8.3 ARITHMETIC AND GEOMETRIC SEQUENCES

[Pages:11]8.3 Airthmetic and Geometric Sequences

451

8.3 A R I T H M E T I C A N D G E O M E T R I C S E Q U E N C E S

Whenever you tell me that mathematics is just a human invention like the game of chess I would like to believe you. But I keep returning to the same problem. Why does the mathematics we have discovered in the past so often turn out to describe the workings of the Universe?

John Barrow

I remember that when I

was about twelve I learned

from [my uncle] that by the distributive law 1 times 1 equals 1. I thought that was great.

Peter Lax

Two kinds of regular sequences occur so often that they have specific names, arithmetic and geometric sequences. We treat them together because some obvious parallels between these kinds of sequences lead to similar formulas. This also makes it easier to learn and work with the formulas. The greatest value in this association is understanding how the ideas are related and how to derive the formulas from fundamental concepts. Anyone learning the formulas this way can recover them whenever needed.

Both arithmetic and geometric sequences begin with an arbitrary first term, and the sequences are generated by regularly adding the same number (the common difference in an arithmetic sequence) or multiplying by the same number (the common ratio in a geometric sequence). Definitions emphasize the parallel features, which examples will clarify.

Definition: arithmetic and geometric sequences

Arithmetic Sequence

a1 a and an an1 d for n 1

The sequence an is an arithmetic sequence with first term a and common difference d. Geometric Sequence

a1 a and an r ? an1 for n 1

The sequence an is a geometric sequence with first term a and common ratio r.

The definitions imply convenient formulas for the nth term of both kinds of sequences. For an arithmetic sequence we get the nth term by adding d to the first term n 1 times; for a geometric sequence, we multiply the first term by r, n 1 times.

Formulas for the nth terms of arithmetic and geometric sequences

For an arithmetic sequence, a formula for the nth term of the sequence is

an a n 1d.

(1)

For a geometric sequence, a formula for the nth term of the sequence is

an a ? r n1.

(2)

The definitions allow us to recognize both arithmetic and geometric sequences. In an arithmetic sequence the difference between successive terms, an1 an, is always the same, the constant d; in a geometric sequence the ratio of successive

terms, an1 , is always the same. an

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Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

Strategy: Calculate the differences and /or ratios of successive terms.

EXAMPLE 1 Arithmetic or geometric? The first three terms of a sequence are given. Determine if the sequence could be arithmetic or geometric. If it is an arithmetic sequence, find d; for a geometric sequence, find r.

(a) 2, 4, 8, . . .

(b) ln 2, ln 4, ln 8, . . .

(c)

1 ,

1,

1,

.

.

.

234

Solution

(a) a2 a1 4 2 2, and a3 a2 8 4 4. Since the differences are not the same, the sequence cannot be arithmetic. Checking ratios, a2 4 2, a1 2 and a3 8 2, so the sequence could be geometric, with a common ratio a2 4 r 2. Without a formula for the general term, we cannot say anything more

about the sequence.

(b)

a2

a1

ln

4

ln

2

ln

4 2

ln

2,

and

a3 a2 ln 8 ln 4

ln

8 4

ln

2,

so

the

sequence

could

be

arithmetic,

with

ln

2

as

the

common

difference. As in part (a), we cannot say more because no general term is given.

(c)

a2

a1

1 3

1 2

1 6

,

and

a3

a2

1 4

1 3

1 12

.

The

differences

are

not

the

same,

so

the

sequence

is

not

arithmetic.

a2 a1

1 3

1 2

2 ,

3

and

a3 a2

1 4

1 3

3 4

,

so

the

sequence

is

not

geometric.

Note

that

the

sequence

in

part

(a) could be geometric and the sequence in part (b) could be arithmetic, but

in part (c) you can conclude unequivocally that the sequence cannot be either

arithmetic or geometric.

EXAMPLE 2 Arithmetic or geometric? Determine whether the sequence is arithmetic, geometric, or neither.

(a) 3 1.6n (b) 2n (c) an ln n

Solution

(a) a2 a1 3 1.6 ? 2 3 1.6 ? 1 0.2 1.4 1.6, and a3 a2 3 1.6 ? 3 3 1.6 ? 2 1.6. From the first three terms, this could be an arithmetic sequence with d 1.6. Check the difference an1 an.

an1 an 3 1.6n 1 3 1.6n 1.6.

