Section 25–2
◆
703
Arithmetic Progressions
495
496
497
498
499
500
0.002 020
0.002 016
0.002 012
0.002 008
0.002 004
0.002 000
6.782 784
6.784 800
6.786 813
6.788 821
6.790 825
6.792 825
0.997 980
0.997 984
0.997 988
0.997 992
0.997 996
0.998 000
Does the series converge or diverge?
Solution: Again we apply the three tests.
1. We see that the terms are getting smaller, so this test is not conclusive.
2. The ratio of the terms appears to approach 1, so this test also is not conclusive.
3. The partial sum Sn does not seem to approach a limit, but appears to keep growing even
after 495 terms.
It thus appears that this series diverges.
Exercise 1
◆
◆◆◆
Sequences and Series
Write the first five terms of each series, given the general term.
1. un 3n
2. un 2n 3
n1
3. un n2
n
2
4. un n
Deduce the general term of each series. Use it to predict the next two terms.
5. 2 4 6 6. 1 8 27 2 4 8 16
7. 4 5 6
7
8. 2 5 10 Deduce a recursion relation for each series. Use it to predict the next two terms.
9. 1 5 9 10. 5 15 45 11. 3 9 81 12. 5 8 14 26 25–2 Arithmetic Progressions
Recursion Formula
We stated earlier that an arithmetic progression (or AP) is a sequence of terms in which each term
after the first equals the sum of the preceding term and a constant, called the common difference,
d. If an is any term of an AP, the recursion formula for an AP is as follows:
AP:
Recursion
Formula
an an1 d
236
Each term of an AP after the first equals the sum of the preceding term and the common
difference.
704
Chapter 25
◆
Sequences, Series, and the Binomial Theorem
◆◆◆ Example 10: The following sequences are arithmetic progressions. The common difference for each is given in parentheses:
(a) 1, 5, 9, 13, . . .
(d 4)
(b) 20, 30, 40, 50, . . .
(d 10)
(c) 75, 70, 65, 60, . . .
(d 25)
We see that each series is increasing when d is positive and decreasing when d is negative.
◆◆◆
General Term
For an AP whose first term is a and whose common difference is d, the terms are
a, a d, a 2d, a 3d, a 4d, . . .
We see that each term is the sum of the first term and a multiple of d, where the coefficient of d
is one less than the number n of the term. So the nth term an is given by the following equation:
The general term is sometimes
called the last term, but this is
not an accurate name since the
AP continues indefinitely.
AP:
General
Term
an a (n 1)d
237
The nth term of an AP is found by adding the first term and (n 1) times the common
difference.
◆◆◆ Example 11: Find the twentieth term of an AP that has a first term of 5 and common difference of 4.
Solution: By Eq. 237, with a 5, n 20, and d 4,
a20 5 19(4) 81
◆◆◆
Of course, Eq. 237 can be used to find any of the four quantities (a, n, d, or an) given the
other three.
◆◆◆
Example 12: Write the AP whose eighth term is 19 and whose fifteenth term is 33.
Solution: Applying Eq. 237 twice gives
33 a 14d
19 a 7d
We now have two equations in two unknowns, which we solve simultaneously. Subtracting the
second from the first gives 14 7d, so d 2. Also, a 33 14d 5, so the general term of
our AP is
an 5 (n 1)2
and the AP is then
5, 7, 9, . . .
◆◆◆
AP: Sum of n Terms
Let us derive a formula for the sum sn of the first n terms of an AP. Adding term by term gives
sn a (a d) (a 2d ) (an d ) an
(1)
Section 25–2
◆
705
Arithmetic Progressions
or, written in reverse order,
sn an (an d) (an 2d) (a d) a
(2)
Adding Eqs. (1) and (2) term by term gives
2sn (a an) (a an) (a an) n(a an)
Dividing both sides by 2 gives the following formula:
AP: Sum of
n Terms
n
sn (a an)
2
238
The sum of n terms of an AP is half the product of n and the sum of the first and nth terms.
◆◆◆
Example 13: Find the sum of 10 terms of the AP
2, 5, 8, 11, . . .
