# 7.7.2: Finding the nth Term Given Two Terms for a Geometric Sequence

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# Finding the nth Term Given the Common Ratio and any Term or Two Terms

A bacteria sample doubles every hour. After four hours there are 64 bacteria in the sample. What is the \(\ n^{t h}\) term rule for the geometric sequence represented by this situation?

# Finding the n^{th} Term

We will be using the general rule for the \(\ n^{t h}\) term in a geometric sequence and the given term(s) to determine the first term and write a general rule to find any other term.

Let's consider the geometric sequence in which the common ratio is \(\ -\frac{4}{5}\) and \(\ a_{5}=1280\). We'll find the first term in the sequence and write the general rule for the sequence.

We will start by using the term we know, the common ratio and the general rule, \(\ a_{n}=a_{1} r^{n-1}\).

By plugging in the values we know, we can then solve for the first term, \(\ a_{1}\).

\(\ \begin{aligned}

a_{5} &=a_{1}\left(-\frac{4}{5}\right)^{4} \\

1280 &=a_{1}\left(-\frac{4}{5}\right)^{4} \\

\frac{1280}{\left(-\frac{4}{5}\right)^{4}} &=a_{1} \\

3125 &=a_{1}

\end{aligned}\)

Now, the \(\ n^{t h}\) term rule is \(\ a_{n}=3125\left(-\frac{4}{5}\right)^{n-1}\).

Now, let's find the \(\ n^{t h}\) term rule for a sequence in which \(\ a_{1}=16\) and \(\ a_{7}=\frac{1}{4}\).

Since \(\ a_{7}=\frac{1}{4}\) and we know the first term, we can write the equation \(\ \frac{1}{4}=16 r^{6}\) and solve for the common ratio:

\(\ \begin{aligned}

\frac{1}{4} &=16 r^{6} \\

\frac{1}{64} &=r^{6} \\

\sqrt[6]{\frac{1}{64}} &=\sqrt[6]{r^{6}} \\

\frac{1}{2} &=r

\end{aligned}\)

The \(\ n^{t h}\) term rule is \(\ a_{n}=16\left(\frac{1}{2}\right)^{n-1}\).

Finally, let's find the \(\ n^{t h}\) term rule for the geometric sequence in which \(\ a_{5}=8\) and \(\ a_{10}=\frac{1}{4}\).

Using the same method at the previous problem, we can solve for \(\ r\) and \(\ a_{1}\). Then, write the general rule.

Equation 1: \(\ a_{5}=8\), so \(\ 8=a_{1} r^{4}\), solving for \(\ a_{1}\) we get \(\ a_{1}=\frac{8}{r^{4}}\).

Equation 2: \(\ a_{10}=\frac{1}{4}\), so \(\ \frac{1}{4}=a_{1} r^{9}\), solving for \(\ a_{1}\) we get \(\ a_{1}=\frac{\frac{1}{4}}{r^{9}}\).

\(\ \begin{aligned}

\frac{8}{r^{4}} &=\frac{\frac{1}{4}}{r^{9}} \\

8 r^{9} &=\frac{1}{4} r^{4} \\

\frac{8 r^{9}}{8 r^{4}} &=\frac{\frac{1}{4} r^{4}}{8 r^{4}} \\

r^{5} &=\frac{1}{32} \\

\sqrt[5]{r^{5}} &=\sqrt[5]{\frac{1}{32}} \\

r &=\frac{1}{2}

\end{aligned}\)

Thus, \(\ a_{1}=\frac{8}{\left(\frac{1}{2}\right)^{4}}=\frac{8}{\frac{1}{16}}=\frac{8}{1} \cdot \frac{16}{1}=128\).

The \(\ n^{t h}\) term rule is \(\ a_{n}=\left(\frac{3}{8}\right)(2)^{n-1}\).

Note

In solving the equation above for \(\ r\) we divided both sides by \(\ r^{4}\). In general it is not advisable to divide both sides of an equation by the variable because we may lose a possible solution, \(\ r=0\). However, in this case, \(\ r \neq 0\) since it is the common ratio in a geometric sequence.

# Examples

Example 1

Earlier, you were asked to find the \(\ n^{t h}\) term rule of the geometric sequence represented by a bacteria sample that doubles every hour.

**Solution**

We are given that \(\ a_{4}=64\) and because the sample doubles every hour we know that the common ratio is \(\ 2\). We can therefore plug the known values into the equation \(\ a_{n}=a_{1} r^{n-1}\) to get \(\ a_{1}\).

\(\ \begin{aligned}

a_{4}&=a_{1} r^{n-1} & \\

64 &=a_{1}(2)^{3} \\

64 &=8 a_{1} \\

8 &=a_{1}

\end{aligned}\)

Therefore, there are \(\ 8\) bacteria in the sample to begin with and the \(\ n^{t h}\) term rule is \(\ a_{n}=8 \cdot 2^{n-1}\).