The sequence is arithmetic, with d 1.6. (b) a2 a1 4 2 2, and a3 a2 8 4 4, so the sequence is not

arithmetic. Using the formula for the general term,

an1 an

2n1 2n

2.

The sequence 2n is geometric, with 2 as the common ratio.

8.3 Airthmetic and Geometric Sequences

453

(c)

an1

an

lnn

1

ln

n

ln

n

n

1

.

The

difference

depends

on

n,

so

the sequence is not arithmetic. Checking ratios, an1 lnn 1 , so the ratio

an

ln n

also changes with n. The sequence is neither arithmetic nor geometric.

EXAMPLE 3 Arithmetic sequences Show that the sequence is arithmetic; find the common difference and the twentieth term.

(a) an 2n 1 (b) 50, 45, 40, . . . , 55 5n, . . .

Solution

(a) The first few terms of an are 1, 3, 5, 7, . . . , from which it is apparent that each term is 2 more than the preceding term; this is an arithmetic sequence with first term and common difference a 1 and d 2. Check to see that an1 an 2. To find a20, use either the defining formula for the sequence or Equation (1) for the nth term:

a20 2 ? 20 1 39 or a20 a 19d 1 19 ? 2 39.

(b) If bn 55 5n, then bn1 bn 55 5n 1 55 5n 5. This is an arithmetic sequence with a 50, d 5, and so b20 55 5 ? 20 45.

Given the structure of arithmetic and geometric sequences, any two terms completely determine the sequence. Using Equation (1) or (2), two terms of the sequence give us a pair of equations from which we can find the first term and either the common difference or common ratio, as illustrated in the next example.

EXAMPLE 4 Arithmetic sequences Suppose an is an arithmetic sequence with a8 6 and a12 4. Find a, d, and the three terms between a8

and a12.

Solution

From Equation (1), a8 a 7d, and a12 a 11d, from which the difference

is given by a12 a8 4d. Use the given values for a8 and a12 to get 4 6 4d,

or

d

5 2

.

Substitute

5 2

for

d

in

6

a

7d

and

solve

for

a,

a

47 2

.

Find

the

three

terms

between

a8

and

a12

by

successively

adding

5 2

:

a9

a8

5 2

7 2

,

a10

a9

5 2

1,

a11

a10

5 2

3. 2

Therefore,

a9

is

7 2

,

a10

is

1,

and

a11

is

3 2

.

EXAMPLE 5 Geometric sequences Determine whether the sequence is

geometric. If it is geometric, then find the common ratio and the terms a1, a3, and a10.

(a) 2n

(b)

2,

2 ,

2 ,

.

.

.

,

2

1

n1

,...

39

3

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Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

Strategy: The property that identifies a geometric sequence is the common ratio: the values a2 , a3 , a4 , . . .

a1 a2 a3 must all be the same. For a geometric sequence, use Equation (2).

Solution

(a) The first few terms are 2, 4, 8, 16, . . . , each of which is twice the preceding

term. This is a geometric sequence with first term a 2, and common ratio

given

by

r

an1 an

2n1 2n

2.

Using

an

2n,

a1 2 a3 23 8 and a10 210 1024.

(b) Consider the ratio

2 1 n

an1

3

1 ,

an

2 1 n1

3

3

so

the

sequence

is

geometric

with

a

2

and

r

1 3

.

Using

an

2

1 3

n1,

we

get

a1

2,

a3

ar 2

2 9

,

and

a10

ar 9

2

1 3

9

2 19683

.

Partial Sums of Arithmetic Sequences

There is a charming story told about Carl Freidrich Gauss, one of the greatest mathematicians of all time. Early in Gauss' school career, the schoolmaster assigned the class the task of summing the first hundred positive integers, 1 2 3 ? ? ? 99 100. That should have occupied a good portion of the morning, but while other class members busied themselves at their slates calculating 1 2 3, 3 3 6, 6 4 10, and so on, Gauss sat quietly for a few moments, wrote a single number on his slate, and presented it to the teacher. Young Gauss observed that 1 and 100 add up to 101, as do the pair 2 and 99, 3 and 98, and so on up to 50 and 51. There are fifty such pairs, each with a sum of 101, for a total of 50 ? 101 5050, the number he wrote on his slate.

