# What Is The General Term Of An Arithmetic Sequence? (Question)

An arithmetic sequence is a sequence where the difference d between successive terms is constant. The general term of an arithmetic sequence can be written in terms of its first term a1, common difference d, and index n as follows: an=a1+(n−1)d. An arithmetic series is the sum of the terms of an arithmetic sequence.

## What is general term in arithmetic?

An arithmetic sequence is a string of numbers where each number is the previous number plus a constant. This constant difference between each pair of successive numbers in our sequence is called the common difference. The general term is the formula that is used to calculate any number in an arithmetic sequence.

## What is the general term of sequences?

The nth (or general) term of a sequence is usually denoted by the symbol an. a1=2, the second term is a2=6 and so forth. A term is multiplied by 3 to get the next term. If you know the formula for the nth term of a sequence in terms of n, then you can find any term.

## What does the general term mean?

Definition of general term: a mathematical expression composed of variables and constants that yields the successive terms of a sequence or series when integers are substituted for one of the variables often denoted by k.

## How do you find the arithmetic sequence?

The arithmetic sequence formula is given as, an=a1+(n−1)d a n = a 1 + ( n − 1 ) d where, an a n = a general term, a1 a 1 = first term, and and d is the common difference. This is to find the general term in the sequence.

## What is general term example?

Definition and examples of a general term: Here is an example of how to substitute a general term for a list of items in order to summarize this sentence: details: “John bought some milk, bread, fruit, cheese, potato chips, butter, hamburger and buns.” general term: ” John bought some groceries.”

## Introduction to Arithmetic Progressions

For example, Eric W. Weisstein’s “Arithmetic series” is included in the Encyclopedia of Mathematics (EMS Press, 2001); “Arithmetic progression” is also included in the Encyclopedia of Mathematics (EMS Press, 2001). ; Weisstein, Eric W. “Arithmetic series.” MathWorld; Weisstein, Eric W. “Arithmetic series.”

1. Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP) are all types of progression.

AP, also known as Arithmetic Sequence, is a sequence or series of integers in which the common difference between two subsequent numbers in the series is always the same. As an illustration: Series 1: 1, 3, 5, 7, 9, and 11. Every pair of successive integers in this sequence has a common difference of 2, which is always true. Series 2: numbers 28, 25, 22, 19, 16, 13,. There is no common difference between any two successive numbers in this series; instead, there is a strict -3 between any two consecutive numbers.

### Terminology and Representation

• Common difference, d = a 2– a 1= a 3– a 2=. = a n– a n – 1
• A n= n thterm of Arithmetic Progression
• S n= Sum of first n elements in the series
• A n= n

### General Form of an AP

Given that ais treated as a first term anddis treated as a common difference, the N thterm of the AP may be calculated using the following formula: As a result, using the above-mentioned procedure to compute the n terms of an AP, the general form of the AP is as follows: Example: The 35th term in the sequence 5, 11, 17, 23,. is to be found. Solution: When looking at the given series,a = 5, d = a 2– a 1= 11 – 5 = 6, and n = 35 are the values. As a result, we must use the following equations to figure out the 35th term: n= a + (n – 1)da n= 5 + (35 – 1) x 6a n= 5 + 34 x 6a n= 209 As a result, the number 209 represents the 35th term.

### Sum of n Terms of Arithmetic Progression

The arithmetic progression sum is calculated using the formula S n= (n/2)

### Derivation of the Formula

Allowing ‘l’ to signify the n thterm of the series and S n to represent the sun of the first n terms of the series a, (a+d), (a+2d),., a+(n-1)d S n = a 1 plus a 2 plus a 3 plus .a n-1 plus a n S n= a + (a + d) + (a + 2d) +. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. (1) When we write the series in reverse order, we obtain S n= l + (l – d) + (l – 2d) +. + (a + 2d) + (a + d) + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a … (2) Adding equations (1) and (2) results in equation (2).

+ (a + l) + (a + l) + (a + l) +.

(3) As a result, the formula for calculating the sum of a series is S n= (n/2)(a + l), where an is the first term of the series, l is the last term of the series, and n is the number of terms in the series.

d S n= (n/2)(a + a + (n – 1)d)(a + a + (n – 1)d) S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) Observation: The successive terms in an Arithmetic Progression can alternatively be written as a-3d, a-2d, a-d, a, a+d, a+2d, a+3d, and so on.

