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5
MATHEMATICS
OF FINANCE
Copyright © Cengage Learning. All rights reserved.
5.4
Arithmetic and Geometric
Progressions
Copyright © Cengage Learning. All rights reserved.
Arithmetic Progressions
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Arithmetic Progressions
An arithmetic progression is a sequence of numbers in
which each term after the first is obtained by adding a
constant d to the preceding term. The constant d is called
the common difference. For example, the sequence
2, 5, 8, 11, . . .
is an arithmetic progression with common difference equal
to 3.
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Arithmetic Progressions
Observe that an arithmetic progression is completely
determined if the first term and the common difference are
known. In fact, if
a1, a2, a3, . . . , an, . . .
is an arithmetic progression with the first term given by a
and common difference given by d, then by definition we
have
a1 = a
a2 = a1 + d = a + d
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Arithmetic Progressions
a3 = a2 + d = (a + d) + d = a + 2d
a4 = a3 + d = (a + 2d) + d = a + 3d
an = an – 1 + d = a + (n – 2)d + d = a + (n – 1)d
Thus, we have the following formula for finding the nth term
of an arithmetic progression with first term a and common
difference:
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Example 2
Write the first five terms of an arithmetic progression whose
third and eleventh terms are 21 and 85, respectively.
Solution:
Using the equation 16, we obtain
a3 = a + 2d = 21
a11 = a + 10d = 85
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Example 2 – Solution
cont’d
Subtracting the first equation from the second gives
8d = 64, or d = 8. Substituting this value of d into the first
equation yields a + 16 = 21, or a = 5.
Thus, the required arithmetic progression is given by the
sequence
5, 13, 21, 29, 37, . . .
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Arithmetic Progressions
Let Sn denote the sum of the first n terms of an arithmetic
progression with first term a1 = a and common difference d.
Then
Sn = a + (a + d ) + (a + 2d ) +    + [a + (n – 1)d]
(17)
Rewriting the expression for Sn with the terms in reverse
order gives
Sn = [a + (n – 1)d] + [a + (n – 2)d] +    + (a + d) + a
(18)
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Arithmetic Progressions
Adding Equations (17) and (18), we obtain
2Sn = [2a + (n – 1)d] + [2a + (n – 1)d]
+    + [2a + (n – 1)d]
= n[2a + (n – 1)d]
Sn =
[2a + (n – 1)d]
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Arithmetic Progressions
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Applied Example 4 – Company Sales
Madison Electric Company had sales of $200,000 in its first
year of operation.
If the sales increased by $30,000 per year thereafter, find
Madison’s sales in the fifth year and its total sales over the
first 5 years of operation.
Solution:
Madison’s yearly sales follow an arithmetic progression,
with the first term given by a = 200,000 and the common
difference given by d = 30,000. The sales in the fifth year
are found by using the equation 16 with n = 5.
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Applied Example 4 – Solution
cont’d
Thus,
a5 = 200,000 + (5 – 1)30,000 = 320,000
or $320,000.
Madison’s total sales over the first 5 years of operation are
found by using the equation 19 with n = 5. Thus,
S5 =
[2(200,000) + (5 – 1)30,000]
= 1,300,000
or $1,300,000.
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Practice
p. 317 Exercises #1, 5, 11
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