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Math Lesson 12.2.6 - Combined Series

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Welcome to our Math lesson on Combined Series, this is the sixth lesson of our suite of math lessons covering the topic of Working with Arithmetic and Geometric Series. How to find the Sum of the First n-Terms of a Series., you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Combined Series Explained

Knowing how to deal with arithmetic and geometric series allows finding many more things related to combined series, such as for example in series with rational terms, where numerators form a different pattern from denominators. Let's consider an example to clarify this point.

Example 8

Calculate the sum of the first 30 terms of the series deriving from the sequence below.

1/2; 3/4; 5/2; 7/4

Solution 8

It is evident that numerators form an arithmetic sequence where the first term is x1 = 1 and the common difference is d = 2. On the other hand, denominators form an alternate sequence with terms 2, 4, 2, 4, etc. Therefore, it is better to form two separate arithmetic sequences: one containing only fractions with denominator 2 and the other, only fractions with denominator 4. Therefore, finding the sum of the first 30 terms of this combined sequence means finding the total sum of two separate series with 15 terms each, i.e.

Stotal = SA + SB = (1/2 + 5/2 + ⋯)15 terms + (3/4 + 7/4 + ⋯)15 terms

Thus, for the numerators of the first series SA, we have x1 = 1 and d = 4. Therefore, the 15th term (n = 15) of this series of numerators is

xnA = x1A + (n - 1) ∙ dA
x15A = 1 + (15 - 1) ∙ 4
= 1 + 14 ∙ 4
= 1 + 56
= 57

This means the true values of the first and the 15th terms of SA are

y1A = 3/2 and y15A = 57/2

Moreover, the true difference of this series is dA = .4/2 = 2.

As for the numerators of the second series SB, we have x1 = 3 and d = 4. Therefore, the 15th term (n = 15) of this series is

xnB = x1B + (n - 1) ∙ dB
x15B = 3 + (15 - 1) ∙ 4
= 3 + 14 ∙ 4
= 3 + 56
= 59

This means the true values of the first and the 15th terms of SB are

y1B = 3/4 and y15B = 59/4

Moreover, the true difference of this series is dB = 4/4 = 1.

Therefore, the values of the corresponding series are:

SA = (x1A + xnA ) ∙ dA/2
= (1/2 + 57/2) ∙ 2/2
= 58/2
= 29

and

SB = (x1B + xnB ) ∙ dB/2
= (3/4 + 59/4) ∙ 1/2
= 62/4/2
= 62/41/2
= 62/8
= 31/4

Hence, the sum of the first 30 terms of the original series is

Stotal = SA + SB
= 29 + 31/4
= 116/4 + 31/4
= 147/4

Sometimes, the situations involving series are given in the form of the question below.

Example 9

The sum of the first n terms Sn of a particular arithmetic progression is given by Sn = 2n + 5n2. Find the first term and the common difference.

Solution 9

Since the sum of the first n terms in an arithmetic series is

Sn = (x1 + xn ) ∙ n/2

we can write for the specific case

(x1 + xn ) ∙ n/2 = 2n + 5n2

Doing some equivalent transformations on the right side yields

(x1 + xn ) ∙ n/2 = (2 + 5n) ∙ n
(x1 + xn ) ∙ n/2 = (4 + 10n) ∙ n/2

Therefore, comparing the terms on both sides we obtain x1 = 4 (the first term) and xn = 10n (the general term). Hence, since in an arithmetic progression

xn = x1 + (n - 1) ∙ d

then, the expression for common difference d is

d = xn - x1/n - 1
= 10n - 4/n - 1

More Working with Arithmetic and Geometric Series. How to find the Sum of the First n-Terms of a Series. Lessons and Learning Resources

Sequences and Series Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
12.2Working with Arithmetic and Geometric Series. How to find the Sum of the First n-Terms of a Series.
Lesson IDMath Lesson TitleLessonVideo
Lesson
12.2.1Series versus Sequences
12.2.2The Gauss Method and Arithmetic Series
12.2.3An Alternative Formula for the Calculation of Sn in Arithmetic Series
12.2.4Geometric Series
12.2.5The Combination of Sequences and Series
12.2.6Combined Series
12.2.7The Practical Applications of Number Series and Sequences

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