Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

Section 8.3 Geometric Sequences and Series

Before Starting this Section, Review

1 The general term of a sequence (Section 8.1 , page 716)
2 Graphing exponential functions (Section 4.1 , page 425)
3 Partial sums (Section 8.1 , page 724)
4 Compound interest (Section 4.1 , page 431)

Objectives

1 Identify a geometric sequence and find its common ratio.
2 Find the sum of a finite geometric sequence.
3 Solve annuity problems.
4 Find the sum of an infinite geometric sequence.

Spreading the Wealth

Cities across the nation compete to host a Super Bowl. Cities across the world compete to host the Olympic Games. What is the incentive? The multiplier effect is often listed among the benefits for hosting one of these events. The effect refers to the phenomenon that occurs when money spent directly on an event has a ripple effect on the economy that is many times the amount spent directly on the event.

Suppose, for example, that $10 million is spent directly by tourists for hotels, meals, tickets, cabs, and so on. Some portion of that amount will be respent by its recipients on groceries, clothes, gas, and so on.

This process is repeated over and over, affecting the economy much more than the initial $10 million spent. Under certain assumptions, the spending just described results in an infinite geometric series, whose sum is called the multiplier. Example 10 illustrates this effect.

Geometric Sequence

1 Identify a geometric sequence and find its common ratio.

When the ratio (an+1an) $(\frac{a_{n + 1}}{a_{n}})$ of any two consecutive terms of a sequence is always the same number, the sequence is called a geometric sequence. In other words, the sequence a1, a2, a3, a4, … $a_{1}, a_{2}, a_{3}, a_{4}, \dots$ is a geometric sequence if

a 2 a 1 = a 3 a 2 = a 4 a 3 = . \dots

$\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{a_{4}}{a_{3}} = . \dots$

Example 1 Finding the Common Ratio

Find the common ratio for each geometric sequence.

2, 10, 50, 250, 1250, … $2, 10, 50, 250, 1250, \dots$
−162, −54, −18, −6, −2, … $- 162, - 54, - 18, - 6, - 2, \dots$

Solution

The common ratio is 102=5 (or 5010, 25050, or 1250250). $\frac{10}{2} = 5 (or \frac{50}{10}, \frac{250}{50}, or \frac{1250}{250}) .$
The common ratio is −54−162=13(or −18−54, −6−18, or −2−6). $\frac{- 54}{- 162} = \frac{1}{3} (or \frac{- 18}{- 54}, \frac{- 6}{- 18}, or \frac{- 2}{- 6}) .$

Practice Problem 1

Find the common ratio for the geometric sequence 6, 18, 54,162, 486, … .

A geometric sequence can be completely specified by giving the first term a1 $a_{1}$ and the common ratio r. Suppose, for example, that we are told that the sequence a1, a2, a3, a4, … $a_{1}, a_{2}, a_{3}, a_{4}, \dots$ is geometric and that a1=3 $a_{1} = 3$ and r=4 $r = 4$ . Then we can find a2: $a_{2} :$

a 2 a 1 = 4, so a 2 = 4 a 1 = 4 \cdot 3 = 12 .

$\frac{a_{2}}{a_{1}} = 4, {so a}_{2} = 4 a_{1} = 4 \cdot 3 = 12 .$

Now we continue:

a 3 a 2 a 4 a 3 = = 4, so a 3 = 4 a 2 = 4 \cdot 12 = 48; 4, so a 4 = 4 a 3 = 4 \cdot 48 = 192; and so on .

$\begin{array}{rcl} \frac{a_{3}}{a_{2}} & = & 4, {so a}_{3} = 4 a_{2} = 4 \cdot 12 = 48; \\ \frac{a_{4}}{a_{3}} & = & 4, {so a}_{4} = 4 a_{3} = 4 \cdot 48 = 192; and so on . \end{array}$

By rewriting an+1an=r $\frac{a_{n + 1}}{a_{n}} = r$ as an+1=ran, $a_{n + 1} = r a_{n},$ we have a recursive definition of a geometric sequence.

Example 2 Determining Whether a Sequence Is Geometric

Determine whether each sequence is geometric. If it is, find the first term and the common ratio.

an=4n $a_{n} = 4^{n}$
bn=32n $b_{n} = \frac{3}{2^{n}}$
cn=1−5n $c_{n} = 1 - 5^{n}$

Solution

The sequence 4, 16, 64, 256, …, 4n, … $4, 16, 64, 256, \dots, 4^{n}, \dots$ appears to be geometric, with a1=41=4 $a_{1} = 4^{1} = 4$ and r=4. $r = 4 .$ To verify this, we check that an+1an=4 $\frac{a_{n + 1}}{a_{n}} = 4$ for all n≥1. $n \geq 1 .$