This approach works for the partial sum of any arithmetic sequence, and we will use the method to derive some useful formulas. However, the ideas are more valuable than memorizing formulas. If you understand the idea, you can recreate the formula when needed.

To find a formula for the nth partial sum of an arithmetic sequence, that is, the sum of n consecutive terms, pair the first and last terms, the second and next-to-last, and so on; each pair has the same sum. In fact, it is easier to pair all terms twice, as illustrated with Gauss' sum:

S100 1 2 ? ? ? 99 100 S100 100 99 ? ? ? 2 1 2S100 101 101 ? ? ? 101 101

The sum on the right has 100 terms, so 2S100 100101. Dividing by 2, S100 50101 5050.

For the general case, pairing the terms in Sn and adding gives 2Sn na1 an because there are n pairs, each with the same sum. Dividing by 2 yields the desired formula.

8.3 Airthmetic and Geometric Sequences

455

Partial sums of an arithmetic sequence

Suppose an is an arithmetic sequence. The sum Sn of the first n terms is given by

Sn

na1

2

an

(3)

The formula is probably most easily remembered as n times the average of the first and last terms.

Strategy: Let an 2n 1. To find S25 from Equation (3) requires a1 and a25, which the formula for an can provide. For (b), substitute 1 for a1 and 2n 1 for an in Equation (3) and simplify.

EXAMPLE 6 Partial sums For the sequence an 2n 1,

(a)

evaluate

the

sum

S25

25 k1

2k

1 and

(b) find a formula for Sn.

Solution Follow the strategy.

(a)

By

Equation

(3),

S25

25a1 2

a25 .

Now,

find

a1 and

a25 .

a1 2 ? 1 1 1 and a25 2 ? 25 1 49

Thus,

S25

251

2

49

625.

(b) In general,

Sn

na1

2

an

n1

2n 2

1

n2n 2

n 2.

Hence, Sn n2.

EXAMPLE 7 Arithmetic sequence The sum of the first eight terms of an arithmetic sequence an is 24; the sixth term is 0. Find a formula for an.

Solution For an, first find a and d. Since a6 a 5d, a 5d 0. Express S8 in terms of a and d,

S8

8a

a 2

7d

42a

7d.

Since we are given S8 24, Equation (3) states that 42a 7d 24. This gives a pair of equations to solve for a and d.

a 5d 0 2a 7d 6

We find d 2 and a 10. Therefore, the nth term is

an a n 1d 10 n 12 12 2n.

Partial Sums of Geometric Sequences

The idea of pairing terms, which works so well for arithmetic sequences, does not help with a geometric sequence. Another idea does make the sum easy to calculate

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Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

though. Multiply both sides by r and subtract:

Sn a ar ar 2 ? ? ? ar n1

rSn

ar ar 2 ? ? ? ar n1 ar n

Sn rSn a ar n

Thus,

Sn1 r a1 r n. If r 1, dividing both sides by 1 r yields a formula for Sn.

Partial sums of a geometric sequence

Suppose an is a geometric sequence with r 1. The sum of the first n terms is

Sn

a1 1

r n r

(4)

In the special case where r 1, the geometric sequence is also an arithmetic sequence, and Sn a a a ? ? ? a na.

Strategy: Since it is given that the sequence is geometric, find the common ratio r a2 and then use

a1 Equations (2) and (4).

EXAMPLE 8

Partial sum

Find

an

and

Sn

for

the

geometric

sequence

1 3

,

1 6

,

1 12

,

.

.

.

.

Solution

Follow the strategy. We know

a2 a1

1 6

1 3

1 2

.

From

Equation

that (2),

a1

1 3

and

a2

is

1 6

.

The

common

ratio

is

r

an arn1

1 3

1 2

n1

1 3 ? 2n1 .

Since

r

1 2

,

1

r

1 2

and

1

rn

1

1 2

n.

Applying

Equation

(4)

gives

Sn

a1 1

r n r

1 1 1 n

3

2

1 1

2 3

1

1 2n

.

2

Therefore,

an

3

?

1 2n1

and

Sn

2 3

1

1 2n

EXAMPLE 9 Limit of a sum (a) Find the sum of the first 5, 10, and 100

terms of the geometric sequence from Example 8. (b) Draw a graph of Sn

2 3

1

1

2n

in

0,

c

0,

2,

where

c

is

the

number

of

pixel

columns

of

your

calculator (see inside front cover). Trace to find the smallest integer n for which the

y-value

is

displayed

as

2 3

.

