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Addition and subtraction Multiplication and division Absolute value Substitution SequencesHere you will learn what geometric sequences are, how to continue a geometric sequence, how to generate a geometric sequence formula and how to translate between recursive and explicit formulas.

Students will first learn about geometric sequence formula as part of algebra in high school.

**Geometric sequences** are ordered sets of numbers that progress by multiplying or dividing each term by a common ratio.

If you **multiply** or **divide** by the **same** number each time to make the sequence, it is a geometric **sequence.**

The **common ratio** is the same for any two consecutive terms.

For example,

The geometric sequence recursive formula is:

a_{n+1}=r \cdot a_nWhere,

a_{n} is the n th term (general term)

a_{n+1} is the term after n

n is the term position

r is the common ratio

The recursive formula calculates the next term of a geometric sequence, n+1, based on the previous term, n.

The geometric sequence explicit formula is:

a_{n}=a_{1}(r)^{n-1}Where,

a_{n} is the n th term (general term)

a_{1} is the first term

n is the term position

r is the common ratio

The explicit formula calculates the n th term of a geometric sequence, given the term number, n.

You create both geometric sequence formulas by looking at the following example:

You can see the common ratio (r) is 2, so r=2.

The recursive formula is a_{n+1}=2 \cdot a_n, for a_1=5.

\begin{aligned}& a_2=2 \cdot a_1=2 \cdot 5=10 \\\\ & a_3=2 \cdot a_2=2 \cdot 10=20 \\\\ & a_4=2 \cdot a_3=2 \cdot 20=40 \\\\ & a_5=2 \cdot a_4=2 \cdot 40=80 \end{aligned}The explicit formula is a_n=5(2)^{n-1}.

\begin{aligned}& a_2=5(2)^{2-1}=5(2)^1=5 \cdot 2=10 \\\\ & a_3=5(2)^{3-1}=5(2)^2=5 \cdot 4=20 \\\\ & a_4=5(2)^{4-1}=5(2)^3=5 \cdot 8=40 \\\\ & a_5=5(2)^{5-1}=5(2)^4=5 \cdot 16=80 \end{aligned}How does this relate to high school math?

**Functions – Building Functions (HSF.BF.A.2)**Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms.

**Functions – Linear, Quadratic and Exponential Models (HSF.LE.A.2)**

Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).

Use this quiz to check your grade 4 to 5 students’ understanding of number patterns. 10+ questions with answers covering a range of 4th and 5th grade number patterns topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 4 to 5 students’ understanding of number patterns. 10+ questions with answers covering a range of 4th and 5th grade number patterns topics to identify areas of strength and support!

DOWNLOAD FREETo continue a geometric sequence:

**To identify \textbf{r}, take two consecutive terms from the sequence.****Divide the second term by the first term to find the common ratio \textbf{r}.****Multiply the last term in the sequence by the common ratio to find the next term.****Repeat Step \bf{3} for each new term.**

Calculate the next three terms for the geometric sequence shown in the graph below.

**To identify \textbf{r}, take two consecutive terms from the sequence.**

There are 5 terms shown. Any two consecutive terms of a geometric sequence can be used. Let’s use the third term and the fourth term: (3,4) and (4,8).

The x coordinate is the term position (first, second, third, etc.). The y coordinate is the actual term value.

So (3,4) is a_3=4 and (4,8) is the a_4=8.

2**Divide the second term by the first term to find the common ratio \textbf{r}. **

3**Multiply the last term in the sequence by the common ratio to find the next term.**

4**Repeat Step \bf{3} for each new term.**

The next three terms in the sequence are 32, 64, and 128.

Calculate the next three terms for the sequence in the table below.

**To identify \textbf{r}, take two consecutive terms from the sequence.**

There are 5 terms shown. Any two consecutive terms of a geometric sequence can be used. Let’s use the second term and the third term: (2,–10) and (3,–50).

In the table, the x value is the term position (first, second, third, etc.). The y value is the actual term value.

**Divide the second term by the first term to find the common ratio \textbf{r}. **

\begin{aligned}r&=-50 \div -10 \\\\ r&=5 \end{aligned}

**Multiply the last term in the sequence by the common ratio to find the next term.**

-1250 \times 5=-6,250

**Repeat Step \bf{3} for each new term.**

\begin{aligned}& -6250 \times 5=-31,250 \\\\ & -31250 \times 5=-156,250\end{aligned}

The next three terms are -6,250, -31,250, and -156,250.

In order to find a formula for a geometric sequence:

**Identify the first term.****Divide the second term by the first term to find the common ratio, \textbf{r} .****Write the recursive formula, \bf{\textbf{a}}_{\textbf{n}+1}={\textbf{r}} {\textbf{a}}_{\textbf{n}} .****Write the explicit formula, \bf{\textbf{a}_\textbf{n}=\textbf{a}_1 (\textbf{r})^{\textbf{n}-1}} .**

Write the recursive and explicit formula for the sequence below.