$a n + 1 a n + 1 a n = = 4 n + 1 4 n + 1 4 n = 4 Replace n with n + 1 in the general term a n = 4 n . 4 n + 1 4 n = 4 n 4 4 n = 4$ $\begin{array}{rcll} a_{n + 1} & = & 4^{n + 1} & Replace n with n + 1 in the general term a_{n} = 4^{n} . \\ \frac{a_{n + 1}}{a_{n}} & = & \frac{4^{n + 1}}{4^{n}} = 4 & \frac{4^{n + 1}}{4^{n}} = \frac{4^{n} 4}{4^{n}} = 4 \end{array}$

Consequently, $\frac{a_{n + 1}}{a_{n}} = 4$ for all $n \geq 1,$ and the sequence is geometric.
The sequence $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \dots$ appears to be geometric, with $b_{1} = \frac{3}{2}$ and $r = \frac{1}{2} .$

To verify this, we check that $\frac{b_{n + 1}}{b_{n}} = \frac{1}{2}$ for all $n \geq 1 .$

$\begin{array}{rcll} b_{n + 1} & = & \frac{3}{2^{n + 1}} & Replace n with n + 1 in the general term b_{n} = \frac{3}{2^{n}} . \\ \frac{b_{n + 1}}{b_{n}} & = & \frac{3}{2^{n + 1}} \div \frac{3}{2^{n}} & \frac{b_{n + 1}}{b_{n}} = b_{n + 1} \div b_{n} \\ = & \frac{3}{2^{n + 1}} \cdot \frac{2^{n}}{3} = \frac{1}{2} & 2^{n + 1} = 2^{n} \cdot 2 \end{array}$

Consequently, $\frac{b_{n + 1}}{b_{n}} = \frac{1}{2}$ for all $n \geq 1,$ and the sequence is geometric.
The sequence $- 4, - 24, - 124, - 624, \dots, 1 - 5^{n}, \dots$ is not a geometric sequence because not all pairs of consecutive terms have identical quotients, $\frac{a_{n + 1}}{a_{n}} .$ The quotient of the first two terms is $\frac{- 24}{- 4} = 6,$ but the quotient of the third and second terms is $\frac{- 124}{- 24} = \frac{31}{6} .$

Practice Problem 2

Determine whether the sequence $a_{n} = {(\frac{3}{2})}^{n}$ is geometric. If so, find the first term and the common ratio.

From the equation $a_{n + 1} = a_{n} r$ used in the recursive definition of a geometric sequence, we see that each term after the first is obtained by multiplying the preceding term by the common ratio r.

Example 3 Finding Terms in a Geometric Sequence

For the geometric sequence 1, 3, 9, 27, …, find each of the following:

$a_{1}$
r
$a_{n}$

Solution

The first term of the sequence is given: $a_{1} = 1 .$
The sequence is geometric, so we can take the ratio of any two consecutive terms, $\frac{a_{n + 1}}{a_{n}} .$ The ratio of the first two terms gives us

$r = \frac{3}{1} = 3 .$
$\begin{array}{rcll} a_{n} & = & a_{1} r^{n - 1} & Formula for the n th term \\ = & (1) (3^{n - 1}) & Replace a_{1} with 1 and r with 3 . \\ = & 3^{n - 1} \end{array}$

Practice Problem 3

For the geometric sequence $2, \frac{6}{5}, \frac{18}{25}, \frac{54}{125}, \dots,$ find the following:
1. $a_{1}$ .
2. r.
3. $a_{n}$ .

We can rewrite the formula for the nth term of a geometric sequence as follows:

$\begin{array}{rcll} a_{n} & = & a_{1} r^{n - 1} \\ a_{n} & = & a_{1} \frac{r^{n}}{r} & Rules of exponents \\ a_{n} & = & \frac{a_{1}}{r} r^{n} & Simplify . \\ a_{n} & = & A \cdot B^{n}, where A = \frac{a_{1}}{r} and B = r \end{array}$

We conclude that the nth term of this geometric sequence (with ratio $r > 0$ ) is the value of the exponential (type) function $f (x) = A \cdot B^{x}$ for $x = n$ . That is, $a_{n} = f (n)$ .

This means that the points on the graph of the geometric sequence (with ratio $r > 0$ ) lie on the graph of the exponential (type) function $f (x) = A \cdot B^{x}$

When $r < 0$ , the geometric sequence alternates its sign. Setting $r = - s$ , we see that the points representing our sequence alternate between the graphs of two exponential (type) functions $f (x) = \frac{a_{1}}{r} \cdot s^{x}$ and $g (x) = - \frac{a_{1}}{r} \cdot s^{x}$ (we explore this further in Exercise 116).

Example 4 Graphing a Geometric Sequence

Graph the geometric sequence

$8, 4, 2, 1, \frac{1}{2}, \dots$

and use the graph to find an expression for the nth term.

