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For each of the following number patterns, determine the general term \( \left(n^{\text {th }}\right. \) term) and hence the \( 10^{\text {th }} \) term. \( \begin{array}{lll}\text { (a) } 2 ; 4 ; 8 ; 16 ; \ldots & \text { (b) } 1 ; 3 ; 9 ; 27 ; \ldots & \text { (c) } 4 ; 12 ; 36 ; \ldots \\ \text { (d) } 5 ; 10 ; 20 ; 40 ; \ldots & \text { (e) } 2 ; 8 ; 32 ; \ldots & \text { (f) } 3 ; 9 ; 27 ; \ldots\end{array} \)

Ask by Ramos Vaughn. in South Africa
Mar 08,2025

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**(a) \( a_n = 2^n \), \( a_{10} = 1024 \)** **(b) \( a_n = 3^{n-1} \), \( a_{10} = 19683 \)** **(c) \( a_n = 4 \cdot 3^{n-1} \), \( a_{10} = 78732 \)** **(d) \( a_n = 5 \cdot 2^{n-1} \), \( a_{10} = 2560 \)** **(e) \( a_n = 2 \cdot 4^{n-1} \), \( a_{10} = 524288 \)** **(f) \( a_n = 3^n \), \( a_{10} = 59049 \)**

Solution

**(a) Sequence: \(2,\,4,\,8,\,16,\,\ldots\)** The sequence is geometric with first term \[ a=2 \quad \text{and common ratio} \quad r=2. \] The general term is: \[ a_n = 2 \cdot 2^{n-1} = 2^n. \] The \(10^{\text{th}}\) term is: \[ a_{10} = 2^{10} = 1024. \] **(b) Sequence: \(1,\,3,\,9,\,27,\,\ldots\)** This is a geometric sequence with \[ a=1 \quad \text{and} \quad r=3. \] The general term is: \[ a_n = 1 \cdot 3^{n-1} = 3^{n-1}. \] Thus, the \(10^{\text{th}}\) term is: \[ a_{10} = 3^{9} = 19683. \] **(c) Sequence: \(4,\,12,\,36,\,\ldots\)** This is a geometric sequence with \[ a=4 \quad \text{and} \quad r=3. \] The general term is: \[ a_n = 4 \cdot 3^{n-1}. \] The \(10^{\text{th}}\) term is: \[ a_{10} = 4 \cdot 3^{9} = 4 \times 19683 = 78732. \] **(d) Sequence: \(5,\,10,\,20,\,40,\,\ldots\)** This is a geometric sequence with \[ a=5 \quad \text{and} \quad r=2. \] The general term is: \[ a_n = 5 \cdot 2^{n-1}. \] The \(10^{\text{th}}\) term is: \[ a_{10} = 5 \cdot 2^{9} = 5 \times 512 = 2560. \] **(e) Sequence: \(2,\,8,\,32,\,\ldots\)** This is a geometric sequence with \[ a=2 \quad \text{and} \quad r=4. \] The general term is: \[ a_n = 2 \cdot 4^{n-1}. \] The \(10^{\text{th}}\) term is: \[ a_{10} = 2 \cdot 4^{9}. \] Since \[ 4^{9} = 262144, \] we have \[ a_{10} = 2 \times 262144 = 524288. \] **(f) Sequence: \(3,\,9,\,27,\,\ldots\)** This sequence is geometric with \[ a=3 \quad \text{and} \quad r=3. \] The general term is: \[ a_n = 3 \cdot 3^{n-1} = 3^n. \] Thus, the \(10^{\text{th}}\) term is: \[ a_{10} = 3^{10} = 59049. \]

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For each pattern, we can find the general term and specifically the \(10^{\text{th}}\) term: (a) The pattern is \(2, 4, 8, 16, \ldots\) which can be expressed as \(2^n\) where \(n\) starts from 1. The \(n^{\text{th}}\) term is \(2^n\) and the \(10^{\text{th}}\) term is \(2^{10} = 1024\). (b) The sequence \(1, 3, 9, 27, \ldots\) is \(3^{n-1}\) where \(n\) starts from 1. Therefore, the \(n^{\text{th}}\) term is \(3^{n-1}\) and the \(10^{\text{th}}\) term is \(3^{9} = 19683\). (c) For \(4, 12, 36, \ldots\), we can see this is \(4 \cdot 3^{n-1}\) where \(n\) starts from 1. Thus, the \(n^{\text{th}}\) term becomes \(4 \cdot 3^{n-1}\) and the \(10^{\text{th}}\) term is \(4 \cdot 3^{9} = 88512\). (d) The series \(5, 10, 20, 40, \ldots\) follows \(5 \cdot 2^{n-1}\) for \(n\) starting from 1. Hence, the \(n^{\text{th}}\) term can be detailed as \(5 \cdot 2^{n-1}\) and the \(10^{\text{th}}\) term is \(5 \cdot 2^{9} = 2560\). (e) The pattern \(2, 8, 32, \ldots\) relates to \(2^{2n}\) where \(n\) starts from 1. Thus, the \(n^{\text{th}}\) term is \(2^{2n}\) and the \(10^{\text{th}}\) term evaluates to \(2^{20} = 1048576\). (f) For \(3, 9, 27, \ldots\), we have the general term \(3^{n}\) starting from \(n=1\). So, the \(n^{\text{th}}\) term is \(3^{n}\) and the \(10^{\text{th}}\) term is \(3^{10} = 59049\).

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