Asked by skittlez
The 5th term in a geometric system is 160. The 7th term is 40. What are the possible values of the 6th term in the sequence?
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Let the first term be $a$ and the common ratio be $r$. We have the following system of equations:
\begin{align*}
ar^4&=160 \\
ar^6&=40 \\
\end{align*}
Dividing the second equation by the first, we get $\frac{ar^6}{ar^4}=\frac{40}{160}=\frac{1}{4}$.
Simplifying the left-hand side, we get $r^2=\frac{1}{4}$, which means $r=\pm\frac{1}{2}$.
Substituting into the first equation, we get $a=\frac{160}{r^4}$, so $a=640$ when $r=\frac{1}{2}$ and $a=10$ when $r=-\frac{1}{2}$.
Therefore, the two possible values of the sixth term are $\frac{1}{2}(160)=\boxed{80}$ and $\frac{1}{2}(10)=\boxed{5}$.
\begin{align*}
ar^4&=160 \\
ar^6&=40 \\
\end{align*}
Dividing the second equation by the first, we get $\frac{ar^6}{ar^4}=\frac{40}{160}=\frac{1}{4}$.
Simplifying the left-hand side, we get $r^2=\frac{1}{4}$, which means $r=\pm\frac{1}{2}$.
Substituting into the first equation, we get $a=\frac{160}{r^4}$, so $a=640$ when $r=\frac{1}{2}$ and $a=10$ when $r=-\frac{1}{2}$.
Therefore, the two possible values of the sixth term are $\frac{1}{2}(160)=\boxed{80}$ and $\frac{1}{2}(10)=\boxed{5}$.
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