Solution

To graph this sequence, we plot the points $(n, a_{n})$ for $n = 1, 2, 3 \dots .$ The first five points are

$(1, 8), (2, 4), (3, 2), (4, 1), (5, \frac{1}{2}), \dots$

Our sequence has a constant (positive) sign, so the ratio r is greater than 0. To fit an exponential (type) function $A \cdot B^{x}$ to our sequence, we first find B as a common ratio of consecutive sequence terms.

$B = r = \frac{a_{2}}{a_{1}} = \frac{4}{8} = \frac{1}{2} .$

Second, we use the fact the graph passes through the point (1, 8) to find the constant A.

$\begin{array}{rcll} y & = & A \cdot {(\frac{1}{2})}^{x} \\ 8 & = & A \cdot {(\frac{1}{2})}^{1} & Substitute x = 1 and y = 8 . \\ 8 & = & A \cdot \frac{1}{2} & Simplify . \\ A & = & 16 & Solve for A . \end{array}$

The exponential (type) function that produces this sequence is $f (x) = 16 \cdot {(\frac{1}{2})}^{x}$ , and the general term for this sequence is $a_{n} = f (n) = 16 \cdot {(\frac{1}{2})}^{n}$ (see Figure 8.5).

Practice Problem 4

Repeat Example 4 for the sequence $\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{8}{3}, \frac{16}{3}, \dots$

Example 5 Finding a Particular Term in a Geometric Sequence

Find the 23rd term of a geometric sequence whose first term is 10 and whose common ratio is 1.2.

Solution

$\begin{array}{rcll} a_{n} & = & a_{1} r^{n - 1} & Formula for the n th term \\ a_{23} & = & 10 {(1.2)}^{23 - 1} & Replace n with 23, a_{1} with 10, and r with 1.2 . \\ = & 10 {(1.2)}^{22} & Exact value \\ \approx & 552.06 & Use a calculator . \end{array}$

Practice Problem 5

Find the 18th term of a geometric sequence whose first term is 7 and whose common ratio is 1.5.

Example 6 Finding the Number of Terms of a Geometric Sequence

Find the number of terms of the geometric sequence $4, 2, 1, \frac{1}{2}, \frac{1}{4}, \dots, \frac{1}{64} .$

Solution

The given geometric sequence has first term $a_{1} = 4$ and common ratio $r = \frac{2}{4} = \frac{1}{2} .$

Let the nth term be $\frac{1}{64} .$ Then

$\begin{array}{rcll} a_{n} & = & \frac{1}{64} \\ a_{1} r^{n - 1} & = & \frac{1}{64} & a_{n} = a_{1} r^{n - 1} \\ 4 {(\frac{1}{2})}^{n - 1} & = & \frac{1}{64} & a_{1} = 4, r = \frac{1}{2} \\ {(\frac{1}{2})}^{n - 1} & = & \frac{1}{4 \cdot 64} = \frac{1}{256} & Divide both sides by 4. \\ {(\frac{1}{2})}^{n - 1} & = & {(\frac{1}{2})}^{8} & \frac{1}{256} = \frac{1}{2^{8}} = {(\frac{1}{2})}^{8} \\ n - 1 & = & 8 & a^{m} = a^{n} implies m = n . \\ n & = & 9 & Solve for n . \end{array}$

There are 9 terms in the given sequence.

Practice Problem 6

Find the number of terms of the geometric sequence $\frac{1}{9}, \frac{1}{3}, 1, 3, \dots, 243 .$

Finding the Sum of a Finite Geometric Sequence

2 Find the sum of a finite geometric sequence.

We can find a formula for the sum $S_{n}$ of the first n terms in a geometric sequence. (We have already found a similar formula for arithmetic sequences.)

$\begin{array}{rcll} S_{n} & = & a_{1} + a_{1} r + a_{1} r^{2} + a_{1} r^{3} + \dots + a_{1} r^{n - 1} & Definition of S_{n} \\ r S_{n} & = & a_{1} r + a_{1} r^{2} + a_{1} r^{3} + a_{1} r^{4} + \dots + a_{1} r^{n} & Multiply both sides by r . \\ S_{n} - r S_{n} & = & a_{1} - a_{1} r^{n} & Subtract the second equation from the first . \\ (1 - r) S_{n} & = & a_{1} (1 - r^{n}) & Factor both sides . \\ S_{n} & = & \frac{a_{1} (1 - r^{n})}{1 - r}, r \neq 1 & Divide both sides by 1 - r . \end{array}$

A geometric sequence with $r = 1$ is a sequence in which all terms are identical; that is, $a_{n} = a_{1} {(1)}^{n - 1} = a_{1} .$ Consequently, the sum of the first n terms, $S_{n}$ , is

$\underset{n terms}{\underset{︸}{a_{1} + a_{1} + \dots + a_{1}}} = n a_{1} .$

Example 7 Finding the Sum of Terms of a Finite Geometric Sequence

Find each sum.

