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Geometric Sequences
Find the common ratio for a geometric sequence.
Geometric Sequences
- Determine if a sequence is geometric (IA 12.3.1).
- Find the general term (nth term) of a geometric sequence (IA 12.3.2).
A sequence is called a geometric sequence if the ratio between consecutive terms is always the same.
The ratio between consecutive terms in a geometric sequence is r, the common ratio, where n is greater than or equal to two.
\[r=\frac{{a}_{n}}{{a}_{n-1}}\]Example
Try it.
Determine if each sequence is geometric. If so, indicate the common ratio.
ⓐ \(4,8,16,32,64,128\text{,}\ \ldots\)
ⓑ \(-2,6,-12,36,-72,216\text{,}\ \ldots\)
Solution
To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.
ⓐ
\(\begin{array}{llllllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}4, & & 8, & & 16, & & 32, & & 64, & & 128,\ldots \\ \\ \\ \frac{8}{4} & & \frac{16}{8} & & \frac{32}{16} & & \frac{64}{32} & & \frac{128}{64} & & \\ 2 & & 2 & & 2 & & 2 & & 2 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is geometric. The common ratio is}\ r=2.\end{array}\)
ⓑ
\(\begin{array}{llllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}-2, & & 6, & & -12, & & 36, & & -72, & & 216,\ldots \\ \\ \\ \frac{6}{-2} & & \frac{-12}{6} & & \frac{36}{-12} & & \frac{-72}{36} & & \frac{216}{-72} \\ -3 & & -2 & & -3 & & -2 & & -3 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is not geometric. There is no common ratio.}\end{array}\)
Determine if each sequence is geometric. If so, indicate the common ratio.
Try it.
\(-150\), \(-30\), \(-15\), \(-5\), \(-\frac{5}{2}\), …
| Find the ratio of consecutive terms. | |
Try it.
8, 4, 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\), …
| Find the ratio of consecutive terms. | |
Example
Try it.
Write the first five terms of the sequence where the first term is 3 and the common ratio is \(r=-2.\)
Solution
We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.
\[\begin{array}{lllllllllllll}{a}_{1} & & & {a}_{2} & & & {a}_{3} & & & {a}_{4} & & & {a}_{5} \\ 3 & & & 3\cdot (-2) & & & -6\cdot (-2) & & & 12\cdot (-2) & & & -24\cdot (-2) \\ & & & -6 & & & 12 & & & -24 & & & 48\end{array}\]The sequence is \(3,-6,12,-24,48\text{,}\ \ldots\)
Try it.
Write the first five terms of the sequence where the first term is 7 and the common ratio is \(r=-3\) .
| \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| 7 | ||||
| The sequence is: ________________________________________ |
Condensed: the full section is in OpenStax College Algebra 2e.
Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
Example
Try it.
Is the sequence geometric? If so, find the common ratio.
- ⓐ \(1\text{,}\ 2\text{,}\ 4\text{,}\ 8\text{,}\ 16\text{,}\ ...\)
- ⓑ \(48\text{,}\ 12\text{,}\ 4\text{, }2\text{,}\ ...\)
Solution
Divide each term by the previous term to determine whether a common ratio exists.
- ⓐ \(\begin{array}{llllllllll}\frac{2}{1}=2 & & & \frac{4}{2}=2 & & & \frac{8}{4}=2 & & & \frac{16}{8}=2\end{array}\)
The sequence is geometric because there is a common ratio. The common ratio is 2.
- ⓑ \(\begin{array}{lllllll}\frac{12}{48}=\frac{1}{4} & & & \frac{4}{12}=\frac{1}{3} & & & \frac{2}{4}=\frac{1}{2}\end{array}\)
The sequence is not geometric because there is not a common ratio.
Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is \({a}_{1}=-2\) and the common ratio is \(r=4,\) we can find subsequent terms by multiplying \(-2⋅4\) to get \(-8\) then multiplying the result \(-8⋅4\) to get \(-32\) and so on.
\[\begin{array}{l}{a}_{1}=-2 \\ {a}_{2}=(-2⋅4)=-8 \\ {a}_{3}=(-8⋅4)=-32 \\ {a}_{4}=(-32⋅4)=-128\end{array}\]The first four terms are \(\{-2\text{, }-8\text{, }-32\text{, }-128\}.\)
Example
Try it.
List the first four terms of the geometric sequence with \({a}_{1}=5\) and \(r=-2.\)
Solution
Multiply \({a}_{1}\) by \(-2\) to find \({a}_{2}.\) Repeat the process, using \({a}_{2}\) to find \({a}_{3},\) and so on.
\[\begin{array}{l}{a}_{1}=5 \\ {a}_{2}=-2{a}_{1}=-10 \\ {a}_{3}=-2{a}_{2}=20 \\ {a}_{4}=-2{a}_{3}=-40\end{array}\]The first four terms are \(\{5,-10,20,-40\}.\)
Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
Example
Try it.
Write a recursive formula for the following geometric sequence.
\[\{6\text{, }9\text{, }13.5\text{, }20.25\text{, }...\}\]Solution
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
\[r=\frac{9}{6}=1.5\]Substitute the common ratio into the recursive formula for geometric sequences and define \({a}_{1}.\)
\[\begin{array}{l}{a}_{n}=r{a}_{n-1} \\ {a}_{n}=1.5{a}_{n-1}\ \text{for }n\ge 2 \\ {a}_{1}=6\end{array}\]Using Explicit Formulas for Geometric Sequences
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.
\[{a}_{n}={a}_{1}{r}^{n-1}\]Let’s take a look at the sequence \(\{18\text{, }36\text{, }72\text{, }144\text{, }288\text{, }...\}.\) This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
\[{a}_{n}=18\cdot {2}^{n-1}\]The graph of the sequence is
Example
Try it.
Given a geometric sequence with \({a}_{1}=3\) and \({a}_{4}=24,\) find \({a}_{2}.\)
Solution
The sequence can be written in terms of the initial term and the common ratio \(r.\)
\[3,3r,3{r}^{2},3{r}^{3},...\]Find the common ratio using the given fourth term.
\[\begin{array}{ll}{a}_{n}={a}_{1}{r}^{n-1} & \\ {a}_{4}=3{r}^{3} & \text{Write the fourth term of sequence in terms of }{\alpha }_{1}\ \text{and }r \\ 24=3{r}^{3} & \text{Substitute }24\ \text{for}\ {a}_{4} \\ \ 8={r}^{3} & \text{Divide} \\ \ r=2 & \text{Solve for the common ratio}\end{array}\]Find the second term by multiplying the first term by the common ratio.
\[\begin{array}{ll}{a}_{2} & =2{a}_{1} \\ & =2(3) \\ & =6\end{array}\]Example
Try it.
Write an explicit formula for the \(n\text{th}\) term of the following geometric sequence.
\[\{2\text{, }10\text{, }50\text{, }250\text{, }...\}\]Solution
The first term is 2. The common ratio can be found by dividing the second term by the first term.
\[\frac{10}{2}=5\]The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
\[\begin{array}{l}{a}_{n}={a}_{1}{r}^{(n-1)} \\ {a}_{n}=2⋅{5}^{n-1}\end{array}\]The graph of this sequence in shows an exponential pattern.
Solving Application Problems with Geometric Sequences
In real-world scenarios involving geometric sequences, we may need to use an initial term of \({a}_{0}\) instead of \({a}_{1}.\) In these problems, we can alter the explicit formula slightly by using the following formula:
\[{a}_{n}={a}_{0}{r}^{n}\]Example
Try it.
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.
- ⓐWrite a formula for the student population.
- ⓑEstimate the student population in 2020.
