# Which Equation Represents The Formula For The General Term An Of An Arithmetic Sequence? (Question)

Finding the nth Term of an Arithmetic Sequence Given an arithmetic sequence with the first term a1 and the common difference d, the nth (or general) term is given by an=a1+(n−1)d. Example 1: Find the 27th term of the arithmetic sequence 5,8,11,54,. a8=60 and a12=48.

## What is the formula for the general term of an arithmetic sequence?

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.

## What is a general term of a sequence?

What is the general term of the sequence? The general term of a sequence a n a_n an​ is a term that can represent every other term in the sequence. It relates each term is the sequence to its place in the sequence. For example, given the sequence. { − 1, − 2, − 3, − 4, − 5,… }

## What is the general term?

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.

## What is an2 bn C?

The word QUADRATIC refers to terms of the second degree (or squared). A sequence which is quadratic in nature will always have the nth term in the form: Tn = an2 + bn + c where a, b and c are constants.

## Introduction to Arithmetic Progressions

Generally speaking, a progression is a sequence or series of numbers in which the numbers are organized in a certain order so that the relationship between the succeeding terms of a series or sequence remains constant. It is feasible to find the n thterm of a series by following a set of steps. There are three different types of progressions in mathematics:

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.

## Formulas for Arithmetic Sequences

• Create a formal formula for an arithmetic series using explicit notation
• Create a recursive formula for the arithmetic series using the following steps:

## Using Explicit Formulas for Arithmetic Sequences

It is possible to think of anarithmetic sequence as a function on the domain of natural numbers; it is a linear function since the rate of change remains constant throughout the series. The constant rate of change, often known as the slope of the function, is the most frequently seen difference. If we know the slope and the vertical intercept of a linear function, we can create the function. = +dleft = +dright For the -intercept of the function, we may take the common difference from the first term in the sequence and remove it from the result.

• Considering that the average difference is 50, the series represents a linear function with an associated slope of 50.
• You may also get the they-intercept by graphing the function and calculating the point at which a line connecting the points would intersect the vertical axis, as shown in the example.
• When working with sequences, we substitute _instead of y and ninstead of n.
• Using 50 as the slope and 250 as the vertical intercept, we arrive at this equation: = -50n plus 250 To create an explicit formula for an arithmetic series, we do not need to identify the vertical intercept of the sequence.

### A General Note: Explicit Formula for an Arithmetic Sequence

For the textterm of an arithmetic sequence, the formula = +dleft can be used to express it explicitly.

### How To: Given the first several terms for an arithmetic sequence, write an explicit formula.

1. Find the common difference between the two sentences, – To solve for = +dleft(n – 1right), substitute the common difference and the first term into the equation

### Example: Writing then th Term Explicit Formula for an Arithmetic Sequence

Create an explicit formula for the arithmetic sequence.left 12text 22text 32text 42text ldots right 12text 22text 32text 42text ldots

### Try It

For the arithmetic series that follows, provide an explicit formula for it. left With the use of a recursive formula, several arithmetic sequences may be defined in terms of the preceding term. The formula contains an algebraic procedure that may be used to determine the terms of the series. We can discover the next term in an arithmetic sequence by utilizing a function of the term that came before it using a recursive formula. In each term, the previous term is multiplied by the common difference, and so on.

The initial term in every recursive formula must be specified, just as it is with any other formula.

### A General Note: Recursive Formula for an Arithmetic Sequence

In the case of an arithmetic sequence with common differenced, the recursive formula is as follows: the beginning of the sentence = +dnge 2 the finish of the sentence

### How To: Given an arithmetic sequence, write its recursive formula.

1. To discover the common difference between two terms, subtract any phrase from the succeeding term. In the recursive formula for arithmetic sequences, start with the initial term and substitute the common difference

### Example: Writing a Recursive Formula for an Arithmetic Sequence

Write a recursive formula for the arithmetic series in the following format: left

### How To: Do we have to subtract the first term from the second term to find the common difference?

No.