$\sum_{i = 1}^{15} 5 {(0.7)}^{i - 1}$
$\sum_{i = 1}^{15} 5 {(0.7)}^{i}$

Solution

We can evaluate $\sum_{i = 1}^{15} 5 {(0.7)}^{i - 1}$ by using the formula for $S_{n} = \sum_{i = 1}^{n} a_{1} r^{i - 1} = \frac{a_{1} (1 - r^{n})}{1 - r},$ where $a_{1} = 5, r = 0.7,$ and $n = 15 .$

$\begin{array}{rcll} S_{15} & = & \sum_{i = 1}^{15} a_{1} r^{i - 1} = 5 [\frac{1 - {(0.7)}^{15}}{1 - 0.7}] & Replace a_{1} with 5, r = 0.7, and n = 15 . \\ \approx & 16.5875 & Use a calculator . \end{array}$
$\begin{array}{rcll} \sum_{i = 1}^{15} 5 {(0.7)}^{i} & = & (0.7) \sum_{i = 1}^{15} 5 {(0.7)}^{i - 1} & Factor out 0.7 from each term . \\ \approx & (0.7) (16.5875) & From part a \\ = & 11.6113 \end{array}$

Practice Problem 7

Find the sum $\sum_{i = 1}^{17} 3 {(0.4)}^{i} .$

Annuities

3 Solve annuity problems.

One of the most important applications of the sum of a finite geometric sequence is the computation of the value of an annuity. An annuity is a sequence of equal periodic payments. When a fixed rate of compound interest applies to all payments, the sum of all of the payments made plus all interest can be found by using the formula for the sum of a finite geometric sequence. The value of an annuity (also called the future value of an annuity) is the sum of all payments and interest. To develop a formula for the value of an annuity, we suppose $P is deposited at the end of each year at an annual interest rate i compounded annually.

The compound interest formula $A = P {(1 + i)}^{t}$ gives the total value after t years when $P earns an annual interest rate i (in decimal form) compounded once a year. At the end of the first year, the initial payment of $P is made and the annuity’s value is $P. At the end of the second year, $P is again deposited. At this time, the first deposit has earned interest during the second year. The value of the annuity after two years is

$\begin{array}{rcl} P & + & P (1 + i) . \\ ↑ & ↑ \\ Payment made at the end of year 2 & First payment plus interest earned for one year \end{array}$

The value of the annuity after three years is

$\begin{array}{l} \begin{array}{l} P & + P (1 + i) & + P {(1 + i)}^{2} \\ ↑ & ↑ & ↑ \end{array} \\ \begin{matrix} \begin{array}{l} Payment made \\ at the end of \\ year 3 \end{array} & \begin{array}{l} Second payment \\ plus interest earned \\ for one year \end{array} & \begin{array}{l} Second payment \\ plus interest earned \\ for one year \end{array} \end{matrix} \end{array}$

The value of the annuity after t years is

$\begin{array}{l} P + P (1 + i) + P {(1 + i)}^{2} + P {(1 + i)}^{3} + \dots + P {(1 + i)}^{t - 1} . \\ ↑ ↑ \\ Payment made at the end of year t First payment plus interest earned for t - 1 year \end{array}$

The value of the annuity after t years is the sum of a geometric sequence with first term $a_{1} = P$ and common ratio $r = 1 + i .$ We compute this sum A as follows:

$\begin{array}{rcll} A & = & S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} & Formula for S_{n} \\ A & = & \frac{P [1 - {(1 + i)}^{t}]}{1 - (1 + i)} & Replace n with t, a_{1} with P, and r with 1 + i . \\ = & \frac{P [1 - {(1 + i)}^{t}]}{- i} & Simplify . \\ = & \frac{P [{(1 + i)}^{t} - 1]}{i} & Multiply numerator and denominator by - 1 . \end{array}$

If interest is compounded n times per year and equal payments are made at the end of each compounding period, we adjust the formula as we did in Section 3.1 to account for the more frequent compounding.

Example 8 Finding the Value of an Annuity

An individual retirement account (IRA) is a common way to save money to provide funds after retirement. Suppose you make payments of $1200 into an IRA at the end of each year at an annual interest rate of 4.5% per year, compounded annually. What is the value of this annuity after 35 years?

Solution

Each annuity payment is $P = $ 1200 .$ The annual interest rate is 4.5% $(so i = 0.045),$ and the number of years is $t = 35 .$ Because interest is compounded annually, $n = 1 .$ The value of the annuity is

$\begin{array}{rcll} A & = & 1200 [\frac{{(1 + \frac{0.045}{1})}^{(1) 35} - 1}{\frac{0.045}{1}}] \\ = & $ 97,795.94 & Use a calculator . \end{array}$

The value of the IRA after 35 years is $97,795.94.

Practice Problem 8

If in Example 7 the end-of-year payments are $1500, the annual interest rate remains 4.5%, compounded annually, and the payments are made for 30 years, what is the value of the annuity?

