QUESTION 1 [17] 1.1 Evaluate: \[ \sum_{n=1}^{20} 3^{n-2} \] 1.2 The following sequence forms a convergent geometric sequence: \( 5 x ; x^{2} ; \frac{x^{3}}{5} ; \ldots \) 1.2.1 Determine the possible value of \( x \). (3) 1.2.2 If \( x=2 \), calculate \( S_{\infty} \). (2) 1.3 The following arithmetic sequence is given: \( 20 ; 23 ; 26 ; 29 ; \ldots ; 101 \) 1.3.1 How many terms are there in this sequence? (2) 1.3.2 The even numbers are removed from the sequence. Calculate the sum of the terms of the remaining sequence; \( \qquad \) .... \( \qquad \) \( \qquad \)

To evaluate the sum \(\sum_{n=1}^{20} 3^{n-2}\), we can rewrite it as \(\sum_{n=0}^{18} 3^n\) since the starting point is shifted. This creates a geometric series with a first term \(a = 1\) (when \(n = 0\)) and a common ratio \(r = 3\). The sum formula for a geometric series is \(S_n = a \frac{1 - r^n}{1 - r}\). Thus, we find: \[ S_{19} = 1 \cdot \frac{1 - 3^{19}}{1 - 3} = \frac{1 - 3^{19}}{-2} = \frac{3^{19} - 1}{2} \quad \text{(Final Value)} \] For the convergent geometric sequence, which is \(5x; x^2; \frac{x^3}{5}; \ldots\), we need the ratio between its successive terms to be less than 1 for convergence. The ratio \(r = \frac{x^2}{5x} = \frac{x}{5}\) and \(r = \frac{\frac{x^3}{5}}{x^2} = \frac{x}{5}\) gives us \( |r| < 1 \), thus \( |x| < 5 \). So, \( x \) can range anywhere in \((-5, 5)\). If we let \(x = 2\), the sum of the infinite series \(S_{\infty}\) for a geometric series is given by \(S_{\infty} = \frac{a}{1 - r}\). Here \(a = 5(2) = 10\) and \(r = \frac{2}{5}\), so: \[ S_{\infty} = \frac{10}{1 - \frac{2}{5}} = \frac{10}{\frac{3}{5}} = \frac{10 \cdot 5}{3} = \frac{50}{3} \quad \text{(Final Value)} \] Next, for the arithmetic sequence \(20; 23; 26; 29; \ldots; 101\), we find the total number of terms. This sequence's \(n\)-th term can be expressed as \(a_n = 20 + (n-1)3\). Setting \(a_n = 101\): \[ 101 = 20 + (n-1)3 \implies 81 = (n-1)3 \implies n - 1 = 27 \implies n = 28 \quad \text{(Total Terms)} \] Finally, we remove the even numbers from this sequence: \(20, 26, 32, \ldots\) all the way up to \(100\), giving us the odd terms \(23, 29, \ldots, 101\). The remaining sequence is thus an arithmetic sequence starting at \(23\) and ending at \(101\) with a common difference of \(6\) (it skips the evens). The last odd term can be framed similarly to the previous count: - There are \(\{23, 29, \ldots, 101\}\) with: \[ a_n = 23 + (n-1)6 = 101 \implies n = \frac{101 - 23}{6} + 1 = \frac{78}{6} + 1 = 13 + 1 = 14 \quad \text{(Total Odd Terms)} \] The sum of an arithmetic series \(S_n = \frac{n}{2}(first term + last term)\) gives us: \[ S = \frac{14}{2}(23 + 101) = 7 \times 124 = 868 \quad \text{(Final Odd Terms Sum)} \]