[0, c] by [0, 2]

FIGURE 2 y (2/ 3)(1 1/ 2x)

8.3 Airthmetic and Geometric Sequences

457

Solution

(a)

In

Example

8

we

found

a

formula

for

the

nth

partial

sum,

Sn

2 3

1

1 2n

.

Substituting 5, 10, and 100 for n,

S5

2 3

1

1 25

31 0.646 , 48

S100

2 3

1

1 2100

.

S10

2 3

1023 1024

1023 0.6660 1536

1 The term 2100 has 30 zeros immediately following the decimal point. That

means that S100 is so near 1 that a calculator cannot display the difference except as 1.

(b) In the window 0, c 0, 2 we see a graph something like Figure 2. Because

calculators display trace coordinates differently, you may see something other

than ours, but somewhere between 25 and 35, you should see the y-value

displayed something like 0.6666666 . . . , the nearest your calculator can come

to

displaying

2 3

.

Looking Ahead to Calculus: Infinite Series

As indicated above, each sequence an is associated with a sequence of partial sums Sn, where Sn a1 a2 ? ? ? an. What happens to Sn as n gets larger and larger, that is, as we add more and more terms? We are considering an "infinite sum" written as a1 a2 a3 ? ? ? , or in summation notation,

an.

n1

This is called an infinite series.

Since we cannot add an infinite set of numbers, we need instead the notion of

a limit. In one sense, calculus is the study of limits. It is beyond the scope of this

book to deal with infinite series in general, but for a geometric sequence an, we

can at least get an intuitive feeling for what happens to Sn as n becomes large.

( ) In

Examples

8

and

9,

where

an

3

?

1 2n1

and

Sn

2 3

1

1 2n

, it is reasonable to

assume

that

1 2n

gets

close

to

0

as

n

becomes

large.

In

calculus

notation

lim

nA

1 2n

0,

from which

lim

nA

Sn

2 3

.

1

2

We say that the infinite series, n1 3 ? 2n1 converges to 3 , and we write

n1

3

?

1 2n1

1 3

1 6

1 12

?

?

?

2 3

.

In general, we associate each geometric sequence ar n1 with an infinite geometric series

ar n1 a ar ar 2 ? ? ? ar n1 ? ? ? .

n1

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Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

The only meaning we give to this infinite sum is the limit of the sequence of partial sums,

lim

nA

Sn

lim

nA

a1 1

r n r

,

which depends on limnA r n. Looking at different values of r, we conclude that if r is any number between 1 and 1, then limnA r n 0, from which

lim

nA

Sn

lim

nA

a1 1

r n r

1

a

. r

Infinite geometric series

Associated with every geometric sequence ar n1 is an infinite geometric series

ar n1 a ar ar 2 ? ? ? ar n1 ? ? ? .

n1

If

1

r

1,

then

the

series

converges

to

1

a

, r

and

we

write

n1

ar n1

a

ar

ar 2

?

?

?

ar n1

?

?

?

1

a

. r

(5)

If r 1, then the infinite series does not have a sum, and it diverges.

Repeating decimals. In Section 1.2 we said that the decimal representation of any rational number is a repeating decimal. The following example illustrates how we can use an infinite geometric series to express a repeating decimal as a fraction of integers.

EXAMPLE 10 Repeating decimal Write 1.2454545 ? ? ? 1.245 in terms of an infinite geometric series, then use Equation (5) to express 1.245 in the

p form , where p and q are integers.

q

Solution

1.2454545 ? ? ? 1.2 0.045 0.00045 ? ? ?

6 5

45 103

45 105

?

?

?

The

terms

following

12 10

form

an

infinite

geometric

series

with

a

0.045

45 1000

and

r

0.01

. 1

100

Since

r

is

between

1

and

1,

we

may

use

Equation

(5)

to

express

the sum as

6 5

1

0.045 0.01

6 5

45 990

137 .

110

Therefore,

137 110

and

1.245

represent

the

same

number.

Functions represented by infinite series. The infinite series 1 x x 2 ? ? ? is geometric (with a 1 and r x), so if x is any number between 1 and 1, the

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