Solution
- ⓐ
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 \(P\) be the student population and \(n\) be the number of years after 2013. Using the explicit formula for a geometric sequence we get
\[{P}_{n}=284⋅{1.04}^{n}\] - ⓑ
We can find the number of years since 2013 by subtracting.
\[2020-2013=7\]We are looking for the population after 7 years. We can substitute 7 for \(n\) to estimate the population in 2020.
\[{P}_{7}=284⋅{1.04}^{7}\approx 374\]The student population will be about 374 in 2020.
Key Equations
| recursive formula for \(nth\) term of a geometric sequence | \({a}_{n}=r{a}_{n-1},n\ge 2\) |
| explicit formula for \(nth\) term of a geometric sequence | \({a}_{n}={a}_{1}{r}^{n-1}\) |
Key Concepts
- A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
- The constant ratio between two consecutive terms is called the common ratio.
- The common ratio can be found by dividing any term in the sequence by the previous term.
- The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly.
- A recursive formula for a geometric sequence with common ratio \(r\) is given by \({a}_{n}=r{a}_{n-1}\) for \(n\ge 2\) .
- As with any recursive formula, the initial term of the sequence must be given.
- An explicit formula for a geometric sequence with common ratio \(r\) is given by \({a}_{n}={a}_{1}{r}^{n-1}.\)
- In application problems, we sometimes alter the explicit formula slightly to \({a}_{n}={a}_{0}{r}^{n}.\)
Geometric Sequences
- Determine if a sequence is geometric (IA 12.3.1).
- Find the general term (nth term) of a geometric sequence (IA 12.3.2).
A sequence is called a geometric sequence if the ratio between consecutive terms is always the same.
The ratio between consecutive terms in a geometric sequence is r, the common ratio, where n is greater than or equal to two.
\[r=\frac{{a}_{n}}{{a}_{n-1}}\]Example
Try it.
Determine if each sequence is geometric. If so, indicate the common ratio.
ⓐ \(4,8,16,32,64,128\text{,}\ \ldots\)
ⓑ \(-2,6,-12,36,-72,216\text{,}\ \ldots\)
Solution
To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.
ⓐ
\(\begin{array}{llllllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}4, & & 8, & & 16, & & 32, & & 64, & & 128,\ldots \\ \\ \\ \frac{8}{4} & & \frac{16}{8} & & \frac{32}{16} & & \frac{64}{32} & & \frac{128}{64} & & \\ 2 & & 2 & & 2 & & 2 & & 2 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is geometric. The common ratio is}\ r=2.\end{array}\)
ⓑ
\(\begin{array}{llllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}-2, & & 6, & & -12, & & 36, & & -72, & & 216,\ldots \\ \\ \\ \frac{6}{-2} & & \frac{-12}{6} & & \frac{36}{-12} & & \frac{-72}{36} & & \frac{216}{-72} \\ -3 & & -2 & & -3 & & -2 & & -3 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is not geometric. There is no common ratio.}\end{array}\)
Determine if each sequence is geometric. If so, indicate the common ratio.
Try it.
\(-150\), \(-30\), \(-15\), \(-5\), \(-\frac{5}{2}\), …
| Find the ratio of consecutive terms. | |
Try it.
8, 4, 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\), …
| Find the ratio of consecutive terms. | |
Example
Try it.
Write the first five terms of the sequence where the first term is 3 and the common ratio is \(r=-2.\)
Solution
We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.
\[\begin{array}{lllllllllllll}{a}_{1} & & & {a}_{2} & & & {a}_{3} & & & {a}_{4} & & & {a}_{5} \\ 3 & & & 3\cdot (-2) & & & -6\cdot (-2) & & & 12\cdot (-2) & & & -24\cdot (-2) \\ & & & -6 & & & 12 & & & -24 & & & 48\end{array}\]The sequence is \(3,-6,12,-24,48\text{,}\ \ldots\)
Try it.