100, \, 10, \, 1, \, 0.1, \, 0.01, …**Identify the first term.**

a_1=100

**Divide the second term by the first term to find the common ratio, \textbf{r}. **

\begin{aligned}& r=10 \div 100 \\\\
& r=0.1\end{aligned}

**Write the recursive formula, \bf{\textbf{a}}_{\textbf{n}+1}={\textbf{r}} {\textbf{a}}_{\textbf{n}}. **

Replace r with 0.1 in the formula.

a_{n+1}=0.1 a_n

**Write the explicit formula, ** \bf{\textbf{a}_\textbf{n}=\textbf{a}_1 (\textbf{r})^{\textbf{n}-1}} **.**

Replace a_1 with 100 and r with 0.1 in the formula.

a_n=100(0.1)^{n-1}

Write the recursive and explicit formula for the sequence below.

10, \, 5, \, 2 \cfrac{1}{2} \, , \, 1 \cfrac{1}{4} \, , \, \cfrac{5}{8} \ldots**Identify the first term.**

a_1=10

**Divide the second term by the first term to find the common ratio, \textbf{r}. **

\begin{aligned}& r=5 \div 10 \\\\
& r=\cfrac{1}{2}\end{aligned}

**Write the recursive formula, \bf{\textbf{a}}_{\textbf{n}+1}={\textbf{r}} {\textbf{a}}_{\textbf{n}}. **

Replace r with \cfrac{1}{2} in the formula.

a_{n+1}=\cfrac{1}{2} \, a_n

**Write the explicit formula, ** \bf{\textbf{a}_\textbf{n}=\textbf{a}_1 (\textbf{r})^{\textbf{n}-1}} **.**

Replace a_1 with 10 and r with \cfrac{1}{2} in the formula.

a_n=10\left(\cfrac{1}{2}\right)^{n-1}

In order to translate between recursive and explicit formulas:

**Identify the common ratio, \textbf{r} and first term, \bf{\textbf{a}_1}.****Rewrite the formula.**

The recursive formula for a geometric sequence is a_{n+1}=-4 a_n and a_1=3. What is the explicit formula?

**Identify the common ratio, \textbf{r} and first term, \bf{\textbf{a}_1}. **

r is the coefficient in the recursive formula, so r=-4 and a_1=3.

**Rewrite the formula.**

The explicit formula of a geometric sequence is a_n=a_1(r)^{n-1}.

The explicit formula for this sequence is a_n=3(-4)^{n-1}.

The explicit formula for a geometric sequence is a_n=-3(2.5)^{n-1}. What is the recursive formula?

**Identify the common ratio, \textbf{r} and first term, \bf{\textbf{a}_1}. **

a_1 is the coefficient in the explicit formula, so a_1=-3.

r is the base of the exponent in the explicit formula, so r=2.5.

**Rewrite the formula.**

The recursive formula of a geometric sequence is a_{n+1}=r a_n.

The recursive formula for this sequence is a_{n+1}=2.5 a_n and a_1=-3.

Note: Always define a_1 with the recursive formula.

- Spend time letting students explore the tables and graphs of geometric sequences to compare and contrast how different values for a_1 and r impact the sequence.

**Mixing up the common ratio of a geometric sequence with the common difference for arithmetic sequences**

Although these two phrases are similar, each type of sequence is different. The successive term in a geometric sequence of numbers is calculated by multiplying the previous term by a common ratio and not by adding a common difference.

**Misunderstanding sequence notation**

It can be easy to confuse sequence notation with other types of mathematical notation, such as exponential notation. The first term of the sequence should always be defined, and is often a_1.

If a_1 is the first term, the successive terms of the geometric sequence follow this same pattern. Such as the 10 th term, which is a_{10}. This is NOT the same as a^{10}, which is a to the power of 10, an exponent.

For example,

a_n is the nth term, a_{n+1} is the next term, and a_{n-1} is the preceding term.

So a sequence looks like: a_1, a_2, \ldots, a_{n-1}, a_n, a_{n+1}, \ldots

**Thinking a negative value for r means that all terms in the sequence are negative.**

This is not always the case as when the constant ratio, r, is raised to an even power, the product of the exponent is always positive.

For example,

If the value of the common ratio, r, is –2 and the first term is 1, the first five terms of the sequence are: 1, –2, 4, –8, 16.

**Incorrect simplifying of the \bf{\textbf{n}}^{\textbf{th}} term**

Unless the first term is 1, it cannot be multiplied by the base of the exponent, following the order of operations.

For example,

6 \times 3^{n-1} is not equivalent to 18^{n-1}.

6 \times 3^{n-1} cannot be simplified any further.

- Number patterns
- Shape patterns
- Input/output tables
- Quadratic sequence
- Sequences
- Arithmetic sequence
- Recursive formula
- Explicit formula

1. Calculate the next three terms for the given sequence below.

1000, 5000, 10000

950, 1400, 1850

5000, 50000, 500000

550, 555, 560

In the table, the x value is the term position (first, second, third, etc.).

The y value is the actual term value.

There are 4 terms shown. Any two consecutive terms of a geometric sequence can be used to find the common ratio.

Let’s use the second term and the third term: (2,5) and (3,50).

