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Check my answer? General formula for nth term?
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I got D
\[\begin{align} \sf\text{geometric series} : a_n = a_1r^{n-1} \\ &: ~ a_5 = a_1r^{n-1} \\ &: \frac{1}{8} = a_1 \left(\frac{1}{2}\right)^{5-1} \\ &: \frac{1}{8} = a_1\left(\frac{1}{2}\right)^4 \\ &: \frac{1}{8} =a_1\left(\frac{1}{16}\right) \\ &: \frac{1}{8} \cdot 16=a_1 \\ &:~ a_1 = 2 \end{align}\] Plug your value in to the general format for a geometric series with:\[a_1 = 2~,~ r=\frac{1}{2}\] And you'll have your answer,
Oh so its A then? @jhannybean
yes ;)
as it is as a form \( a_n = a_1r^{n-1} \)
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thank you so much
ur welcome ;) but pay close attention to the solution
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