The first three terms of a geometric sequence are: \( 16 ; 4 ; 1 ; \ldots \) 1.2 Calculate the value of the \( 12^{\text {th }} \) term (Leave your answe simplified exponential form) 1.2 Calculate the sum of the first 10 terms of the sequence 2. Given the geometric sequence: \( 27 ; 9 ; 3 ; \ldots \) 2.1 Determine the formula for the nth term of the sequence 2.2 Why does the sum to infinity of this sequence exist? 1.3 Determine \( S_{\infty} \)

Let's solve the problems step by step. ### Problem 1: Geometric Sequence \( 16, 4, 1, \ldots \) #### 1.1 Calculate the value of the \( 12^{\text{th}} \) term In a geometric sequence, each term can be expressed as: \[ a_n = a_1 \cdot r^{n-1} \] where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number. 1. **Identify the first term and common ratio:** - First term \( a_1 = 16 \) - Second term \( a_2 = 4 \) - Common ratio \( r = \frac{a_2}{a_1} = \frac{4}{16} = \frac{1}{4} \) 2. **Calculate the \( 12^{\text{th}} \) term:** \[ a_{12} = 16 \cdot \left(\frac{1}{4}\right)^{12-1} = 16 \cdot \left(\frac{1}{4}\right)^{11} \] Now, let's simplify this expression: \[ a_{12} = 16 \cdot \frac{1}{4^{11}} = 16 \cdot \frac{1}{(2^2)^{11}} = 16 \cdot \frac{1}{2^{22}} = \frac{16}{2^{22}} = \frac{2^4}{2^{22}} = \frac{1}{2^{18}} \] Thus, the \( 12^{\text{th}} \) term is: \[ a_{12} = \frac{1}{2^{18}} \] #### 1.2 Calculate the sum of the first 10 terms of the sequence The sum of the first \( n \) terms of a geometric sequence is given by: \[ S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \] For the first 10 terms: - \( n = 10 \) - \( a_1 = 16 \) - \( r = \frac{1}{4} \) Now, substituting these values into the formula: \[ S_{10} = 16 \cdot \frac{1 - \left(\frac{1}{4}\right)^{10}}{1 - \frac{1}{4}} = 16 \cdot \frac{1 - \frac{1}{4^{10}}}{\frac{3}{4}} = 16 \cdot \frac{4}{3} \cdot \left(1 - \frac{1}{4^{10}}\right) \] Calculating \( 4^{10} \): \[ 4^{10} = (2^2)^{10} = 2^{20} \] Thus: \[ S_{10} = \frac{64}{3} \cdot \left(1 - \frac{1}{2^{20}}\right) \] ### Problem 2: Geometric Sequence \( 27, 9, 3, \ldots \) #### 2.1 Determine the formula for the \( n^{\text{th}} \) term of the sequence 1. **Identify the first term and common ratio:** - First term \( a_1 = 27 \) - Common ratio \( r = \frac{9}{27} = \frac{1}{3} \) 2. **Write the formula for the \( n^{\text{th}} \) term:** \[ a_n = 27 \cdot \left(\frac{1}{3}\right)^{n-1} \] #### 2.2 Why does the sum to infinity of this sequence exist? The sum to infinity of a geometric series exists if the absolute value of the common ratio \( r \) is less than 1: \[ |r| < 1 \] In this case, \( r = \frac{1}{3} \), which satisfies this condition. #### 1.3 Determine \( S_{\infty} \) The sum to infinity \( S_{\infty} \) of a geometric series is given by: \[ S_{\infty} = \frac{a_1}{1 - r} \] Substituting the values: \[ S_{\infty} = \frac{27}{1 - \frac{1}{3}} = \frac{27}{\frac{2}{3}} = 27 \cdot \frac{3}{2} = \frac{81}{2} \] ### Summary of Results 1. The \( 12^{\text{th}} \) term of the first sequence is \( \frac{1}{2^{18}} \). 2. The sum of the first 10 terms of the first sequence is \( \frac{64}{3} \cdot \left(1 - \frac{1}{2^{20}}\right) \). 3. The formula for the \( n^{\text{th}} \) term of the second sequence is \( 27 \cdot \left(\frac{1}{3}\right)^{n-1} \). 4. The sum to infinity \( S_{\infty} \) of the second sequence is \( \frac{81}{2} \).