We can take any phrase in the sequence and remove it from the term after it. Generally speaking, though, it is more customary to subtract the first from the second term since it is frequently the quickest and most straightforward technique of determining the common difference.

### Try It

Create a recursive formula for the arithmetic sequence using the information provided. left

## Find the Number of Terms in an Arithmetic Sequence

When determining the number of terms in a finite arithmetic sequence, explicit formulas can be employed to make the determination. Finding the common difference and determining the number of times the common difference must be added to the first term in order to produce the last term of the sequence are both necessary steps.

### How To: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms.

1. Find the common differences between the two
2. To solve for = +dleft(n – 1right), substitute the common difference and the first term into the equation Fill in the blanks with the final word from and solve forn
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### Example: Finding the Number of Terms in a Finite Arithmetic Sequence

The number of terms in the infinite arithmetic sequence is to be determined. left

### Try It

The number of terms in the finite arithmetic sequence has to be determined. 11 text 16 text. text 56 right 11 text 16 text 16 text 56 text 56 text 56 Following that, we’ll go over some of the concepts that have been introduced so far concerning arithmetic sequences in the video lesson that comes after that.

## Solving Application Problems with Arithmetic Sequences

In many application difficulties, it is frequently preferable to begin with the term instead of_ as an introductory phrase. When solving these problems, we make a little modification to the explicit formula to account for the change in beginning terms. The following is the formula that we use: = +dn = = +dn

### Example: Solving Application Problems with Arithmetic Sequences

Every week, a kid under the age of five receives a \$1 stipend from his or her parents. His parents had promised him a \$2 per week rise on a yearly basis.

1. Create a method for calculating the child’s weekly stipend over the course of a year
2. What will be the child’s allowance when he reaches the age of sixteen

### Try It

A lady chooses to go for a 10-minute run every day this week, with the goal of increasing the length of her daily exercise by 4 minutes each week after that. Create a formula to predict the timing of her run after n weeks has passed. In eight weeks, how long will her daily run last on average?

## 7.2 – Arithmetic Sequences

Was there anything you thought might be improved on the page? Would appreciate it if you could provide some feedback. This page could be improved. Obtaining Additional Information

## 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.

1. 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. 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.
3. This would be 5/2 * (16) = 5(8) = 40 as a total.
4. The number 5 refers to the fact that there were five terms, n.
5. In this case, we added the total twice and it will always be a 2.
6. Another formula for the n th partial sum of an arithmetic series is occasionally used in conjunction with the previous one.
7. 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.

## 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)?

## Arithmetic Sequences and Series

The succession of arithmetic operations There is a series of integers in which each subsequent number is equal to the sum of the preceding number and specified constants. orarithmetic progression is a type of progression in which numbers are added together. This term is used to describe a series of integers in which each subsequent number is the sum of the preceding number and a certain number of constants (e.g., 1). an=an−1+d Sequence of Arithmetic Operations Furthermore, becauseanan1=d, the constant is referred to as the common difference.

For example, the series of positive odd integers is an arithmetic sequence, consisting of the numbers 1, 3, 5, 7, 9, and so on.

This word may be constructed using the generic terman=an1+2where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, To formulate the following equation in general terms, given the initial terma1of an arithmetic series and its common differenced, we may write: a2=a1+da3=a2+d=(a1+d)+d=a1+2da4=a3+d=(a1+2d)+d=a1+3da5=a4+d=(a1+3d)+d=a1+4d⋮ As a result, we can see that each arithmetic sequence may be expressed as follows in terms of its initial element, common difference, and index: an=a1+(n−1)d Sequence of Arithmetic Operations In fact, every generic word that is linear defines an arithmetic sequence in its simplest definition.

### Example 1

Identify the general term of the above arithmetic sequence and use that equation to determine the series’s 100th term. For example: 7,10,13,16,19,… Solution: The first step is to determine the common difference, which is d=10 7=3. It is important to note that the difference between any two consecutive phrases is three. The series is, in fact, an arithmetic progression, with a1=7 and d=3. an=a1+(n1)d=7+(n1)3=7+3n3=3n+4 and an=a1+(n1)d=7+(n1)3=7+3n3=3n+4 and an=a1+(n1)d=3 As a result, we may express the general terman=3n+4 as an equation.