### Sample Problems on Arithmetic Progressions

Problem 1: Calculate the sum of the first 35 terms in the sequence 5,11,17,23, and so on. a = 5 in the given series, d = a 2–a in the provided series, and so on. The number 1 equals 11 – 5 = 6, and the number n equals 35. S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) S n= (35/2)(2 x 5 + (35 – 1) x 6)(35/2)(2 x 5 + (35 – 1) x 6) S n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) A = 35214/2A = 3745S n= 35214/2A = 3745 Find the sum of a series where the first term of the series is 5 and the last term of the series is 209, and the number of terms in the series is 35, as shown in Problem 2.

Problem 2.

S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) A = 35214/2A = 3745S n= 35214/2A = 3745 Problem 3: A amount of 21 rupees is divided among three brothers, with each of the three pieces of money being in the AP and the sum of their squares being the sum of their squares being 155.

Solution: Assume that the three components of money are (a-d), a, and (a+d), and that the total amount allocated is in AP.

155 divided by two equals 155 Taking the value of ‘a’ into consideration, we obtain 3(7) 2+ 2d.

2= 4d = 2 = 2 The three portions of the money that was dispersed are as follows:a + d = 7 + 2 = 9a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5 As a result, the most significant portion is Rupees 9 million.

## How to find the formula for the general term of a sequence — Krista King Math

Take, for example, the following sequence:?we must recognize that the first term of the series is -1? What is the second term in the sequence? -2? What is the third word in the sequence? -3? What is the fourth term in the sequence? -4? What is the fifth term in the sequence? -5? To put it another way, when?n=1?, the value of the sequence is?-1? Whenever n=2 is reached, the value of the sequence is equal to?-2? Whenever n=3? is reached, what is the value of the sequence?-3? Whenever n=4 is reached, the value of the sequence is -4.

• is reached, what is the value of the sequence?-5?
• Even though this was a simple example, we’ll always use the same procedure to get the general term of any sequence.
• It is always necessary to pay close attention to the signals of the phrases that appear in the sequence.
• will be positive if and only if all of the words in the sequence are positively oriented.
• The terms will be included in?a n?
• Assuming that the even terms (?n=2, 4, 6, and so on?) are negative, the a n will include?(-1)?
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## 7.2 – Arithmetic Sequences

An arithmetic sequence is a succession of terms in which the difference between consecutive terms is a constant number of terms.

## Common Difference

The common difference is named as such since it is shared by all subsequent pairs of words and is thus referred to as such. It is indicated by the letter d. If the difference between consecutive words does not remain constant throughout time, the sequence is not mathematical in nature. The common difference can be discovered by removing the terms from the sequence that are immediately preceding them. The following is the formula for the common difference of an arithmetic sequence: d = an n+1- a n

## General Term

A linear function is represented as an arithmetic sequence. As an alternative to the equation y=mx+b, we may write a =dn+c, where d is the common difference and c is a constant (not the first term of the sequence, however). Given that each phrase is discovered by adding the common difference to the preceding term, this definition is a k+1 = anagrammatical definition of the term “a k +d.” For each phrase in the series, we’ve multiplied the difference by one less than the number of times the term appears in the sequence.

For the second term, we’ve just included the difference once in the calculation. For the third term, we’ve multiplied the difference by two to get the total. When considering the general term of an arithmetic series, we may use the following formula: 1+ (n-1) d

## Partial Sum of an Arithmetic Sequence

A series is made up of a collection of sequences. We’re looking for the n th partial sum, which is the sum of the first n terms in the series, in this case. The n thpartial sum shall be denoted by the letter S n. Take, for example, the arithmetic series. S 5 = 2 + 5 + 8 + 11 + 14 = S 5 = 2 + 5 + 8 + 11 + 14 = S 5 The sum of an arithmetic series may be calculated in a straightforward manner. S 5 is equal to 2 + 5 + 8 + 11 + 14 The secret is to arrange the words in a different sequence. Because addition is commutative, altering the order of the elements has no effect on the sum.

