11.3 Geometric sequences  (Page 2/6)

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$\begin{array}{l}{a}_{1}=-2\hfill \\ {a}_{2}=\left(-2\cdot 4\right)=-8\hfill \\ {a}_{3}=\left(-8\cdot 4\right)=-32\hfill \\ {a}_{4}=\left(-32\cdot 4\right)-128\hfill \end{array}$

The first four terms are

Given the first term and the common factor, find the first four terms of a geometric sequence.

1. Multiply the initial term, ${a}_{1},$ by the common ratio to find the next term, ${a}_{2}.$
2. Repeat the process, using ${a}_{n}={a}_{2}$ to find ${a}_{3}$ and then ${a}_{3}$ to find ${a}_{4,}$ until all four terms have been identified.
3. Write the terms separated by commons within brackets.

Writing the terms of a geometric sequence

List the first four terms of the geometric sequence with ${a}_{1}=5$ and $r=–2.$

Multiply ${a}_{1}$ by $-2$ to find ${a}_{2}.$ Repeat the process, using ${a}_{2}$ to find ${a}_{3},$ and so on.

$\begin{array}{l}{a}_{1}=5\hfill \\ {a}_{2}=-2{a}_{1}=-10\hfill \\ {a}_{3}=-2{a}_{2}=20\hfill \\ {a}_{4}=-2{a}_{3}=-40\hfill \end{array}$

The first four terms are $\left\{5,–10,20,–40\right\}.$

List the first five terms of the geometric sequence with ${a}_{1}=18$ and $r=\frac{1}{3}.$

$\left\{18,6,2,\frac{2}{3},\frac{2}{9}\right\}$

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.

Recursive formula for a geometric sequence

The recursive formula for a geometric sequence with common ratio $r$ and first term ${a}_{1}$ is

${a}_{n}=r{a}_{n-1},n\ge 2$

Given the first several terms of a geometric sequence, write its recursive formula.

1. State the initial term.
2. Find the common ratio by dividing any term by the preceding term.
3. Substitute the common ratio into the recursive formula for a geometric sequence.

Using recursive formulas for geometric sequences

Write a recursive formula for the following geometric sequence.

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 ${a}_{1}.$

Do we have to divide the second term by the first term to find the common ratio?

No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.

Write a recursive formula for the following geometric sequence.

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 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

${a}_{n}=18·{2}^{n-1}$

The graph of the sequence is shown in [link] .

Explicit formula for a geometric sequence

The n th term of a geometric sequence is given by the explicit formula    :

${a}_{n}={a}_{1}{r}^{n-1}$

Writing terms of geometric sequences using the explicit formula

Given a geometric sequence with $\text{\hspace{0.17em}}{a}_{1}=3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}_{4}=24,\text{\hspace{0.17em}}$ find ${a}_{2}.$

The sequence can be written in terms of the initial term and the common ratio $\text{\hspace{0.17em}}r.$

$3,3r,3{r}^{2},3{r}^{3},...$

Find the common ratio using the given fourth term.

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

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

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