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The concentration of a reactant in a chemical reaction follows a geometric sequence. At time \(0 \)seconds, the concentration is \(10\) mM. One second later, the concentration is \(5\) mM. Another second later, the concentration is \(2.5\) mM. How many seconds does it take for the concentration to reach \(0.156\thinspace 25\) mM?
The sequence is \(10, 5, 2.5, ..., 0. 156\thinspace 25\).
\(t_1 = 10 \)
\(r = 0.5 \) or \(\frac{1}{2} \)
Using the general term, solve for \(n \).
\(t_1 = 10 \)
\(r = 0.5 \) or \(\frac{1}{2} \)
Using the general term, solve for \(n \).
\(\begin{align}
t_n &= t_1 r^{n - 1} \\
0.156\thinspace 25 &= 10\left( {\frac{1}{2}} \right)^{n - 1} \\
0.0156\thinspace 25 &= \left( {\frac{1}{2}} \right)^{n - 1} \\
\frac{1}{{64}} &= \left( {\frac{1}{2}} \right)^{n - 1} \\
\left( {\frac{1}{2}} \right)^6 &= \left( {\frac{1}{2}} \right)^{n - 1} \\
6 &= n - 1 \\
7 &= n \\
\end{align}\)
t_n &= t_1 r^{n - 1} \\
0.156\thinspace 25 &= 10\left( {\frac{1}{2}} \right)^{n - 1} \\
0.0156\thinspace 25 &= \left( {\frac{1}{2}} \right)^{n - 1} \\
\frac{1}{{64}} &= \left( {\frac{1}{2}} \right)^{n - 1} \\
\left( {\frac{1}{2}} \right)^6 &= \left( {\frac{1}{2}} \right)^{n - 1} \\
6 &= n - 1 \\
7 &= n \\
\end{align}\)
Using \(r = 0.5\), you will arrive at the same answer.
\(\begin{align}
t_n &= t_1 r^{n - 1} \\
0.156\thinspace 25 &= 10\left( {0.5} \right)^{n - 1} \\
0.0156\thinspace 25 &= \left( {0.5} \right)^{n - 1} \\
\left( {0.5} \right)^6 &= \left( {0.5} \right)^{n - 1} \\
6 &= n - 1 \\
7 &= n \\
\end{align}\)
\(\begin{align}
t_n &= t_1 r^{n - 1} \\
0.156\thinspace 25 &= 10\left( {0.5} \right)^{n - 1} \\
0.0156\thinspace 25 &= \left( {0.5} \right)^{n - 1} \\
\left( {0.5} \right)^6 &= \left( {0.5} \right)^{n - 1} \\
6 &= n - 1 \\
7 &= n \\
\end{align}\)
Because term 1 is at 0 seconds, term 7 is at 6 seconds. It will take 6 seconds for the concentration to reach \(0.156\thinspace 25\) mM.
| For more examples on geometric sequences, please read through pp. 32 - 39 in Pre-Calculus 11. |