(e) Consider the geometric sequence \( -320 ; 160 ;-80 ; \ldots ;-\frac{5}{64} \). (1) Calculate \( T_{8} \). (2) Determine the value of \( k \) if \( T_{k}=-\frac{5}{16} \). (3) How many terms are there in this sequence?

To solve the geometric sequence \( -320, 160, -80, \ldots, -\frac{5}{64} \), we first need to identify the first term and the common ratio. 1. **Calculate \( T_{8} \)**: The first term \( a = -320 \) and the common ratio \( r \) can be calculated as: \[ r = \frac{160}{-320} = -\frac{1}{2} \] The formula for the \( n \)-th term of a geometric sequence is given by: \[ T_n = a \cdot r^{n-1} \] Therefore, for \( T_8 \): \[ T_8 = -320 \cdot \left(-\frac{1}{2}\right)^{8-1} = -320 \cdot \left(-\frac{1}{2}\right)^7 = -320 \cdot -\frac{1}{128} = \frac{320}{128} = 2.5 \] 2. **Determine the value of \( k \) if \( T_{k}=-\frac{5}{16} \)**: We need to set up the equation: \[ T_k = a \cdot r^{k-1} = -320 \cdot \left(-\frac{1}{2}\right)^{k-1} = -\frac{5}{16} \] Dividing both sides by -1 gives: \[ 320 \cdot \left(\frac{1}{2}\right)^{k-1} = \frac{5}{16} \] Simplifying gives: \[ \left(\frac{1}{2}\right)^{k-1} = \frac{5}{16 \cdot 320} = \frac{5}{5120} \] Now, recognizing that \( 5120 = 2^8 \cdot 5 \) (since \( 5120 = 256 \times 20 = 2^8 \cdot 2^2 \cdot 5 \)), we have: \[ \left(\frac{1}{2}\right)^{k-1} = \frac{5}{2^9} \] The powers of \( 2 \) imply: \[ -k + 1 = -9 \text{ (implying we get to the } 9^{th} \text{ power with \( k-1=9 \) )} \] Thus, \( k = 10 \). 3. **How many terms are in this sequence?**: To find the total number of terms, we will set \( T_n = -\frac{5}{64} \) in our formula: \[ T_n = -320 \cdot \left(-\frac{1}{2}\right)^{n-1} = -\frac{5}{64} \] Dividing both sides by -1 gives: \[ 320 \cdot \left(\frac{1}{2}\right)^{n-1} = \frac{5}{64} \] Simplifying yields: \[ \left(\frac{1}{2}\right)^{n-1} = \frac{5}{20480} \] Since \( 20480 = 2^{11} \cdot 5 \) leads us to deduce: \[ n-1 = 11 \text{, thus } n = 12. \] Therefore, there are a total of **12 terms** in the sequence.