Write the first five terms of the sequence where the first term is 7 and the common ratio is \(r=-3\) .
| \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| 7 | ||||
| The sequence is: ________________________________________ |
Condensed: the full section is in OpenStax Precalculus 2e.
Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
Example
Try it.
Is the sequence geometric? If so, find the common ratio.
- ⓐ \(1\text{,}\ 2\text{,}\ 4\text{,}\ 8\text{,}\ 16\text{,}\ ...\)
- ⓑ \(48\text{,}\ 12\text{,}\ 4\text{, }2\text{,}\ ...\)
Solution
Divide each term by the previous term to determine whether a common ratio exists.
- ⓐ \(\begin{array}{llllllllll}\frac{2}{1}=2 & & & \frac{4}{2}=2 & & & \frac{8}{4}=2 & & & \frac{16}{8}=2\end{array}\)
The sequence is geometric because there is a common ratio. The common ratio is 2.
- ⓑ \(\begin{array}{lllllll}\frac{12}{48}=\frac{1}{4} & & & \frac{4}{12}=\frac{1}{3} & & & \frac{2}{4}=\frac{1}{2}\end{array}\)
The sequence is not geometric because there is not a common ratio.
Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is \({a}_{1}=-2\) and the common ratio is \(r=4,\) we can find subsequent terms by multiplying \(-2⋅4\) to get \(-8\) then multiplying the result \(-8⋅4\) to get \(-32\) and so on.
\[\begin{array}{l}{a}_{1}=-2 \\ {a}_{2}=(-2⋅4)=-8 \\ {a}_{3}=(-8⋅4)=-32 \\ {a}_{4}=(-32⋅4)=-128\end{array}\]The first four terms are \(\{-2\text{, }-8\text{, }-32\text{, }-128\}.\)
Example
Try it.
List the first four terms of the geometric sequence with \({a}_{1}=5\) and \(r=-2.\)
Solution
Multiply \({a}_{1}\) by \(-2\) to find \({a}_{2}.\) Repeat the process, using \({a}_{2}\) to find \({a}_{3},\) and so on.
\[\begin{array}{l}{a}_{1}=5 \\ {a}_{2}=-2{a}_{1}=-10 \\ {a}_{3}=-2{a}_{2}=20 \\ {a}_{4}=-2{a}_{3}=-40\end{array}\]The first four terms are \(\{5,-10,20,-40\}.\)
Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
Example
Try it.
Write a recursive formula for the following geometric sequence.
\[\{6\text{, }9\text{, }13.5\text{, }20.25\text{, }...\}\]Solution
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
\[r=\frac{9}{6}=1.5\]Substitute the common ratio into the recursive formula for geometric sequences and define \({a}_{1}.\)
\[\begin{array}{l}{a}_{n}=r{a}_{n-1} \\ {a}_{n}=1.5{a}_{n-1}\ \text{for }n\ge 2 \\ {a}_{1}=6\end{array}\]Using Explicit Formulas for Geometric Sequences
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.
\[{a}_{n}={a}_{1}{r}^{n-1}\]Let’s take a look at the sequence \(\{18\text{, }36\text{, }72\text{, }144\text{, }288\text{, }...\}.\) This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
\[{a}_{n}=18\cdot {2}^{n-1}\]The graph of the sequence is
Example
Try it.
Given a geometric sequence with \({a}_{1}=3\) and \({a}_{4}=24,\) find \({a}_{2}.\)
Solution
The sequence can be written in terms of the initial term and the common ratio \(r.\)
\[3,3r,3{r}^{2},3{r}^{3},...\]Find the common ratio using the given fourth term.
\[\begin{array}{ll}{a}_{n}={a}_{1}{r}^{n-1} & \\ {a}_{4}=3{r}^{3} & \text{Write the fourth term of sequence in terms of }{\alpha }_{1}\ \text{and }r \\ 24=3{r}^{3} & \text{Substitute }24\ \text{for}\ {a}_{4} \\ \ 8={r}^{3} & \text{Divide} \\ \ r=2 & \text{Solve for the common ratio}\end{array}\]Find the second term by multiplying the first term by the common ratio.
\[\begin{array}{ll}{a}_{2} & =2{a}_{1} \\ & =2(3) \\ & =6\end{array}\]Example
Try it.
