1)(5 + 1) = 24 \)
- First five terms: \( 0, 3, 8, 15, 24 \)

#### f. \( T_n = 3n^2 \)
- \( T_1 = 3(1^2) = 3 \)
- \( T_2 = 3(2^2) = 12 \)
- \( T_3 = 3(3^2) = 27 \)
- \( T_4 = 3(4^2) = 48 \)
- \( T_5 = 3(5^2) = 75 \)
- First five terms: \( 3, 12, 27, 48, 75 \)

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Section 6: Find the nth term (\( T_n \))

#### a. \( 3, 5, 7, 9, 11, \ldots \)
- The sequence increases by 2 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 3 \Rightarrow a(1) + b = 3 \Rightarrow a + b = 3 \).
- From the second term: \( T_2 = 5 \Rightarrow a(2) + b = 5 \Rightarrow 2a + b = 5 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 3 \\
&2a + b = 5
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 5 - 3 \Rightarrow a = 2 \).
Substitute \( a = 2 \) into \( a + b = 3 \): \( 2 + b = 3 \Rightarrow b = 1 \).
- Therefore, \( T_n = 2n + 1 \).

#### b. \( 2, 6, 10, 14, 18, \ldots \)
- The sequence increases by 4 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 2 \Rightarrow a(1) + b = 2 \Rightarrow a + b = 2 \).
- From the second term: \( T_2 = 6 \Rightarrow a(2) + b = 6 \Rightarrow 2a + b = 6 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 2 \\
&2a + b = 6
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 6 - 2 \Rightarrow a = 4 \).
Substitute \( a = 4 \) into \( a + b = 2 \): \( 4 + b = 2 \Rightarrow b = -2 \).
- Therefore, \( T_n = 4n - 2 \).

#### c. \( 2, 5, 8, 11, 14, \ldots \)
- The sequence increases by 3 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 2 \Rightarrow a(1) + b = 2 \Rightarrow a + b = 2 \).
- From the second term: \( T_2 = 5 \Rightarrow a(2) + b = 5 \Rightarrow 2a + b = 5 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 2 \\
&2a + b = 5
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 5 - 2 \Rightarrow a = 3 \).
Substitute \( a = 3 \) into \( a + b = 2 \): \( 3 + b = 2 \Rightarrow b = -1 \).
- Therefore, \( T_n = 3n - 1 \).

#### d. \( 7, 11, 15, 19, 23, \ldots \)
- The sequence increases by 4 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 7 \Rightarrow a(1) + b = 7 \Rightarrow a + b = 7 \).
- From the second term: \( T_2 = 11 \Rightarrow a(2) + b = 11 \Rightarrow 2a + b = 11 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 7 \\
&2a + b = 11
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 11 - 7 \Rightarrow a = 4 \).
Substitute \( a = 4 \) into \( a + b = 7 \): \( 4 + b = 7 \Rightarrow b = 3 \).
- Therefore, \( T_n = 4n + 3 \).

#### e. \( 18, 16, 14, 12, 10, \ldots \)
- The sequence decreases by 2 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 18 \Rightarrow a(1) + b = 18 \Rightarrow a + b = 18 \).
- From the second term: \( T_2 = 16 \Rightarrow a(2) + b = 16 \Rightarrow 2a + b = 16 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 18 \\
&2a + b = 16
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 16 - 18 \Rightarrow a = -2 \).
Substitute \( a = -2 \) into \( a + b = 18 \): \( -2 + b = 18 \Rightarrow b = 20 \).
- Therefore, \( T_n = -2n + 20 \).

#### f. \( 30, 27, 24, 21, 18, \ldots \)
- The sequence decreases by 3 each time. The general form is \( T_n = an + b \).
- From the first term: \( T_1 = 30 \Rightarrow a(1) + b = 30 \Rightarrow a + b = 30 \).
- From the second term: \( T_2 = 27 \Rightarrow a(2) + b = 27 \Rightarrow 2a + b = 27 \).
- Solving the system:
\[
\begin{aligned}
&a + b = 30 \\
&2a + b = 27
\end{aligned}
\]
Subtract the first equation from the second: \( (2a + b) - (a + b) = 27 - 30 \Rightarrow a = -3 \).
Substitute \( a = -3 \) into \( a + b = 30 \): \( -3 + b = 30 \Rightarrow b = 33 \).
- Therefore, \( T_n = -3n + 33 \).

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Final Answers

\[
\boxed{
\begin{array}{ll}
\text{Section 1:} & \text{a. Next term: 22, Type: Linear} \\
& \text{b. Next term: 32, Type: Quadratic} \\
& \text{c. Next term: 32, Type: Geometric} \\
& \text{d. Next term: 5, Type: Linear} \\
\text{Section 3:} & \text{a. 26, b. 1, c. 86, d. -2, e. 40, f. 18, g. 42, h. -10} \\
\text{Section 4:} & \text{Correct: b, d, e, g; Incorrect: a, c, f, h} \\
\text{Section 5:} & \text{a. } 5, 8, 11, 14, 17 \\
& \text{b. } 3, 7, 11, 15, 19 \\
& \text{c. } 2, 7, 12, 17, 22 \\
& \text{d. } -2, 1, 4, 7, 10 \\
& \text{e. } 0, 3, 8, 15, 24 \\
& \text{f. } 3, 12, 27, 48, 75 \\
\text{Section 6:} & \text{a. } 2n + 1 \\
& \text{b. } 4n - 2 \\
& \text{c. } 3n - 1 \\
& \text{d. } 4n + 3 \\
& \text{e. } -2n + 20 \\
& \text{f. } -3n + 33 \\
\end{array}
}
\]