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Find the common ratio of the geometric sequence $64,-16,4,-1, \ldots$ Then express each sequence in the form $a_{n}=a_{1} r^{n-1}$ and find the eighth term of the sequence.
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\begin{aligned} &\text { Since } r=\frac{a_{2}}{a_{1}}=\frac{-16}{64}=-\frac{1}{4} \\ &\text { Then } a_{n}=64\left(-\frac{1}{4}\right)^{n-1} \end{aligned}
Therefore, the eighth term of the sequence is
\begin{aligned} a_{8} &=64\left(-\frac{1}{4}\right)^{8-1} \\ &=64\left(-\frac{1}{4}\right)^{7} \\ &=-\frac{1}{256} \end{aligned}
by Diamond (58,513 points)

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