Infinite Geometric Series

4 Find the sum of an infinite geometric sequence.

The sum $S_{n}$ of the first n terms of a geometric series is given by the formula $S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} .$

If this finite sum $S_{n}$ approaches a number S as $n \to \infty$ (that is, as n gets larger and larger), we say that S is the sum of the infinite geometric series, and we write

$S = \sum_{i = 1}^{\infty} a_{1} r^{i - 1} .$

If r is any real number with $- 1 < r < 1$ (equivalently, $| r | < 1$ ), then the value of $r^{n},$ like the value of ${(\frac{1}{2})}^{n},$ gets closer and closer to 0 as n gets larger and larger. We indicate that the values of $r^{n}$ approach 0 by writing

$\lim_{n \to \infty} r^{n} = 0 if | r | < 1 .$

The expression $\lim_{n \to \infty} r^{n}$ is read “the limit, as n approaches infinity, of r to the n” or “the limit, as n approaches infinity, of r to the nth power.”

When $| r | < 1,$

$S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} \to \frac{a_{1} (1 - 0)}{1 - r} = \frac{a_{1}}{1 - r} as n \to \infty .$

When $| r | \geq 1,$ the infinite geometric series does not have a sum. This is because the value of $r^{n}$ does not approach 0 as $n \to \infty .$ For example, if $r = 2,$ we have

$2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16, 2^{5} = 32, 2^{6} = 64, \dots .$

Example 9 Finding the Sum of an Infinite Geometric Series

Find the sum $2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \dots .$

Solution

The first term is $a_{1} = 2,$ and the common ratio is

$r = \frac{3 / 2}{2} = \frac{3}{4} .$

Because $| r | = \frac{3}{4} < 1,$ we can use the formula for the sum of an infinite geometric series.

$\begin{array}{rcll} S & = & \frac{a_{1}}{1 - r} & Formula for S \\ = & \frac{2}{1 - \frac{3}{4}} & Replace a_{1} with 2 and r with \frac{3}{4} . \\ = & 8 & Simplify . \end{array}$

Practice Problem 9

Find the sum $3 + 2 + \frac{4}{3} + \frac{8}{9} + \dots .$

Example 10 Calculating the Multiplier Effect

The host city for the Super Bowl expects that tourists will spend $10,000,000. Assume that 80% of this money is spent again in the city, then 80% of this second round of spending is spent again, and so on. In the introduction to this section, we said that such a spending pattern results in a geometric series whose sum is called the multiplier. Find this series and its sum.

Solution

We start our series with the $10,000,000 brought into the city and add the subsequent amounts spent.

$\begin{array}{l} 10,000,000 & + & 10,000,000 (0.80) & + & 10,000,000 {(0.80)}^{2} & + & 10,000,000 {(0.80)}^{3} + \dots \\ ↑ & ↑ & ↑ & ↑ \\ original & 80 % of & 80 % of & 80 % of \\ 10,000,000 & 10,000,000 & previous amount & previous amount \end{array}$

Using the formula $\sum_{i = 1}^{\infty} a r^{i - 1} = \frac{a}{1 - r}$ for the sum of an infinite geometric series, we have

$\sum_{i = 1}^{\infty} (10,000,000) {(0.80)}^{i - 1} = \frac{$ 10,000,000}{1 - 0.80} = $ 50,000,000 .$

This amount should shed some light on why there is so much competition to serve as the host city for a major sports event.

Practice Problem 10

Find the series and the sum that results in Example 10 , assuming that 85%, rather than 80%, occurs in the repeated spending.

Section 8.3 Exercises

Concepts and Vocabulary

If 5 is the common ratio of a geometric sequence with general term $a_{n},$ then $\frac{a_{63}}{a_{62}} = \underline{}$ .
If 5 is the common ratio of a geometric sequence with general term $a_{n}$ and $a_{15} = - 2,$ then $a_{16} = \underline{}$ .
If 24 is the term immediately following the sequence term 8 in a geometric sequence, then the common ratio is .
An infinite geometric series $a_{1} + a_{1} r + a_{1} r^{2} + \dots$ does not have a sum if $\underline{} \geq 1$ .
True or False. If the sum of an infinite geometric series having the common ratio $r = \frac{1}{3}$ is $S = 15,$ then $a_{1} = 10 .$
True or False. The sum of an infinite geometric series can never be negative.
True or False. The general term for a geometric sequence with first term $a_{1}$ and common ratio r is $a_{n} = a_{1} r^{n}$ .
True or False. $\sum_{k = 5}^{n} a_{k} = \sum_{k = 1}^{n} a_{k} - \sum_{k = 1}^{4} a_{k}$ .

Building Skills

In Exercises 9–28, determine whether each sequence is geometric. If it is, find the first term and the common ratio.