Common ratio:

50\div 5=10

Use the common ratio to calculate the next three terms.

\begin{aligned}& 500 \times 10=5000 \\\\ & 5000 \times 10=50000 \\\\ & 50000 \times 10=500000\end{aligned}

2. Calculate the next three terms for the sequence in the graph below.

50, 100, 200

125, 625, 3125

50, 1000, 20000

12.5, 6.25, 3.125

In the graph, the x coordinate is the term position (first, second, third, etc.). The y coordinate is the actual term value.

There are 5 terms shown. Any two consecutive terms of a geometric sequence can be used to find the common ratio. Let’s use the fourth term and the fifth term: (4,5) and (5,25).

Common ratio:

25\div 5=5

Use the common ratio to calculate the next three terms.

\begin{aligned}&\begin{aligned}& 25 \times 5=125 \\\\ & 125 \times 5=625\end{aligned} \\\\ &625 \times 5=3125\end{aligned}

3. What is the explicit formula for the sequence -1, -3, -9, -27, -81…?

a_n=(-3)^{n-1}

a_{n+1}=3(-1)^n

a_{n+1}=-1(-3)^n

a_n=-1(3)^{n-1}

Identify the first term.

a_1=-1

Choose two consecutive terms: -27 and -9

Common ratio:

-27 \div-9=3

Write the geometric explicit formula, a_n=a_1(r)^{n-1}.

Replace a_1 with -1 and r with 3 in the formula.

a_n=-1(3)^{n-1}

4. Write the recursive formula for the sequence below.

36, \, 12, \, 4, \, \cfrac{4}{3} \, , \, \cfrac{4}{9} \, , \ldots

a_{n+1}=\cfrac{1}{3} \, a_n and a_1=36

a_n=3 a_{n+1} and a_1=36

a_{n+1}=a_n+\cfrac{1}{3} and a_1=36

a_n=\cfrac{1}{3} \, a_n and a_1=36

Identify the first term.

a_1=36

Choose two consecutive terms: 12 and 4

Common ratio:

4\div 12=\cfrac{1}{3}

Write the geometric recursive formula, a_{n+1}=r a_n.

Replace r with \cfrac{1}{3} in the formula.

a_{n+1}=\cfrac{1}{3} \, a_n

5. The recursive formula for a geometric sequence is a_{n+1}=0.9 a_n and a_1=-88. What is the explicit formula?

a_{n+1}=-88+0.9 a_n

a_{n+1}=0.9(-88)^{n-1}

a_n=-88(0.9)^{n-1}

a_n=\cfrac{1}{3} \, a_n and a_1=36

r is the coefficient in the recursive formula, so r=0.9 and a_1=-88.

The explicit formula of a geometric sequence is a_n=a_1(r)^{n-1}.

The explicit formula for this sequence is a_n=-88(0.9)^{n-1}.

6. The explicit formula for a geometric sequence is a_n=-1\left(\cfrac{1}{2}\right)^{n-1}.

What answer choice does NOT represent the sequence?

a_{n+1}=\cfrac{1}{2} \, a_n and a_1=-1

a_{n+1}=-1\left(\cfrac{1}{2}\right)^n

a_{n+1}=-1\left(\cfrac{1}{2} \, a_n\right)

-1, \,-\cfrac{1}{2} \, , \, -\cfrac{1}{4} \, , \, -\cfrac{1}{8} \, , \, -\cfrac{1}{16}

a_1 is the coefficient in the explicit formula, so a_1=-1.

r is the base of the exponent in the explicit formula, so r=\cfrac{1}{2} \, .

The recursive formula of a geometric sequence is a_{n+1}=r a_n.

The recursive formula for this sequence is a_{n+1}=\cfrac{1}{2} \, a_n and a_1=-1.

AND

a_n=-1\left(\cfrac{1}{2}\right)^{n-1} is equal to a_{n+1}=-1\left(\cfrac{1}{2}\right)^{n}, because the input for both formulas is the previous term and the output is the next term.

AND

The first five terms:

\begin{aligned}& a_1=-1\left(\cfrac{1}{2}\right)^{1-1}=-1\left(\cfrac{1}{2}\right)^0=-1 \\\\ & a_2=-1\left(\cfrac{1}{2}\right)^{2-1}=-1\left(\cfrac{1}{2}\right)^1=-\cfrac{1}{2} \\\\ & a_3=-1\left(\cfrac{1}{2}\right)^{3-1}=-1\left(\cfrac{1}{2}\right)^2=-\cfrac{1}{4} \\\\ & a_4=-1\left(\cfrac{1}{2}\right)^{4-1}=-1\left(\cfrac{1}{2}\right)^3=-\cfrac{1}{8} \\\\ & a_5=-1\left(\cfrac{1}{2}\right)^{5-1}=-1\left(\cfrac{1}{2}\right)^4=-\cfrac{1}{16}\end{aligned}

The recursive formula requires the term before it to calculate the next term. The explicit formula uses the term position to calculate the term value.

The sum of the terms in a geometric sequence (geometric progression) is a geometric series. It can be the sum of an infinite geometric sequence, from n=1 to \infty (know as an infinite geometric series) or a range of terms, such as the sum of the first n terms, where n is the number of terms included in the series.

- Factoring
- Simultaneous equations
- Standard form

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