Arithmetic Series

Contents hide

1 LEARNING OBJECTIVES:

2 MATH CONCEPTS

3 IMPORTANCE

4 ARITHMETIC SERIES

5 ILLUSTRATIVE EXAMPLES

6 ILLUSTRATIVE EXAMPLES

LEARNING OBJECTIVES:

To define and illustrate arithmetic series
To solve for the value of an arithmetic series
To apply the concepts of arithmetic series to solve real-life problems

MATH CONCEPTS

ARITHMETIC SERIES $(S_{n})$ – it is the sum of the terms in an arithmetic sequence/progression.
GENERAL FORMULA FOR ARITHMETIC SERIES – It is a formula to be used if the first term $(t_{1})$ , common difference (d), and the number of terms (n) are given.
ALTERNATE FORMULA FOR ARITHMETIC SERIES– It is the formula to be used if the first term $(t_{1})$ and the last term $(t_{n})$ are given.

IMPORTANCE

Arithmetic series has plenty of applications in real life. Many businesses consider the pattern/behavior of their profit to come up with significant decisions.
In economics, this can be used in predicting and analyzing GNP and GDP of a country.

When we want to get the sum of the terms of an arithmetic sequence, we are actually getting the value of its arithmetic series $S_{n}$ . To find the value, let’s add the terms. Assume that:

We could have also started with the nth term and successively subtracted the common difference, so

$S_{n}=t_{n}+(t_{n}-d)+(t_{n}-2d)+(t_{n}-3d)+ ... + \left [ t_{n}-(n-1)d \right ]$

We could actually get the sum of the arithmetic sequence either way. However, it is noticeable that if you added those two equations together, terms add out.

$S_{n}=t_{1}+(t_{1}+d)+(t_{1}+d)+(t_{1}+d)+ ... + \left [ t_{1}+(n-1)d \right ]$

$+S_{n}=t_{n}+(t_{n}-d)+(t_{n}-2d)+(t_{n}-3d)+ ... + \left [ t_{n}-(n-1)d \right ]$

$2S_{n} = (t_{1}+t_{n})+(t_{1}+t_{n})+(t_{1}+t_{n})+...+(t_{1}+t_{n})$

Obviously, all d’s were cancelled out. Thus, we can factor out the sum as:

$2S_{n} = n(t_{1}+t_{n})$

Dividing both sides by 2, we will obtain:

$S_{n}=n\frac{(t_{1}+t_{n})}{2}$

It can also be expressed as:

$S_{n}=\frac{n}{2}(t_{1}+t_{n})$

By substituting $t_{n}=t_{1}+(n-1)d$ , into the previous formula,

$S_{n} = \frac{n}{2}(t_{1}+\left [ t_{1}+(n-1)d \right ])=S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

In summary,

ILLUSTRATIVE EXAMPLES

A. Find the sum of each arithmetic sequence.
1) 3, 5, 7 ,…. to the 31st term.
2) 30, 21, 12,. …to the 45th term.

SOLUTIONS:

1) The first term is 3 and based on the first terms, the common difference is 2. Thus, it is more convenient to use the alternate formula to get the sum of the terms up to the 31st term.

$t_{1}=3$ , d = 2, n = 31

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$S_{31} = \frac{31}{2}\left [ 2(3)+(31-1)2 \right ]$

$S_{31} = \frac{31}{2}\left [ 6+(30)2 \right ]=\frac{31}{2}\left [ 6+60 \right ]=\frac{31}{2}(66)=(15.5)(66)$

$S_{31} = 93$

2) The first term is 30 and based on the first terms, the common difference is – 9. Thus, it is more convenient to use the alternate formula to get the sum of the terms up to the 31st term.

$t_{1}=30$ , d = -9, n = 45

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$S_{45} = \frac{45}{2}\left [ 2(30)+(45-1)(-9) \right ]$

$S_{45} = \frac{45}{2}\left [ 60+(44)(-9) \right ]=\frac{45}{2}\left [ 60-396 \right ]=(22.5)(-336)$

$S_{45} = -7560$

ILLUSTRATIVE EXAMPLES

B. Solve the following word problems.

Find the sum of the first 15 numbers that are divisible by 3.
Find the sum of the integers between 2 and 100 which are divisible by 4.
If £ 5 is saved on October 1, £ 10 on October 2, £ 15 on October 3, and so on, how much is saved at the end of October?
An arithmetic sequence has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference.

SOLUTIONS:

1. We are asked to find the sum of the first 15 numbers that are divisible by 3. So these numbers are the multiple of 3 ( 3, 6, 9, 12, 15, 18, …) with the first term as 3 and the common difference is also 3. Use the alternate formula:

$t_{1}=3$ , d = 3, n = 15

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$S_{15} =\frac{15}{2}\left [ 2(3)+(15-1)(3) \right ]$

$S_{15} =\frac{15}{2}\left [ 6+(14)(3) \right ]=\frac{15}{2}\left [ 6+42 \right ]=\frac{15}{2}(48)=(7.5)(48)$

$S_{15} =360$

2. We need to find the sum of the integers between 2 and 100 that are divisible by 4. Given this condition, the first integer that is divisible by 4 is 4 and the last number that is divisible by 4 preceding 100 is 96. Thus, the first term is 4, the common difference is also 4, and the last term is 96. With these circumstances, we should use the general formula because both the first and last terms are given. However, the position of 96 as a term is needed to identify first. Let’s use the formula of finding the nth term of the arithmetic sequence to solve for n.

$t_{n}=t_{1}+(n-1)d$

$t_{n}=96$ , $t_{1}=4$ , d = 4

$96=4+(n-1) 4 = 4 + 4n - 4$

96 = 4n

24 = n

Since we have the value n = 24, we can already solve for the sum. Let’s use the general formula.

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$S_{24} = 24\frac{(4+96)}{2}=24\times \frac{100}{2}=24\times 50$

$S_{24} = 1200$

3. We need to find the total savings for the entire month of October. Remember that October has 31 days. Given that the first term is 5 and the common difference is also 5. Use the alternate formula:

$t_{1}=5$ , d = 5, n = 31

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$S_{31} = \frac{31}{2}\left [ 2(5)+(31-1)(5) \right ]$

$S_{31} = \frac{31}{2}\left [ 10+(30)(5) \right ]=\frac{31}{2}\left [ 10+150 \right ]=\frac{31}{2}(160)=(15.5)(160)$

$S_{31} = 2480$

Thus, the total savings for the entire month of October is £ 2, 480.

4. We are given that a = 3. We are also given some information about the sums S8 and S5, and we want to find the common difference. So we shall use the alternate formula.

$S_{n} = \frac{n}{2}\left [ 2t_{1}+(n-1)d \right ]$

$\frac{8}{2}\left [ 6+(7)d \right ]=2\frac{5}{2}\left [ 6+(4)d \right ]<br>$