3, 6, 12, 24, ….
2, 4, 8, 16, …
1, 5, 10, 20, …
1, 1, 3, 3, 9, 9, …
$1, - 3, 9, - 27, \dots$
$- 1, 2, - 4, 8, \dots$
$7, - 7, 7, - 7, \dots$
$1, - 2, 4, - 8, \dots$
$9, 3, 1, \frac{1}{3}, \dots$
$5, 2, \frac{4}{5}, \frac{8}{25}, \dots$
$a_{n} = {(- \frac{1}{2})}^{n}$
$a_{n} = 5 {(\frac{2}{3})}^{n}$
$a_{n} = 2^{n - 1}$
$a_{n} = - {(1.06)}^{n - 1}$
$a_{n} = 7 n^{2} + 1$
$a_{n} = 1 - {(2 n)}^{2}$
$a_{n} = 3^{- n}$
$a_{n} = 50 {(0.1)}^{- n}$
$a_{n} = 5^{n / 2}$
$a_{n} = 7^{\sqrt{n}}$

In Exercises 29–36, find the first term $a_{1},$ the common ratio r, and the nth term $a_{n}$ for each geometric sequence.

2, 10, 50, 250, …
$- 3, - 6, - 12, - 24, \dots$
$5, \frac{10}{3}, \frac{20}{9}, \frac{40}{27}, \dots$
$1, \sqrt{3}, 3, 3 \sqrt{3}, \dots$
$0.2, - 0.6, 1.8, - 5.4, \dots$
$1.3, - 0.26, 0.052, - 0.0104, \dots$
$π^{4}, π^{6}, π^{8}, π^{10}, \dots$
$e^{2}, 1, e^{- 2}, e^{- 4}, \dots$

In Exercises 37–42, graph the geometric sequence and use the graph to find an expression for the nth term.

6, 12, 24, 48, 96,…
$\frac{1}{2}, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}, \frac{81}{2}, \dots$
$36, 12, 4, \frac{4}{3}, \frac{4}{9}, \dots$
$12, 6, 3, \frac{3}{2}, \frac{3}{4}, \dots$
$- \frac{4}{5}, - \frac{16}{5}, - \frac{64}{5}, - \frac{256}{5}, - \frac{1024}{5}, \dots$
$- 45, - 15, - 5, - \frac{5}{3}, - \frac{5}{9}, \dots$

In Exercises 43–52, find the indicated term of each geometric sequence.

$a_{7}$ when $a_{1} = 5$ and $r = 2$
$a_{7}$ when $a_{1} = 8$ and $r = 3$
$a_{10}$ when $a_{1} = 3$ and $r = - 2$
$a_{10}$ when $a_{1} = 7$ and $r = - 2$
$a_{6}$ when $a_{1} = \frac{1}{16}$ and $r = 3$
$a_{6}$ when $a_{1} = \frac{1}{81}$ and $r = 3$
$a_{9}$ when $a_{1} = - 1$ and $r = \frac{5}{2}$
$a_{9}$ when $a_{1} = - 4$ and $r = \frac{3}{4}$
$a_{20}$ when $a_{1} = 500$ and $r = - \frac{1}{2}$
$a_{20}$ when $a_{1} = 1000$ and $r = - \frac{1}{10}$

In Exercises 53–58, find the number of terms of the given geometric sequence.

5, 10, 20, …, 5120
2, 6, 18, …, 4374
$16, 8, 4, \dots, \frac{1}{32}$
$27, 9, 3, \dots, \frac{1}{27}$
$18, - 12, 8, \dots, \frac{512}{729}$
$25, - 15, 9, \dots, \frac{729}{625}$

In Exercises 59–64, find the sum $S_{n}$ of the first n terms of each geometric sequence.

$\frac{1}{10}, \frac{1}{2}, \frac{5}{2}, \frac{25}{2}, \dots; n = 10$
$6, 2, \frac{2}{3}, \frac{2}{9}, \dots; n = 10$
$\frac{1}{25}, - \frac{1}{5}, 1, - 5, \dots; n = 12$
$- 10, \frac{1}{10}, \frac{- 1}{1000}, \frac{1}{100,000}, \dots; n = 12$
$5, \frac{5}{4}, \frac{5}{4^{2}}, \frac{5}{4^{3}}, \dots; n = 8$
$2, \frac{2}{5}, \frac{2}{5^{2}}, \frac{2}{5^{3}}, \dots; n = 8$

In Exercises 65–72, find each sum.

$\sum_{i = 1}^{5} {(\frac{1}{2})}^{i - 1}$
$\sum_{i = 1}^{5} {(\frac{1}{5})}^{i - 1}$
$\sum_{i = 1}^{8} 3 {(\frac{2}{3})}^{i - 1}$
$\sum_{i = 1}^{8} 2 {(\frac{3}{5})}^{i - 1}$
$\sum_{i = 3}^{10} \frac{2^{i - 1}}{4}$
$\sum_{i = 3}^{10} \frac{5^{i - 1}}{2}$
$\sum_{i = 1}^{20} (- \frac{3}{5}) {(- \frac{5}{2})}^{i - 1}$
$\sum_{i = 1}^{20} (- \frac{1}{4}) (3^{2 - i})$

In Exercises 73–78, use a calculator to find each sum. Round your answers to three decimal places.

