Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.
[latex]{a}_{n}={a}_{1}{r}^{n - 1}[/latex]
Let’s write the first few terms of the sequence where the first term is [latex]a_1[/latex] and the common ratio is[latex]r[/latex] to see this pattern.
explicit formula for a geometric sequence
The explicit formula of a geometric sequence with first term [latex]a_1[/latex] and the common ratio [latex]r[/latex] is
[latex]{a}_{n}={a}_{1}{r}^{n - 1}[/latex]
Let’s take a look at the sequence [latex]\left\{18\text{, }36\text{, }72\text{, }144\text{, }288\text{, }...\right\}[/latex].
[latex]\\[/latex]
This is a geometric sequence with a common ratio of [latex]2[/latex] and an exponential function with a base of [latex]2[/latex]. An explicit formula for this sequence is
[latex]{a}_{n}=18\cdot {2}^{n - 1}[/latex]
Given a geometric sequence with [latex]{a}_{1}=3[/latex] and [latex]{a}_{4}=24[/latex], find [latex]{a}_{2}[/latex].
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},\dots[/latex]
Find the common ratio using the given fourth term.
[latex]\begin{align}&{a}_{n}={a}_{1}{r}^{n - 1} \\ &{a}_{4}=3{r}^{3} && \text{Write the fourth term of sequence in terms of }{a}_{1}\text{ and }r \\ &24=3{r}^{3} && \text{Substitute }24\text{ for }{a}_{4} \\ &8={r}^{3} && \text{Divide} \\ &r=2 && \text{Solve for the common ratio} \end{align}[/latex]
Find the second term by multiplying the first term by the common ratio.
The graph of this sequence shows an exponential pattern.
Find the ninth term of the sequence [latex]6, 18, 54, 162, 486, 1458, \dots[/latex] Then find the general term for the sequence.
[latex]\begin{align*} \text{Let's first determine } a_1 \text{ and the common ratio } r: & \\ \text{The first term is } 6, \text{ so } a_1 &= 6 & \\ \text{The ratio is: } \frac{18}{6} &= \frac{54}{18} = \frac{162}{54} = \frac{486}{162} = \frac{1458}{486} = 3 & r &= 3 \end{align*}[/latex][latex]\begin{align*} \text{To find the 9th term, use the formula with } a_1 = 6, \, r = 3, \text{ and } n=9: & \\ \text{Substitute these values and simplify:} & & \\ a_n &= a_1 r^{n-1} & \\ a_9 &= 6(3)^{9-1} & \\ a_9 &= 6(3)^8 & \\ a_9 &= 39366 & \end{align*}[/latex][latex]\begin{align*} \text{To find the general term, substitute } a_1 = 6 \text{ and } r = 3 \text{ into the formula:} & \\ a_n &= a_1 r^{n-1} & \\ a_n &= 6(3)^{n-1} & \end{align*}[/latex]
