In a geometric sequence, the terms are found by multiplying the previous term by a constant factor called the common ratio. Given the fifth through the eighth terms:
- \( a_5 = 80 \)
- \( a_6 = 160 \)
- \( a_7 = 320 \)
- \( a_8 = 640 \)
We can see the common ratio \( r \) can be calculated from the consecutive terms:
\[ r = \frac{a_6}{a_5} = \frac{160}{80} = 2 \] \[ r = \frac{a_7}{a_6} = \frac{320}{160} = 2 \] \[ r = \frac{a_8}{a_7} = \frac{640}{320} = 2 \]
Since the common ratio \( r = 2 \), the sequence can be described by the formula for geometric sequences:
\[ a_n = a_1 \cdot r^{n-1} \]
Where \( a_1 \) is the first term of the sequence. However, since we only know the fifth term (and subsequent terms), we can express the \( n \)-th term in relation to the fifth term.
Using the relationship: \[ a_n = a_5 \cdot r^{n-5} \]
Substituting the known values: \[ a_n = 80 \cdot 2^{n-5} \]
To match the structure of the options given, we can rewrite \( 2^{n-5} \) as \( \frac{2^n}{2^5} = \frac{2^n}{32} \).
Thus: \[ a_n = 80 \cdot \frac{2^n}{32} = 2.5 \cdot 2^n \]
However, to match with our options and perceptually see the specific multipliers: \[ a_n = 80 \cdot 2^{n-5} = 80 \cdot 2^{n-1} \cdot \frac{1}{16} = 80 \cdot 2^{n-5} = 80 \times 2^{n-1} \] This option best reflects the geometric sequence nature from the fifth term provided all computations hold as valid; thus:
The appropriate answer from the options provided is:
D. \( a_n = 80 \times 2^{n-5} \)
(Note: The choices may have implied different indices; it's best to check the refinements over wording for assurance of correctness, but naming choices delineate \( n-5 \) as correctly stated).