Write an explicit formula for the \(n\text{th}\) term of the following geometric sequence.
\[\{2\text{, }10\text{, }50\text{, }250\text{, }...\}\]Solution
The first term is 2. The common ratio can be found by dividing the second term by the first term.
\[\frac{10}{2}=5\]The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
\[\begin{array}{l}{a}_{n}={a}_{1}{r}^{(n-1)} \\ {a}_{n}=2⋅{5}^{n-1}\end{array}\]The graph of this sequence in shows an exponential pattern.
Solving Application Problems with Geometric Sequences
In real-world scenarios involving geometric sequences, we may need to use an initial term of \({a}_{0}\) instead of \({a}_{1}.\) In these problems, we can alter the explicit formula slightly by using the following formula:
\[{a}_{n}={a}_{0}{r}^{n}\]Example
Try it.
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.
- ⓐWrite a formula for the student population.
- ⓑEstimate the student population in 2020.
Solution
- ⓐ
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 \(P\) be the student population and \(n\) be the number of years after 2013. Using the explicit formula for a geometric sequence we get
\[{P}_{n}=284⋅{1.04}^{n}\] - ⓑ
We can find the number of years since 2013 by subtracting.
\[2020-2013=7\]We are looking for the population after 7 years. We can substitute 7 for \(n\) to estimate the population in 2020.
\[{P}_{7}=284⋅{1.04}^{7}\approx 374\]The student population will be about 374 in 2020.
Key Equations
| recursive formula for \(nth\) term of a geometric sequence | \({a}_{n}=r{a}_{n-1},n\ge 2\) |
| explicit formula for \(nth\) term of a geometric sequence | \({a}_{n}={a}_{1}{r}^{n-1}\) |
Key Concepts
- A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
- The constant ratio between two consecutive terms is called the common ratio.
- The common ratio can be found by dividing any term in the sequence by the previous term.
- The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly.
- A recursive formula for a geometric sequence with common ratio \(r\) is given by \({a}_{n}=r{a}_{n-1}\) for \(n\ge 2\) .
- As with any recursive formula, the initial term of the sequence must be given.
- An explicit formula for a geometric sequence with common ratio \(r\) is given by \({a}_{n}={a}_{1}{r}^{n-1}.\)
- In application problems, we sometimes alter the explicit formula slightly to \({a}_{n}={a}_{0}{r}^{n}.\)
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-
Determine if each sequence is geometric. If so, indicate the common ratio.
ⓐ \(4,8,16,32,64,128\text{,}\ \ldots\)
ⓑ \(-2,6,-12,36,-72,216\text{,}\ \ldots\)
Antwort aufdecken
To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.
ⓐ
\(\begin{array}{llllllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}4, & & 8, & & 16, & & 32, & & 64, & & 128,\ldots \\ \\ \\ \frac{8}{4} & & \frac{16}{8} & & \frac{32}{16} & & \frac{64}{32} & & \frac{128}{64} & & \\ 2 & & 2 & & 2 & & 2 & & 2 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is geometric. The common ratio is}\ r=2.\end{array}\)ⓑ
\(\begin{array}{llllllllllllllllllllllllllllllll}\begin{array}{l} \\ \\ \\ \\ \text{Find the ratio of} \\ \text{the consecutive terms.}\end{array} & & & \ \begin{array}{lllllllllll}-2, & & 6, & & -12, & & 36, & & -72, & & 216,\ldots \\ \\ \\ \frac{6}{-2} & & \frac{-12}{6} & & \frac{36}{-12} & & \frac{-72}{36} & & \frac{216}{-72} \\ -3 & & -2 & & -3 & & -2 & & -3 & & \end{array} \\ \\ \\ & & & \ \text{The sequence is not geometric. There is no common ratio.}\end{array}\) -
\(-150\), \(-30\), \(-15\), \(-5\), \(-\frac{5}{2}\), …
Find the ratio of consecutive terms. -
8, 4, 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\), …
Find the ratio of consecutive terms. -
Write the first five terms of the sequence where the first term is 3 and the common ratio is \(r=-2.\)
Antwort aufdecken
We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.