Example 2

Find the first term and the \(\ n^{t h}\) term rule for the geometric sequence given that \(\ r=-\frac{1}{2}\) and \(\ a_{6}=3\).

**Solution**

Use the known quantities in the general form for the \(\ n^{t h}\) term rule to find \(\ a_{1}\).

\(\ \begin{aligned}

3 &=a_{1}\left(-\frac{1}{2}\right)^{5} \\

\left(-\frac{32}{1}\right) \cdot 3 &=a_{1}\left(-\frac{1}{32}\right) \cdot\left(-\frac{32}{1}\right) \\

a_{1} &=-96

\end{aligned}\)

Thus, \(\ a_{n}=-96\left(-\frac{1}{2}\right)^{n-1}\)

Example 3

Find the common ratio and the \(\ n^{t h}\) term rule for the geometric sequence given that \(\ a_{1}=-\frac{16}{625}\) and \(\ a_{6}=-\frac{5}{2}\).

**Solution**

Again, substitute in the known quantities to solve for \(\ r\).

\(\ \begin{aligned}

-\frac{5}{2} &=\left(-\frac{16}{625}\right) r^{5} \\

-\frac{5}{2}\left(-\frac{625}{16}\right) &=r^{5} \\

\frac{3125}{32} &=r^{5} \\

\sqrt[5]{\frac{3125}{32}} &=\sqrt[5]{r^{5}} \\

r &=\frac{5}{2}

\end{aligned}\)

So, \(\ a_{n}=-\frac{16}{625}\left(\frac{5}{2}\right)^{n-1}\)

Example 4

Find the \(\ n^{t h}\) term rule for the geometric sequence in which \(\ a_{5}=6\) and \(\ a_{13}=1536\).

**Solution**

This time we have two unknowns, the first term and the common ratio. We will need to solve a system of equations using both given terms.

Equation 1: \(\ a_{5}=6\), so \(\ 6=a_{1} r^{4}\), solving for \(\ a_{1}\) we get \(\ a_{1}=\frac{6}{r^{4}}\).

Equation 2: \(\ a_{13}=1536\), so \(\ 1536=a_{1} r^{12}\), solving for \(\ a_{1}\) we get \(\ a_{1}=\frac{1536}{r^{12}}\).

Now that both equations are solved for \(\ a_{1}\) we can set them equal to each other and solve for \(\ r\).

\(\ \begin{aligned}

\frac{6}{r^{4}} &=\frac{1536}{r^{12}} \\

6 r^{12} &=1536 r^{4} \\

\frac{6 r^{12}}{6 r^{4}} &=\frac{1536 r^{4}}{6 r^{4}} \\

r^{8} &=256 \\

\sqrt[8]{r^{8}} &=\sqrt[8]{256} \\

r &=2

\end{aligned}\)

Now use \(\ r\) to find \(\ a_{1}\): \(\ a_{1}=\frac{6}{\left(2^{4}\right)}=\frac{6}{16}=\frac{3}{8}\).

The \(\ n^{t h}\) term rule is \(\ a_{n}=\left(\frac{3}{8}\right)(2)^{n-1}\).

# Review

Use the given information to find the \(\ n^{t h}\) term rule for each geometric sequence.

- \(\ r=\frac{2}{3}\) and \(\ a_{8}=\frac{256}{81}\)
- \(\ r=-\frac{3}{4}\) and \(\ a_{5}=\frac{405}{8}\)
- \(\ r=\frac{6}{5}\) and \(\ a_{4}=3\)
- \(\ r=-\frac{1}{2}\) and \(\ a_{7}=5\)
- \(\ r=\frac{6}{7}\) and \(\ a_{0}=1\)
- \(\ a_{1}=\frac{11}{8}\) and \(\ a_{7}=88\)
- \(\ a_{1}=24\) and \(\ a_{4}=81\)
- \(\ a_{1}=48\) and \(\ a_{4}=\frac{3}{4}\)
- \(\ a_{1}=\frac{343}{216}\) and \(\ a_{5}=\frac{6}{7}\)
- \(\ a_{6}=486\) and \(\ a_{10}=39366\)
- \(\ a_{5}=648\) and \(\ a_{10}=\frac{19683}{4}\)
- \(\ a_{3}=\frac{2}{3}\) and \(\ a_{5}=\frac{3}{2}\)
- \(\ a_{5}=\frac{4}{3}\) and \(\ a_{10}=-\frac{128}{3}\)

Use a geometric sequence to solve the following word problems.

- Ricardo’s parents want to have $100,000 saved up to pay for college by the time Ricardo graduates from high school (16 years from now). If the investment vehicle they choose to invest in claims to yield 7% growth per year, how much should they invest today? Give your answer to the nearest one thousand dollars.
- If a piece of machinery depreciates (loses value) at a rate of 6% per year, what was its initial value if it is 10 years old and worth $50,000? Give your answer to the nearest one thousand dollars.

# Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.9.