• 2*S 5= (2+14) + (5+11) + (8+8) + (11+5) + (14+2) = (2+14) + (5+11) + (8+8) + (11+5) + (14+2) Take note that each of the amounts on the right-hand side is a multiple of 16.
• 2*S 5 = 5*(2 + 14) = 2*S 5 Finally, divide the total item by two to obtain the amount, not double the sum as previously stated.
• This would be 5/2 * (16) = 5(8) = 40 as a total.
• The number 5 refers to the fact that there were five terms, n.
• In this case, we added the total twice and it will always be a 2.
• Another formula for the n th partial sum of an arithmetic series is occasionally used in conjunction with the previous one.
• Instead of trying to figure out the n thterm, it is preferable to find out what it is and then enter that number into the formula.

### Example

Find the sum of the numbers k=3 to 17 using the given information (3k-2). 7 is obtained by putting k=3 into 3k-2 and obtaining the first term. The last term is 3(17)-2 = 49, which is an integer. There are 17 – 3 + 1 = 15 words in the sentence. As a result, 15 / 2 * (7 + 49) = 15 / 2 * 56 = 420 is the total. Take note of the fact that there are 15 words in all. When the lower limit of the summation is 1, there is minimal difficulty in determining the number of terms in the equation. When the lower limit is any other number, on the other hand, it appears to cause confusion among individuals.

No one would dispute the fact that if you counted from 1 to 10, there are a total of ten numbers. The difference between 10 and 1 is, on the other hand, merely 9. As a result, when calculating the number of words, the formula is as follows: higher limit minus lower limit plus one.

## Arithmetic Sequences and Series

 HomeLessonsArithmetic Sequences and Series Updated July 16th, 2020
 Introduction Sequences of numbers that follow a pattern of adding a fixed number from one term to the next are called arithmetic sequences. The following sequences are arithmetic sequences:Sequence A:5, 8, 11, 14, 17,.Sequence B:26, 31, 36, 41, 46,.Sequence C:20, 18, 16, 14, 12,.Forsequence A, if we add 3 to the first number we will get the second number.This works for any pair of consecutive numbers.The second number plus 3 is the third number: 8 + 3 = 11, and so on.Forsequence B, if we add 5 to the first number we will get the second number.This also works for any pair of consecutive numbers.The third number plus 5 is the fourth number: 36 + 5 = 41, which will work throughout the entire sequence.Sequence Cis a little different because we need to add -2 to the first number to get the second number.This too works for any pair of consecutive numbers.The fourth number plus -2 is the fifth number: 14 + (-2) = 12.Because these sequences behave according to this simple rule of addiing a constant number to one term to get to another, they are called arithmetic sequences.So that we can examine these sequences to greater depth, we must know that the fixed numbers that bind each sequence together are called thecommon differences. Mathematicians use the letterdwhen referring to these difference for this type of sequence.Mathematicians also refer to generic sequences using the letteraalong with subscripts that correspond to the term numbers as follows:This means that if we refer to the fifth term of a certain sequence, we will label it a 5.a 17is the 17th term.This notation is necessary for calculating nth terms, or a n, of sequences.Thed -value can be calculated by subtracting any two consecutive terms in an arithmetic sequence.where n is any positive integer greater than 1.Remember, the letterdis used because this number is called thecommon difference.When we subtract any two adjacent numbers, the right number minus the left number should be the same for any two pairs of numbers in an arithmetic sequence. To determine any number within an arithmetic sequence, there are two formulas that can be utilized.Here is therecursive rule.The recursive rule means to find any number in the sequence, we must add the common difference to the previous number in this list.Let us say we were given this arithmetic sequence.
First, we would identify the common difference.We can see the common difference is 4 no matter which adjacent numbers we choose from the sequence.To find the next number after 19 we have to add 4.19 + 4 = 23.So, 23 is the 6th number in the sequence.23 + 4 = 27; so, 27 is the 7th number in the sequence, and so on.What if we have to find the 724th term?This method would force us to find all the 723 terms that come before it before we could find it.That would take too long.So, there is a better formula.It is called theexplicit rule.So, if we want to find the 724th term, we can use thisexplicit rule.Our n-value is 724 because that is the term number we want to find.The d-value is 4 because it is thecommon difference.Also, the first term, a 1, is 3.The rule gives us a 724= 3 + (724 – 1)(4) = 3 + (723)(4) = 3 + 2892 = 2895.
 Each arithmetic sequence has its own unique formula.The formula can be used to find any term we with to find, which makes it a valuable formula.To find these formulas, we will use theexplicit rule.Let us also look at the following examples.Example 1 : Let’s examinesequence Aso that we can find a formula to express its nth term.If we match each term with it’s corresponding term number, we get:

The fixed number, which is referred to as the common differenceor d-value, is three. We may use this information to replace the explicit rule in the code. As an example, a n= a 1+ (n – 1)d. a n = a 1 + a (n – 1) the value of da n= 5 + (n-1) (3) the number 5 plus 3n – 3a the number 3n + 2a the number 3n + 2 When asked to identify the 37th term in this series, we would compute for a 37 in the manner shown below. the product of 3n and 2a 37 is 3(37) + 2a 37 is 111 + 2a 37 is 113. Exemple No. 2: For sequence B, find a formula that specifies the nth term in the series.

We can identify a few facts about it.Its first term, a 1, is 26.Itscommon differenceor d-value is 5.We can substitute this information into theexplicit rule.a n= a 1+ (n – 1)da n= 26 + (n – 1)(5)a n= 26 + 5n – 5a n= 5n + 21Now, we can use this formula to find its 14th term, like so. a n= 5n + 21a 14= 5(14) + 21a 14= 70 + 21a 14= 91ideo:Finding the nth Term of an Arithmetic Sequence uizmaster:Finding Formula for General Term
 It may be necessary to calculate the number of terms in a certain arithmetic sequence. To do so, we would need to know two things.We would need to know a few terms so that we could calculate the common difference and ultimately the formula for the general term.We would also need to know the last number in the sequence.Once we know the formula for the general term of a sequence and the last term, the procedure involves the use of algebra.Use the two examples below to see how it is done.Example 1 : Find the number of terms in the sequence 5, 8, 11, 14, 17,., 47.This issequence A.In theprevious section, we found the formula to be a n= 3n + 2 for this sequence.We will use this along with the fact the last number, a n, is 47.We will plug this into the formula, like so.a n= 3n + 247 = 3n + 245 = 3n15 = nn = 15This means there are 15 numbers in this arithmetic sequence.Example 2 : Find the number of terms in the arithmetic sequence 20, 18, 16, 14, 12,.,-26.Our first task is to find the formula for this sequence given a 1= 20 and d = -2.We will substitute this information into theexplicit rule, like so.a n= a 1+ (n – 1)da n= 20 + (n – 1)(-2)a n= 20- 2n + 2a n= -2n + 22Now we can use this formula to find the number of terms in the sequence.Keep in mind, the last number in the sequence, a n, is -26.Substituting this into the formula gives us.a n= -2n + 22-26 = -2n + 22-48 = -2n24 = nn = 24This means there are 24 numbers in the arithmetic sequence. Given our generic arithmeticsequence.we can add the terms, called aseries, as follows.There exists a formula that can add such a finite list of these numbers.It requires three pieces of information.The formula is.where S nis the sum of the first n numbers, a 1is the first number in the sequence and a nis the nth number in the sequence.If you would like to see a derivation of this arithmetic series sum formula, watch this video.ideo:Arithmetic Series: Deriving the Sum FormulaUsually problems present themselves in either of two ways.Either the first number and the last number of the sequence are known or the first number in the sequence and the number of terms are known.The following two problems will explain how to find a sum of a finite series.Example 1 : Find the sum of the series 5 + 8 + 11 + 14 + 17 +. + 128.In order to use the sum formula.We need to know a few things.We need to know n, the number of terms in the series.We need to know a 1, the first number, and a n, the last number in the series.We do not know what the n-value is.This is where we must start.To find the n-value, let’s use the formula for the series.We already determined the formula for the sequence in a previous section.We found it to be a n= 3n + 2.We will substitute in the last number of the series and solve for the n-value.a n= 3n + 2128 = 3n + 2126 = 3n42 = nn = 42There are 42 numbers in the series.We also know the d = 3, a 1= 5, and a 42= 128.We can substitute these number into the sum formula, like so.S n= (1/2)n(a 1+ a n)S 42= (1/2)(42)(5 + 128)S 42= (21)(133)S 42= 2793This means the sum of the first 42 terms of the series is equal to 2793.Example 2 : Find the sum of the first 205 multiples of 7.First we have to figure out what our series looks like.We need to write multiples of seven and add them together, like this.7 + 14 + 21 + 28 +. +?To find the last number in the series, which we need for the sum formula, we have to develop a formula for the series.So, we will use theexplicit ruleor a n= a 1+ (n – 1)d.We can also see that d = 7 and the first number, a 1, is 7.a n= a 1+ (n – 1)da n= 7 + (n – 1)(7)a n= 7 + 7n – 7a n= 7nNow we can find the last term in the series.We can do this because we were told there are 205 numbers in the series.We can find the 205th term by using the formula.a n= 7na n= 7(205)a n= 1435This means the last number in the series is 1435.It means the series looks like this.7 + 14 + 21 + 28 +. + 1435To find the sum, we will substitute information into the sum formula. We will substitute a 1= 7, a 205= 1435, and n = 205.S n= (1/2)n(a 1+ a n)S 42= (1/2)(205)(7 + 1435)S 42= (1/2)(205)(1442)S 42= (1/2)(1442)(205)S 42= (721)(205)S 42= 147805This means the sum of the first 205 multiples of 7 is equal to 147,805.