\[\begin{array}{lllllllllllll}{a}_{1} & & & {a}_{2} & & & {a}_{3} & & & {a}_{4} & & & {a}_{5} \\ 3 & & & 3\cdot (-2) & & & -6\cdot (-2) & & & 12\cdot (-2) & & & -24\cdot (-2) \\ & & & -6 & & & 12 & & & -24 & & & 48\end{array}\]The sequence is \(3,-6,12,-24,48\text{,}\ \ldots\)
-
Write the first five terms of the sequence where the first term is 7 and the common ratio is \(r=-3\) .
\({a}_{1}\) \({a}_{2}\) \({a}_{3}\) \({a}_{4}\) \({a}_{5}\) 7 The sequence is: ________________________________________ -
- ⓐ Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r=1/3.
- ⓑ Find the ninth term of the sequence 6, 18, 54, 162, 486, 1458, … Then find the general term for the sequence.
Antwort aufdecken
- ⓐ
Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r=1/3.
To find the 13th term, use the formula with \({a}_{1}=81,r=1/3andn=13\) \({a}_{n}={a}_{1}{r}^{n-1}\) Substitute \({a}_{13}=81{(\frac{1}{3})}^{13-1}\) Simplify \(\begin{array}{l}{a}_{13}=81{(\frac{1}{3})}^{13-1} \\ {a}_{13}=81{(\frac{1}{3})}^{12} \\ {a}_{13}=\frac{1}{729}\end{array}\) - ⓑ
Find the ninth term of the sequence 6, 18, 54, 162, 486, 1458, … Then find the general term for the sequence.
Let’s first determine \({a}_{1}\) and the common ratio \(r\) \(\begin{array}{l}\text{The}\text{first}\text{term}\text{is}6,\text{so}{a}_{1}=6 \\ \text{The}\text{ratio}\text{is}:\frac{18}{6}=\frac{54}{18}=\frac{162}{54}=\frac{482}{162}=\frac{1458}{486}=3\end{array}\) To find the 9th term, use the formula with \({a}_{1}=6,r=3\) and n=9.
Substitute these values and simplify\(\begin{array}{l}{a}_{n}={a}_{1}{r}^{n-1} \\ {a}_{9}=6{(3)}^{9-1} \\ {a}_{9}=6{(3)}^{8} \\ {a}_{9}=39366\end{array}\) To find the general term, substitute \({a}_{1}=6\) and \(r=3\) into the formula \(\begin{array}{l}{a}_{n}={a}_{1}{r}^{n-1} \\ {a}_{n}=6{(3)}^{n-1}\end{array}\)
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Find the sixteenth term of a sequence where the first term is 11 and the common ratio is −6.
-
Find the 10th term of the sequence 9, 18, 36, 72, 144, 288, …. Then give the formula for the general term.
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Is the sequence geometric? If so, find the common ratio.
- ⓐ \(1\text{,}\ 2\text{,}\ 4\text{,}\ 8\text{,}\ 16\text{,}\ ...\)
- ⓑ \(48\text{,}\ 12\text{,}\ 4\text{, }2\text{,}\ ...\)
Antwort aufdecken
Divide each term by the previous term to determine whether a common ratio exists.
- ⓐ \(\begin{array}{llllllllll}\frac{2}{1}=2 & & & \frac{4}{2}=2 & & & \frac{8}{4}=2 & & & \frac{16}{8}=2\end{array}\)
The sequence is geometric because there is a common ratio. The common ratio is 2.
