Example: Using Recursive Formulas for Geometric Sequences

Write a recursive formula for the following geometric sequence.

[latex]\left\{6\text{, }9\text{, }13.5\text{, }20.25\text{, }...\right\}[/latex]

Answer: The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

[latex]r=\frac{9}{6}=1.5[/latex]

Substitute the common ratio into the recursive formula for geometric sequences and define [latex]{a}_{1}[/latex].

[latex]\begin{array}{l}{a}_{n}=r{a}_{n - 1}\\ {a}_{n}=1.5{a}_{n - 1}\text{ for }n\ge 2\\ {a}_{1}=6\end{array}[/latex]

Analysis of the Solution

The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function.

Example: Writing Terms of Geometric Sequences Using the Explicit Formula

Given a geometric sequence with [latex]{a}_{1}=3[/latex] and [latex]{a}_{4}=24[/latex], find [latex]{a}_{2}[/latex].

Answer: The sequence can be written in terms of the initial term and the common ratio [latex]r[/latex].

[latex]3,3r,3{r}^{2},3{r}^{3},..[/latex].

Find the common ratio using the given fourth term.

[latex]\begin{array}{ll}{a}_{n}={a}_{1}{r}^{n - 1}\hfill & \hfill \\ {a}_{4}=3{r}^{3}\hfill & \text{Write the fourth term of sequence in terms of }{\alpha }_{1}\text{and }r\hfill \\ 24=3{r}^{3}\hfill & \text{Substitute }24\text{ for}{a}_{4}\hfill \\ 8={r}^{3}\hfill & \text{Divide}\hfill \\ r=2\hfill & \text{Solve for the common ratio}\hfill \end{array}[/latex]

Find the second term by multiplying the first term by the common ratio.

[latex]\begin{array}{ll}{a}_{2}\hfill & =2{a}_{1}\hfill \\ \hfill & =2\left(3\right)\hfill \\ \hfill & =6\hfill \end{array}[/latex]

Analysis of the Solution

The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.

Example: Solving Application Problems with Geometric Sequences

In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.

1. Write a formula for the student population.
2. Estimate the student population in 2020.

Answer:

1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.Let [latex]P[/latex] be the student population and [latex]n[/latex] be the number of years after 2013. Using the explicit formula for a geometric sequence we get [latex]{P}_{n} =284\cdot {1.04}^{n}[/latex]
2. We can find the number of years since 2013 by subtracting. [latex-display]2020 - 2013=7[/latex-display] We are looking for the population after 7 years. We can substitute 7 for [latex]n[/latex] to estimate the population in 2020. [latex-display]{P}_{7}=284\cdot {1.04}^{7}\approx 374[/latex-display] The student population will be about 374 in 2020.