QUESTION 2 Given the geometric sequence: \( -\frac{1}{4} ; b ;-1 ; \ldots \ldots \) \( 2.1 \quad \) Calculate the possible values of \( b \). \( 2.2 \quad \) If \( b=\frac{1}{2} \), calculate the \( 19^{\text {th }} \) term \( \left(T_{19}\right) \) of the sequence. \( 2.3 \quad \begin{array}{l}\text { If } b=\frac{1}{2}, \text { write the sum of the first } 20 \text { positive terms of the sequence in sigma } \\ \text { notation. } \\ 2.4 \quad \begin{array}{l}\text { Is the geometric series formed in QUESTION } 2.3 \text { convergent? Give reasons for your } \\ \text { answer. }\end{array}\end{array} \) [12)

**2.1 Calculation of \( b \):** Let the first term be \[ a = -\frac{1}{4} \] and the common ratio be \( r \). In a geometric sequence, the second term is \[ b = a \cdot r \quad \Rightarrow \quad b = -\frac{1}{4} \cdot r. \] The third term is given by \[ -1 = b \cdot r = \left(-\frac{1}{4} \cdot r\right) \cdot r = -\frac{1}{4} \cdot r^2. \] Solve for \( r \): \[ -\frac{1}{4}\,r^2 = -1 \quad \Rightarrow \quad r^2 = 4 \quad \Rightarrow \quad r = 2 \quad \text{or} \quad r = -2. \] Now, substitute back to find \( b \): - If \( r = 2 \): \[ b = -\frac{1}{4} \cdot 2 = -\frac{1}{2}. \] - If \( r = -2 \): \[ b = -\frac{1}{4} \cdot (-2) = \frac{1}{2}. \] Thus, the possible values of \( b \) are \[ b = -\frac{1}{2} \quad \text{or} \quad b = \frac{1}{2}. \] --- **2.2 Calculation of the \( 19^\text{th} \) Term when \( b = \frac{1}{2} \):** If \( b = \frac{1}{2} \), then from \[ b = -\frac{1}{4}\,r, \] we have \[ r = \frac{b}{a} = \frac{\frac{1}{2}}{-\frac{1}{4}} = -2. \] The \( n^\text{th} \) term of a geometric sequence is given by \[ T_n = a \cdot r^{\,n-1}. \] Thus, the \( 19^\text{th} \) term is \[ T_{19} = -\frac{1}{4} \cdot (-2)^{18}. \] Since \( (-2)^{18} = 2^{18} \) (because the exponent is even), we have \[ T_{19} = -\frac{1}{4} \cdot 2^{18}. \] Noting that \( 2^{18} = 262144 \), \[ T_{19} = -\frac{262144}{4} = -65536. \] --- **2.3 Sigma Notation for the Sum of the First 20 Positive Terms (with \( b = \frac{1}{2} \)):** For \( b = \frac{1}{2} \), the sequence is given by \[ T_n = -\frac{1}{4}\,(-2)^{\,n-1}. \] Examine the sign pattern: - \( T_1 = -\frac{1}{4} \) (negative), - \( T_2 = \frac{1}{2} \) (positive), - \( T_3 = -1 \) (negative), - \( T_4 = 2 \) (positive), and so on. Thus, the positive terms occur at even-indexed positions: \( T_2, T_4, T_6, \ldots \) For even-indexed term \( T_{2k} \): \[ T_{2k} = -\frac{1}{4}\,(-2)^{2k-1}. \] Simplify \( (-2)^{2k-1} \) by writing it as: \[ (-2)^{2k-1} = \frac{(-2)^{2k}}{-2} = \frac{4^k}{-2}. \] Then, \[ T_{2k} = -\frac{1}{4} \cdot \frac{4^k}{-2} = \frac{4^k}{8}. \] Thus, the first positive term is when \( k=1 \) (giving \( T_2=\frac{4^1}{8}=\frac{4}{8}=\frac{1}{2} \)). The sum of the first 20 positive terms (i.e., \( T_2, T_4, \ldots, T_{40} \)) can be written in sigma notation as: \[ \sum_{k=1}^{20} \frac{4^k}{8}. \] Alternatively, factor out \( \frac{1}{8} \): \[ \frac{1}{8} \sum_{k=1}^{20} 4^k. \] --- **2.4 Convergence of the Geometric Series in Question 2.3:** The series in 2.3 is the infinite series formed by the positive terms of the sequence if we were to extend beyond the first 20 terms. The general term of the positive subsequence is \[ \frac{4^k}{8}, \] with common ratio \( 4 \) (since \( 4^{k+1} = 4 \cdot 4^k \)). A geometric series converges if and only if the absolute value of the common ratio is less than 1: \[ |r| < 1. \] Here, \( |4| = 4 \), which is larger than 1. Therefore, if we consider the infinite geometric series formed by the positive terms, it is divergent. Thus, the geometric series in Question 2.3 is not convergent because its common ratio \( 4 \) satisfies \( |4| > 1 \).