Solution: From Eq. 237, with a 2, d 3, and n 10,
a10 2 9(3) 29
Then, from Eq. 238,
10 (2 29)
s10 155
2
◆◆◆
We get another form of Eq. 238 by substituting the expression for an from Eq. 237, as
follows:
AP: Sum of
n Terms
n
sn [2a (n 1)d]
2
239
This form is useful for finding the sum without first computing the nth term.
◆◆◆
Example 14: We repeat Example 13 without first having to find the tenth term. From Eq.
239,
10 [2 (2) 9 (3)]
s10 155
2
Sometimes the sum is given and we must find one of the other quantities in the AP.
◆◆◆
706
Chapter 25
◆◆◆
◆
Sequences, Series, and the Binomial Theorem
Example 15: How many terms of the AP 5, 9, 13, . . . give a sum of 275?
Solution: We seek n so that sn 275. From Eq. 239, with a 5 and d 4,
n
275 [2 (5) (n 1)4]
2
5n 2n (n 1)
Removing parentheses and collecting terms gives the quadratic equation
2n2 3n 275 0
From the quadratic formula,
3 兹9 4 (2) (275)
n 11 or
2 (2)
We discard the negative root and get 11 terms as our answer.
12.5
◆◆◆
Arithmetic Means
The first term a and the last (nth) term an of an AP are sometimes called the extremes, while the
intermediate terms, a2, a3, . . ., an1, are called arithmetic means. We now show, by example,
how to insert any number of arithmetic means between two extremes.
◆◆◆
Example 16: Insert five arithmetic means between 3 and 9.
Solution: Our AP will have seven terms, with a first term of 3 and a seventh term of 9. From
Eq. 237,
9 3 6d
from which d 2. The progression is then
3, 1, 1, 3, 5, 7, 9
and the five arithmetic means are 1, 1, 3, 5, and 7.
◆◆◆
Average Value
Let us insert a single arithmetic mean between two numbers.
◆◆◆
Example 17: Find a single arithmetic mean m between the extremes a and b.
Solution: The sequence is a, m, b. The common difference d is m a b m. Solving for m gives
2m b a
Dividing by 2 gives
ab
m 2
which agrees with the common idea of an average of two numbers as the sum of those numbers
◆◆◆
divided by 2.
Harmonic Progressions
A sequence is called a harmonic progression if the reciprocals of its terms form an arithmetic
progression.
◆◆◆
Example 18: The sequence
1
1, ,
3
is a harmonic progression because the
an AP.
1 1 1
1
, , , . . .
5 7 9 11
reciprocals of the terms, 1, 3, 5, 7, 9, 11, . . ., form
,
◆◆◆
Section 25–2
◆
707
Arithmetic Progressions
It is not possible to derive an equation for the nth term or for the sum of a harmonic
progression. However, we can solve problems involving harmonic progressions by taking the
reciprocals of the terms and using the formulas for the AP.
◆◆◆
Example 19: Find the tenth term of the harmonic progression
2
2, ,
3
2
5
, . . .
Solution: We write the reciprocals of the terms,
3
2
1
2
,
5
2
,
, . . .
and note that they form an AP with a 1 and d 1. The tenth term of the AP is then
2
an a (n 1)d
1
a10 (10 1)(1)
2
19
2
19
2
The tenth term of the harmonic progression is the reciprocal of , or .
2
19
◆◆◆
Harmonic Means
To insert harmonic means between two terms of a harmonic progression, we simply take the
reciprocals of the given terms, insert arithmetic means between those terms, and take reciprocals again.
◆◆◆
Example 20: Insert three harmonic means between 29 and 2.
Solution: Taking reciprocals, our AP is
9
2
1
2
, , , , In this AP, a 92 , n 5, and a5 12 . We can find the common difference d from the equation
an a (n 1)d
1
2
9
2
4d
4 4d
d 1
Having the common difference, we can fill in the missing terms of the AP,
9
2
,
7
2
,
5
2
,
3
2
,
1
2
Taking reciprocals again, our harmonic progression is
2
9
,
2
7
,
2
5
,
2
3
,
2
◆◆◆
708
Chapter 25
◆
Sequences, Series, and the Binomial Theorem
Exercise 2
◆
Arithmetic Progressions
General Term
1.
