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RCISE 2 Determine the general term for each of the following geometrio sequences \( \begin{array}{lll}\text { (a) } 2 ;-1 ; \frac{1}{2} ; \ldots \ldots & \text { (b) } 2 ; 8 ; 32 ; \ldots \ldots & \text { (c) }-\frac{2}{3} ; 2 ;-6 ; \ldots \ldots \\ \text { (d) } 1 ; 0,2 ; 0,04 ; \ldots \ldots \\ \text { Determine the } 9 \text { th term for each of the following geometric sequences: } \\ \text { (a) } 128 ; 64 ; 32 ; \ldots . & \text { (b) } 0,25 ; 0,5 ; 1 ; \ldots \ldots & \text { (c) } \frac{4}{9} ; 1 \frac{1}{3}\end{array} \)

Ask by Romero Ellis. in South Africa
Jan 23,2025

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**General Terms:** - **(a)** \( a_n = 2 \cdot \left(-\frac{1}{2}\right)^{n-1} \) - **(b)** \( a_n = 2 \cdot 4^{n-1} \) - **(c)** \( a_n = -\frac{2}{3} \cdot (-3)^{n-1} \) - **(d)** \( a_n = (0.2)^{n-1} \) **9th Terms:** - **(a)** \( a_9 = 0.5 \) - **(b)** \( a_9 = 64 \) - **(c)** \( a_9 = 2916 \)

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Bonus Knowledge

For the geometric sequences provided, let's first derive the general terms. (a) The first term (a₁) is 2, and the common ratio (r) is \(-\frac{1}{2}\) (since \(-1 \div 2 = -\frac{1}{2}\)). Thus the general term can be expressed as: \[ a_n = 2 \left(-\frac{1}{2}\right)^{n-1} \] (b) The first term is 2, and the common ratio is 4 (since \(8 \div 2 = 4\)). Thus the general term is: \[ a_n = 2 \cdot 4^{n-1} \] (c) The first term is \(-\frac{2}{3}\), and the common ratio is -3 (since \(2 \div -\frac{2}{3} = -3\)). Hence, the general term would be: \[ a_n = -\frac{2}{3} \cdot (-3)^{n-1} \] (d) The first term is 1, and the common ratio is 0.2 (since \(0.2 \div 1 = 0.2\)). The general term is: \[ a_n = 1 \cdot (0.2)^{n-1} \] Now for the 9th term of the following geometric sequences: (a) The first term is 128, and the common ratio is \(\frac{1}{2}\) (as \(64 \div 128 = \frac{1}{2}\)). The 9th term is: \[ a_9 = 128 \cdot \left(\frac{1}{2}\right)^{8} = 128 \cdot \frac{1}{256} = 0.5 \] (b) The first term is 0.25 and the common ratio is 2 (as \(0.5 \div 0.25 = 2\)). Therefore, the 9th term is: \[ a_9 = 0.25 \cdot 2^{8} = 0.25 \cdot 256 = 64 \] (c) The first term is \(\frac{4}{9}\), and the common ratio is \(\frac{3}{4}\) (as \(1 \frac{1}{3} \div \frac{4}{9} = 1 \cdot \frac{9}{4} = \frac{3}{4}\)). The 9th term is: \[ a_9 = \frac{4}{9} \cdot \left(\frac{3}{4}\right)^{8} = \frac{4 \cdot 6561}{9 \cdot 65536} = \frac{6561}{147456} \] There you have it! The general terms and the 9th terms for the given geometric sequences. Happy calculating!

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