- ⓑ \(\begin{array}{lllllll}\frac{12}{48}=\frac{1}{4} & & & \frac{4}{12}=\frac{1}{3} & & & \frac{2}{4}=\frac{1}{2}\end{array}\)
The sequence is not geometric because there is not a common ratio.
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Is the sequence geometric? If so, find the common ratio.
\[5,10,15,20,...\]Antwort aufdecken
The sequence is not geometric because \(\frac{10}{5}\ne \frac{15}{10}\) .
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Is the sequence geometric? If so, find the common ratio.
\[100,20,4,\frac{4}{5},...\]Antwort aufdecken
The sequence is geometric. The common ratio is \(\frac{1}{5}\) .
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List the first four terms of the geometric sequence with \({a}_{1}=5\) and \(r=-2.\)
Antwort aufdecken
Multiply \({a}_{1}\) by \(-2\) to find \({a}_{2}.\) Repeat the process, using \({a}_{2}\) to find \({a}_{3},\) and so on.
\[\begin{array}{l}{a}_{1}=5 \\ {a}_{2}=-2{a}_{1}=-10 \\ {a}_{3}=-2{a}_{2}=20 \\ {a}_{4}=-2{a}_{3}=-40\end{array}\]The first four terms are \(\{5,-10,20,-40\}.\)
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List the first five terms of the geometric sequence with \({a}_{1}=18\) and \(r=\frac{1}{3}.\)
Antwort aufdecken
\(\{18,6,2,\frac{2}{3},\frac{2}{9}\}\)
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Write a recursive formula for the following geometric sequence.
\[\{6\text{, }9\text{, }13.5\text{, }20.25\text{, }...\}\]Antwort aufdecken
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
\[r=\frac{9}{6}=1.5\]Substitute the common ratio into the recursive formula for geometric sequences and define \({a}_{1}.\)
\[\begin{array}{l}{a}_{n}=r{a}_{n-1} \\ {a}_{n}=1.5{a}_{n-1}\ \text{for }n\ge 2 \\ {a}_{1}=6\end{array}\] -
Write a recursive formula for the following geometric sequence.
\[\{2\text{, }\frac{4}{3}\text{, }\frac{8}{9}\text{, }\frac{16}{27}\text{, }...\}\]Antwort aufdecken
\(\begin{array}{l}{a}_{1}=2 \\ {a}_{n}=\frac{2}{3}{a}_{n-1}\ \text{for }n\ge 2\end{array}\)
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Given a geometric sequence with \({a}_{1}=3\) and \({a}_{4}=24,\) find \({a}_{2}.\)
Antwort aufdecken
The sequence can be written in terms of the initial term and the common ratio \(r.\)
\[3,3r,3{r}^{2},3{r}^{3},...\]Find the common ratio using the given fourth term.
\[\begin{array}{ll}{a}_{n}={a}_{1}{r}^{n-1} & \\ {a}_{4}=3{r}^{3} & \text{Write the fourth term of sequence in terms of }{\alpha }_{1}\ \text{and }r \\ 24=3{r}^{3} & \text{Substitute }24\ \text{for}\ {a}_{4} \\ \ 8={r}^{3} & \text{Divide} \\ \ r=2 & \text{Solve for the common ratio}\end{array}\]Find the second term by multiplying the first term by the common ratio.
\[\begin{array}{ll}{a}_{2} & =2{a}_{1} \\ & =2(3) \\ & =6\end{array}\] -
Given a geometric sequence with \({a}_{2}=4\) and \({a}_{3}=32\) , find \({a}_{6}.\)
Antwort aufdecken
\({a}_{6}=16,384\)
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Write an explicit formula for the \(n\text{th}\) term of the following geometric sequence.
\[\{2\text{, }10\text{, }50\text{, }250\text{, }...\}\]Antwort aufdecken
The first term is 2. The common ratio can be found by dividing the second term by the first term.
\[\frac{10}{2}=5\]The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
\[\begin{array}{l}{a}_{n}={a}_{1}{r}^{(n-1)} \\ {a}_{n}=2⋅{5}^{n-1}\end{array}\]The graph of this sequence in shows an exponential pattern.