$\frac{1}{5} + \frac{1}{10} + \frac{1}{20} + \dots + \frac{1}{320}$
$2 + 6 + 18 + \dots + 1458$
$\sum_{n = 1}^{10} 3^{n - 2}$
$\sum_{n = 1}^{15} {(\frac{1}{7})}^{n - 1}$
$\sum_{n = 1}^{10} {(- 1)}^{n - 1} 3^{2 - n}$
$\sum_{n = 1}^{20} {(- 4)}^{3 - n}$

In Exercises 79–88, find each sum.

$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots$
$\frac{5}{2} + \frac{5}{4} + \frac{5}{8} + \frac{5}{16} + \dots$
$- \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \dots$
$- \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + \frac{3}{16} + \dots$
$8 - 2 + \frac{1}{2} - \frac{1}{8} + \dots$
$1 - \frac{3}{5} + \frac{9}{25} - \frac{27}{125} + \dots$
$\sum_{n = 0}^{\infty} 5 {(\frac{1}{3})}^{n}$
$\sum_{n = 0}^{\infty} 3 {(\frac{1}{4})}^{n}$
$\sum_{n = 0}^{\infty} {(- \frac{1}{4})}^{n}$
$\sum_{n = 0}^{\infty} {(- \frac{1}{3})}^{n}$

Applying the Concepts

Population growth. The population in a small town is increasing at the rate of 3% per year. If the present population is 20,000, what will the population be at the end of five years?
Growth of a zoo collection. The butterfly collection at a local zoo started with only 25 butterflies. If the number of butterflies doubled each month, how many butterflies were in the collection after 12 months?
Savings growth. Ramón deposits $100 on the last day of each month into a savings account that pays 6% annually, compounded monthly. What is the balance in the account after 36 compounding periods?
Savings growth. Kat deposits $300 semiannually into a savings account for her son. If the account pays 8% annually, compounded semiannually, what is the balance in the account after 20 compounding periods?
Ancestors. Every person has two parents, four grandparents, eight great-grandparents, and so on. Find the number of ancestors a person has in the tenth generation back.
Bacterial growth. A colony of bacteria doubles in number every day. If there are 1000 bacteria now, how many will there be
1. On the 7th day?
2. On the nth day?
Investment. An actuarial firm budgets a new position at $36,000 for the first year and a 5% raise each year thereafter. Find the total compensation a successful applicant for the job would receive during the first 20 years of employment.
Investment. A realtor offers two choices of payment for a small condominium:
1. Pay $4000 per month for the next 25 months.
2. Pay $1 ¢$ the first month, $2 ¢$ the second month, $4 ¢$ the third month, and so on for 25 months.
  
  Which option, a or b, is the better choice for the buyer? Explain your reasoning.
Pendulum motion. The distance traveled by any point on a certain pendulum is 20% less than in the preceding swing. If the length traveled by the pendulum bob during the first swing is 56.25 centimeters, find the total distance the bob has traveled at the end of the fifth swing.
Depreciating a tractor. Every year a tractor loses 20% of the value it had at the beginning of the year. If the tractor now has a value of $120,000, what will its value be in five years?
Rebounding ball. A particular ball always rebounds $\frac{3}{5}$ the distance it falls. If the ball is dropped from a height of 5 meters, how far will it travel before coming to a stop?
Rebounding ball. Repeat Exercise 99, assuming that the ball is dropped from a height of 9 meters.

In Exercises 101 and 102, at each stage the objects are being generated following the depicted pattern. Verify that the total number of shaded objects at each stage is a term of a geometric sequence and determine the total number of shaded objects at the 10th stage.

In Exercises 103 and 104, the accompanying figure shows an infinite sequence of successively smaller objects. At each stage the size of the sides of the objects is half the size of the sides from the preceding stage. The area of the largest object is 1. Verify that the total area of the shaded objects at each stage forms a term of a geometric sequence and find the sum of all terms in this infinite sequence.

The accompanying figure shows the first six squares in an infinite sequence of successively smaller squares formed by connecting the midpoints of the sides of the preceding larger square. The area of the largest square is 1. Find the total area of the indicated sections that are shaded as the number of squares increases to infinity.
The accompanying figure shows the first three equilateral triangles in an infinite sequence of successively smaller equilateral triangles formed by connecting the midpoints of the sides of the preceding larger triangle. The largest triangle has sides of length 4. Find the total perimeter of all of the triangles.