2.
3.
4.
5.
Find the fifteenth term of an AP with first term 4 and common difference 3.
Find the tenth term of an AP with first term 8 and common difference 2.
Find the twelfth term of an AP with first term 1 and common difference 4.
Find the ninth term of an AP with first term 5 and common difference 2.
Find the eleventh term of the AP
9, 13, 17, . . .
6. Find the eighth term of the AP
5, 8, 11, . . .
7. Find the ninth term of the AP
x, x 3y, x 6y, . . .
8. Find the fourteenth term of the AP
1,
6
7
,
5
7
, . . .
For problems 9 through 12, write the first five terms of each AP.
9. First term is 3 and thirteenth term is 55.
10. First term is 5 and tenth term is 32.
11. Seventh term is 41 and fifteenth term is 89.
12. Fifth term is 7 and twelfth term is 42.
13. Find the first term of an AP whose common difference is 3 and whose seventh term is 11.
14. Find the first term of an AP whose common difference is 6 and whose tenth term is 77.
Sum of an Arithmetic Progression
15. Find the sum of the first 12 terms of the AP
3, 6, 9, 12, . . .
16. Find the sum of the first five terms of the AP
1, 5, 9, 13, . . .
17. Find the sum of the first nine terms of the AP
5, 10, 15, 20, . . .
18. Find the sum of the first 20 terms of the AP
1, 3, 5, 7, . . .
19. How many terms of the AP 4, 7, 10, . . . will give a sum of 375?
20. How many terms of the AP 2, 9, 16, . . . will give a sum of 270?
Arithmetic Means
21.
22.
23.
24.
Insert two arithmetic means between 5 and 20.
Insert five arithmetic means between 7 and 25.
Insert four arithmetic means between 6 and 9.
Insert three arithmetic means between 20 and 56
Section 25–3
◆
709
Geometric Progressions
Harmonic Progressions
25. Find the fourth term of the harmonic progression
3 3 3
, , , . . .
5 8 11
26. Find the fifth term of the harmonic progression
4 4 4
, , , . . .
19 15 11
Harmonic Means
7
27. Insert two harmonic means between 79 and .
15
6
28. Insert three harmonic means between and 65 .
21
Applications
29. Loan Repayment: A person agrees to repay a loan of $10,000 with an annual payment of
$1,000 plus 8% of the unpaid balance.
(a) Show that the interest payments alone form the AP: $800, $720, $640, . . . .
(b) Find the total amount of interest paid.
30. Simple Interest: A person deposits $50 in a bank on the first day of each month, at the same
time withdrawing all interest earned on the money already in the account.
(a) If the rate is 1% per month, computed monthly, write an AP whose terms are the
amounts withdrawn each month.
(b) How much interest will have been earned in the 36 months following the first deposit?
31. Straight-Line Depreciation: A certain milling machine has an initial value of $150,000 and
a scrap value of $10,000 twenty years later. Assuming that the machine depreciates the
same amount each year, find its value after 8 years.
32. Salary or Price Increase: A person is hired at a salary of $40,000 and receives a raise of
$2,500 at the end of each year. Find the total amount earned during 10 years.
33. Freely Falling Body: A freely falling body falls gⲐ2 metres during the first second, 3gⲐ2 m
during the next second, 5gⲐ2 m during the third second, and so on, where g ⬵ 9.807 mⲐs2.
Find the total distance the body falls during the first 10 s.
34. Using the information of problem 33, show that the total distance s fallen in t seconds is
s 12 gt 2.
25–3 Geometric Progressions
Recursion Formula
A geometric sequence or geometric progression (GP) is one in which each term after the first is
formed by multiplying the preceding term by a factor r, called the common ratio. Thus if an is
any term of a GP, the recursion relation is as follows:
GP:
Recursion
Formula
an ran1
240
Each term of a GP after the first equals the product of the preceding term and the common ratio.
To find the amount of
depreciation for each year, divide
the total depreciation (initial
value – scrap value) by the
number of years of depreciation.