(3) How many terms are there in 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 determine how many terms there are in the geometric sequence \( -320, 160, -80, \ldots, -\frac{5}{64} \), we first need to find the common ratio \( r \). The ratio is found by dividing the second term by the first term: \[ r = \frac{160}{-320} = -\frac{1}{2} \] Now we can express the \( n \)-th term of the geometric sequence using the formula: \[ T_n = T_1 \cdot r^{n-1} \] Here, \( T_1 = -320 \). We set \( T_n = -\frac{5}{64} \) and solve for \( n \): \[ -\frac{5}{64} = -320 \cdot \left(-\frac{1}{2}\right)^{n-1} \] This simplifies to: \[ \frac{5}{64} = 320 \cdot \left(\frac{1}{2}\right)^{n-1} \] Now we simplify \( 320 \): \[ 320 = \frac{5}{64} \cdot 2^{n-1} \] Converting \( 320 \) into powers of \( 2 \): \[ 320 = 5 \cdot 64 = 5 \cdot 2^6 \quad \implies \quad \frac{5}{64} = \frac{5 \cdot 2^6}{2^{n-1}} \quad \implies \quad 2^{n-1} = 2^6 \cdot 64 \] Equating the powers of 2 gives: \[ n - 1 = 6 \quad \implies \quad n = 7 \] So, there are 7 terms in the sequence. For the specific calculations requested: (1) To calculate \( T_8 \): We use the formula again: \[ T_8 = -320 \cdot \left(-\frac{1}{2}\right)^{8-1} = -320 \cdot \left(-\frac{1}{2}\right)^{7} = -320 \cdot -\frac{1}{128} = \frac{320}{128} = \frac{5}{2} = 2.5 \] (2) To find \( k \) for which \( T_k = -\frac{5}{16} \): \[ -\frac{5}{16} = -320 \cdot \left(-\frac{1}{2}\right)^{k-1} \quad \implies \quad \frac{5}{16} = 320 \cdot \left(\frac{1}{2}\right)^{k-1} \] As before, solving yields: \[ \frac{5 \cdot 2^{k-1}}{16} = 5 \cdot 2^6 \quad \implies \quad 2^{k-1} = 2^6 \cdot 2^4 \quad \implies \quad k - 1 = 10 \quad \implies \quad k = 11 \] So, there are 7 terms in the sequence.