Beyond the Basics

Prove that if $a_{1}, a_{2}, a_{3}, \dots$ is a geometric sequence of positive terms, then the sequence ${\ln a}_{1}, {\ln a}_{2}, {\ln a}_{3}, \dots$ is an arithmetic sequence.
Prove that if $a_{1}, a_{2}, a_{3}, \dots$ is a geometric sequence, then $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, \dots$ is also a geometric sequence.
Show that if $a_{1}, a_{2}, a_{3}, \dots$ is a geometric sequence with common ratio r, then $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \dots$ is also a geometric sequence. Also find its common ratio.
Show that if c is any nonzero real number, then $c^{2}, c, 1, \frac{1}{c}, \dots$ is a geometric sequence.
Show that if x is any nonzero real number, then $x, 2, \frac{4}{x}, \frac{8}{x^{2}}, \dots$ is a geometric sequence.
Show that if a and x are any two nonzero real numbers, then $\frac{x}{a}, - 1, \frac{a}{x}, - \frac{a^{2}}{x^{2}}, \dots$ is a geometric sequence.
Show that if a, x, and y are any three nonzero real numbers, then $\frac{a}{x}, - \frac{a}{x y}, \frac{a}{x y^{2}}, \frac{a}{x y^{3}}, \dots$ is a geometric sequence.
Show that if $a_{1}, a_{2}, a_{3}, \dots$ is an arithmetic sequence with common difference d, then $2^{a_{1}}, 2^{a_{2}}, 2^{a_{3}}, \dots$ is a geometric sequence. Also find its common ratio.

Critical Thinking / Discussion / Writing

Consider two sequences given by the general terms

$a_{n} = 2^{4 - n}, b_{n} = \frac{4}{2^{n - 2}} .$

Verify that each sequence is geometric. Write the general term of each sequence in the form $A \cdot B^{n}$ . Discuss the result.
Graph the geometric sequence

$4, - 2, 1, - \frac{1}{2}, \frac{1}{4}, \dots .$

Note that the points representing this sequence alternate between the graphs of two functions. What are these two functions? Identify the number that the values of $a_{n}$ approach as $n \to \infty$ .
If $a_{n}$ denotes the nth term of an arithmetic sequence with $a_{1} = 10$ and $d = 2.7$ and if $b_{n}$ denotes the nth term of a geometric sequence with $b_{1} = 10$ and $r = 2,$ which number is larger, $a_{1001}$ or $b_{1001}$ ?
If the sum $\frac{a_{1} (1 - r^{n})}{1 - r}$ of the first n terms of a geometric sequence is 1023, find $a_{1}$ when $n = 10$ and $r = \frac{1}{2} .$
Find n and k so that the sum $5 + 5 \cdot 2 + 5 \cdot 2^{2} + \dots + 5 \cdot 2^{15}$ can be written as $\sum_{i = 0}^{n} 5 \cdot 2^{i}$ or as $\sum_{i = 1}^{k} 5 \cdot 2^{i - 1} .$
The sum of three consecutive terms of a geometric sequence is 35, and their product is 1000. Find the numbers.

[Hint: Let $\frac{a}{r}, a,$ and ar be the three terms of the geometric sequence.]
The sum of three consecutive terms of an arithmetic sequence is 15. If 1, 4, and 19 are respectively added to the three terms, the resulting numbers form three consecutive terms of a geometric sequence. Find the numbers.
The product of the first three terms of a geometric sequence is 1000. If 6 is added to the second term and 7 is added to the third term, the resulting numbers form the three consecutive terms of an arithmetic sequences. Find the numbers.

Getting Ready for the Next Section

In Exercises 123–126, a statement $P_{n}$ involving positive integers n is given. Write the statements $P_{3}$ and $P_{4}$ . State whether $P_{3}$ and $P_{4}$ are true or false.

$P_{n} : n (n + 1)$ is even.
$P_{n} : n^{3} + n$ is divisible by 3.
$P_{n} : 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$
$P_{n} : 2^{n} > 3 n$

In Exercises 127–130, a statement $P_{n}$ involving a positive integer n is given. Write the statement $P_{n + 1}$ .

$P_{n} : n^{2} + n$ is divisible by 2.
$P_{n} : 2^{3 n} - 1$ is divisible by 7.
$P_{n} : 3^{n} > 5 n$
$P_{n} : 1 + 4 + 7 + \dots + (3 n - 2) = \frac{1}{2} n (3 n - 1)$
Show that $\frac{k (k + 1)}{2} + (k + 1) = \frac{(k + 1) (k + 2)}{2}$ .
Show that $\frac{1}{6} k (k + 1) (2 k + 1) + {(k + 1)}^{2} = \frac{1}{6} (k + 1) (k + 2) (2 k + 3)$ .

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for Section 8.3 Geometric Sequences and Series

Create new playlist

Sign In

Sign Up

Table of Contents for
Section 8.3 Geometric Sequences and Series