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Write an explicit formula for the following geometric sequence.
\[\{-1\text{, }3\text{, }-9\text{, }27\text{, }...\}\]Antwort aufdecken
\({a}_{n}=-{(-3)}^{n-1}\)
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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.
- ⓐWrite a formula for the student population.
- ⓑEstimate the student population in 2020.
Antwort aufdecken
- ⓐ
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 \(P\) be the student population and \(n\) be the number of years after 2013. Using the explicit formula for a geometric sequence we get
\[{P}_{n}=284⋅{1.04}^{n}\] - ⓑ
We can find the number of years since 2013 by subtracting.
\[2020-2013=7\]We are looking for the population after 7 years. We can substitute 7 for \(n\) to estimate the population in 2020.
\[{P}_{7}=284⋅{1.04}^{7}\approx 374\]The student population will be about 374 in 2020.
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A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
- ⓐWrite a formula for the number of hits.
- ⓑEstimate the number of hits in 5 weeks.
Antwort aufdecken
- ⓐ \({P}_{n}=293⋅1.026{a}^{n}\)
- ⓑThe number of hits will be about 333.
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What is a geometric sequence?
Antwort aufdecken
A sequence in which the ratio between any two consecutive terms is constant.
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How is the common ratio of a geometric sequence found?
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What is the procedure for determining whether a sequence is geometric?
Antwort aufdecken
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
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What is the difference between an arithmetic sequence and a geometric sequence?
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Describe how exponential functions and geometric sequences are similar. How are they different?
Antwort aufdecken
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
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\(1,3,9,27,81,...\)
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\(-0.125,0.25,-0.5,1,-2,...\)
Antwort aufdecken
The common ratio is \(-2\)
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\(-2,-\frac{1}{2},-\frac{1}{8},-\frac{1}{32},-\frac{1}{128},...\)
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\(-6,-12,-24,-48,-96,...\)
Antwort aufdecken
The sequence is geometric. The common ratio is 2.
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\(5,5.2,5.4,5.6,5.8,...\)
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\(-1,\frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...\)
Antwort aufdecken
The sequence is geometric. The common ratio is \(-\frac{1}{2}.\)
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\(6,8,11,15,20,...\)
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\(0.8,4,20,100,500,...\)
Antwort aufdecken
The sequence is geometric. The common ratio is \(5.\)
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\(\begin{array}{ll}{a}_{1}=8, & r=0.3\end{array}\)
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\(\begin{array}{ll}{a}_{1}=5, & r=\frac{1}{5}\end{array}\)
Antwort aufdecken
\(5,1,\frac{1}{5},\frac{1}{25},\frac{1}{125}\)
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\(\begin{array}{ll}{a}_{7}=64, & {a}_{10}\end{array}=512\)
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\(\begin{array}{ll}{a}_{6}=25, & {a}_{8}\end{array}=6.25\)
Antwort aufdecken
\(800,400,200,100,50\)
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The first term is \(2,\) and the common ratio is \(3.\) Find the 5th term.
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The first term is 16 and the common ratio is \(-\frac{1}{3}.\) Find the 4th term.
Antwort aufdecken
\({a}_{4}=-\frac{16}{27}\)
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Wie man: Geometric Sequences
- Find the common ratio for a geometric sequence.
- List the terms of a geometric sequence.
- Use a recursive formula for a geometric sequence.
- Use an explicit formula for a geometric sequence.
- Determine if a sequence is geometric (IA 12.3.1).
- Find the general term (nth term) of a geometric sequence (IA 12.3.2).
- Divide each term by the previous term.
- Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.
Fragen, die die Leute stellen
What is a function, really?
A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.
Why do we need complex numbers?
Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.
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Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates