27)
The given sequence is,
160, -80, 40, -20.
The first term of the sequence is a=160.
The common ratio of a sequence can be found by dividing a term by the previous term.
Hence, the common ratio is,
[tex]\begin{gathered} r=\frac{-80}{160}=\frac{-1}{2} \\ r=\frac{40}{-80}=\frac{-1}{2} \\ r=\frac{-20}{40}=\frac{-1}{2} \end{gathered}[/tex]
So, the common ratio is -1/2.
The n th term of a geometric sequence can be expressed as,
[tex]a_n=ar^{n-1}[/tex]
We have to find the 5th, 6th and 7th term of the given sequence.
[tex]\begin{gathered} a_5=160\times(\frac{-1}{2})^{5-1}=160\times(\frac{-1}{2})^4=10 \\ a_6=160\times(\frac{-1}{2})^{6-1}=160\times(\frac{-1}{2})^5=-5 \\ a_7=160\times(\frac{-1}{2})^{7-1}=160\times(\frac{-1}{2})^6=2.5 \end{gathered}[/tex]
Therefore, the next three terms of the sequence 160, -80, 40, -20 is 10, -5, 2.5.
Hence, option a is correct.