## Arithmetic Sequences and Sums

A sequence is a collection of items (typically numbers) that are arranged in a specific order. Each number in the sequence is referred to as aterm (or “element” or “member” in certain cases); for additional information, see Sequences and Series.

## Arithmetic Sequence

An Arithmetic Sequence is characterized by the fact that the difference between one term and the next is a constant. In other words, we just increase the value by the same amount each time. endlessly.

### Example:

1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25. Each number in this series has a three-digit gap between them. Each time the pattern is repeated, the last number is increased by three, as seen below: As a general rule, we could write an arithmetic series along the lines of

• There are two words: Ais the first term, and dis is the difference between the two terms (sometimes known as the “common difference”).

### Example: (continued)

1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25. Has:

• In this equation, A = 1 represents the first term, while d = 3 represents the “common difference” between terms.

And this is what we get:

### Rule

It is possible to define an Arithmetic Sequence as a rule:x n= a + d(n1) (We use “n1” since it is not used in the first term of the sequence).

### Example: Write a rule, and calculate the 9th term, for this Arithmetic Sequence:

3, 8, 13, 18, 23, 28, 33, and 38 are the numbers three, eight, thirteen, and eighteen. Each number in this sequence has a five-point gap between them. The values ofaanddare as follows:

• A = 3 (the first term)
• D = 5 (the “common difference”)
• A = 3 (the first term).

Making use of the Arithmetic Sequencerule, we can see that_xn= a + d(n1)= 3 + 5(n1)= 3 + 3 + 5n 5 = 5n 2 xn= a + d(n1) = 3 + 3 + 3 + 5n n= 3 + 3 + 3 As a result, the ninth term is:x 9= 5 9 2= 43 Is that what you’re saying? Take a look for yourself! Arithmetic Sequences (also known as Arithmetic Progressions (A.P.’s)) are a type of arithmetic progression.

## Advanced Topic: Summing an Arithmetic Series

To summarize the terms of this arithmetic sequence:a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d) + (a+6d) + (a+7d) + (a+8d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + ( make use of the following formula: What exactly is that amusing symbol? It is referred to as The Sigma Notation is a type of notation that is used to represent a sigma function. Additionally, the starting and finishing values are displayed below and above it: “Sum upnwherengoes from 1 to 4,” the text states. 10 is the correct answer.

### Example: Add up the first 10 terms of the arithmetic sequence:

The values ofa,dandnare as follows:

• In this equation, A = 1 represents the first term, d = 3 represents the “common difference” between terms, and n = 10 represents the number of terms to add up.

As a result, the equation becomes:= 5(2+93) = 5(29) = 145 Check it out yourself: why don’t you sum up all of the phrases and see whether it comes out to 145?

## Footnote: Why Does the Formula Work?

Let’s take a look at why the formula works because we’ll be employing an unusual “technique” that’s worth understanding. First, we’ll refer to the entire total as “S”: S = a + (a + d) +. + (a + (n2)d) +(a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n1)d) + After that, rewrite S in the opposite order: S = (a + (n1)d)+ (a + (n2)d)+.

+(a + d)+a. +(a + d)+a. +(a + d)+a. Now, term by phrase, add these two together:

 S = a + (a+d) + . + (a + (n-2)d) + (a + (n-1)d) S = (a + (n-1)d) + (a + (n-2)d) + . + (a + d) + a 2S = (2a + (n-1)d) + (2a + (n-1)d) + . + (2a + (n-1)d) + (2a + (n-1)d)

Each and every term is the same! Furthermore, there are “n” of them. 2S = n (2a + (n1)d) = n (2a + (n1)d) Now, we can simply divide by two to obtain the following result: The function S = (n/2) (2a + (n1)d) is defined as This is the formula we’ve come up with:

## Using the Formula for Arithmetic Series

In the same way that we looked at different sorts of sequences, we will look at different forms of series. Remember that anarithmetic sequence is a series in which the difference between any two successive terms is equal to the common difference,d, and thus The sum of the terms of an arithmetic sequence is referred to as an anarithmetic series in mathematical jargon. It is possible to represent the sum of the firstnterms of an arithmetic series as: = +left( +dright)+left( +2dright)+.+left( -dright)+ +left( -dright).

The sum of the firstnterms of an arithmetic series may be calculated by adding these two expressions for the sum of the firstnterms of an arithmetic series together.

Fractal = +left( +dright)+left( +2dright)+.+left( -dright)+ hfill + = +left( -dright)+left( -2dright)+.+left( +dright)+ hfill = +left( +dright)+left( -2dright)+.

To determine the formula for the sum of the firstnterms of an arithmetic series, we divide the number by two.

### A General Note: Formula for the Sum of the FirstnTerms of an Arithmetic Series

The sum of the terms of an arithmetic sequence is known as an anarithmetic series. It is written as =frac + right) for the sum of the firstnterms of an arithmetic sequence:

### How To: Given terms of an arithmetic series, find the sum of the firstnterms.

1. Identify and
2. Determinen
3. Substitute values for text , andninto the formula =frac + right)n
4. Substitute values for text , andninto the formula =frac + right)n Make it easier to find_

### Example 2: Finding the FirstnTerms of an Arithmetic Series

Calculate the sum of each arithmetic series in the given time frame.

### Solution

1. We are given the numbers_ =5 and_ =32. To findn=10, count the number of phrases in the sequence to get at n=10. Simplify the formula by substituting values for, text, andninto the equation. begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin Make use of the formula for the general term of an arithmetic series to arrive at the answer. begin = +left begin dhfill -50=20+left(n – 1right)+left(n – 1right)+left(n – 1right)+left(n – 1right)+left(n – 1right)+left(n – 1right)+left(n – 1right)+left (-5right) left(n + 1 right)hfill -70=left(n – 1 right)hfill -70=hfill -70=hfill -70=hfill -70=hfill -70=hfill -70 (-5right) hfill 14=n – 1hfill 15=nhfill nhfill nhfill nhfill end Substitute values for_, _text ninto the formula to make it easier to understand. begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin begin Begin_ =3k – 8hfill text_ =3k – 3hright Fill in the blanks (1right) -8=-5hfill hfill hend We are given the information thatn=12. To find_, enter k=12 into the explicit formula that has been provided. fill in the blanks with text_ =3k – 8hfill in the blanks with text_ =3left(12right)-8=28hfill in the blanks with end text_ =3k – 8hfill in the blanks with end text_ Simplify the formula by substituting values for the variables_, _, andn. hfill =frac + right)hfill =frac=138hfill end
2. Hfill =frac=138hfill end

To get the sum of each arithmetic series, use the formula provided.

### Try It 2

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### Try It 3

Dots in text text Dots in text Solution

### Try It 4

sum 5 – 6k Solution to the problem

### Example 3: Solving Application Problems with Arithmetic Series

A lady is able to walk a half-mile on Sunday after having minor surgery, which she does on Saturday. Every Sunday, she adds an additional quarter-mile to her daily stroll. What do you think the total number of kilometers she has walked will be after 8 weeks?

### Solution

This problem may be represented by an arithmetic series with_ =fracandd=frac as the first and second terms. The total number of kilometers walked after 8 weeks is what we are seeking for; thus, we know thatn=8 and that we are looking for We may use the explicit formula for an arithmetic series to get the value of . commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commencement commence We can now apply the arithmetic series formula to our advantage.

hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill hfill She will have walked a total of 11 miles by the time she is through.

### Try It 5

In the first week of June, a man receives \$100 in pay. The amount he makes each week is \$12.50 greater than the previous week. How much money has he made after 12 weeks of work? Solution