To determine the 100th term, use the following equation: a100=3(100)+4=304 Answer_an=3n+4;a100=304 It is possible that the common difference of an arithmetic series be negative.

### Example 2

Identify the general term of the given arithmetic sequence and use it to determine the 75th term of the series: 6,4,2,0,−2,… Solution: Make a start by determining the common difference, d = 4 6=2. Next, determine the formula for the general term, wherea1=6andd=2 are the variables. an=a1+(n−1)d=6+(n−1)⋅(−2)=6−2n+2=8−2n As a result, an=8nand the 75thterm may be determined as follows: an=8nand the 75thterm a75=8−2(75)=8−150=−142 Answer_an=8−2n;a100=−142 The terms in an arithmetic sequence that occur between two provided terms are referred to as arithmetic means.

### Example 3

Find all of the words that fall between a1=8 and a7=10. in the context of an arithmetic series Or, to put it another way, locate all of the arithmetic means between the 1st and 7th terms. Solution: Begin by identifying the points of commonality. In this situation, we are provided with the first and seventh terms, respectively: an=a1+(n−1) d Make use of n=7.a7=a1+(71)da7=a1+6da7=a1+6d Substitutea1=−8anda7=10 into the preceding equation, and then solve for the common differenced result. 10=−8+6d18=6d3=d Following that, utilize the first terma1=8.

a1=3(1)−11=3−11=−8a2=3 (2)−11=6−11=−5a3=3 (3)−11=9−11=−2a4=3 (4)−11=12−11=1a5=3 (5)−11=15−11=4a6=3 (6)−11=18−11=7} In arithmetic, a7=3(7)11=21=10 means a7=3(7)11=10 Answer: 5, 2, 1, 4, 7, and 8.

### Example 4

Find the general term of an arithmetic series with a3=1 and a10=48 as the first and last terms. Solution: We’ll need a1 and d in order to come up with a formula for the general term. Using the information provided, it is possible to construct a linear system using these variables as variables. andan=a1+(n−1) d:{a3=a1+(3−1)da10=a1+(10−1)d⇒ {−1=a1+2d48=a1+9d Make use of a3=1. Make use of a10=48. Multiplying the first equation by one and adding the result to the second equation will eliminate a1.

an=a1+(n−1)d=−15+(n−1)⋅7=−15+7n−7=−22+7n Answer_an=7n−22 Take a look at this! Identify the general term of the above arithmetic sequence and use that equation to determine the series’s 100th term. For example: 32,2,52,3,72,… Answer_an=12n+1;a100=51

## Arithmetic Series

Series of mathematical operations When an arithmetic sequence is added together, the result is called the sum of its terms (or the sum of its terms and numbers). Consider the following sequence: S5=n=15(2n1)=++++= 1+3+5+7+9+25=25, where S5=n=15(2n1)=++++ = 1+3+5+7+9=25, where S5=n=15(2n1)=++++= 1+3+5+7+9 = 25. Adding 5 positive odd numbers together, like we have done previously, is manageable and straightforward. Consider, on the other hand, adding the first 100 positive odd numbers. This would be quite time-consuming.

When we write this series in reverse, we get Sn=an+(and)+(an2d)+.+a1 as a result.

2.:Sn=n(a1+an) 2 Calculate the sum of the first 100 terms of the sequence defined byan=2n1 by using this formula.

The sum of the two variables, S100, is 100 (1 + 100)2 = 100(1 + 199)2.

### Example 5

The sum of the first 50 terms of the following sequence: 4, 9, 14, 19, 24,. is to be found. The solution is to determine whether or not there is a common difference between the concepts that have been provided. d=9−4=5 It is important to note that the difference between any two consecutive phrases is 5. The series is, in fact, an arithmetic progression, and we may writean=a1+(n1)d=4+(n1)5=4+5n5=5n1 as an anagram of the sequence. As a result, the broad phrase isan=5n1 is used. For this sequence, we need the 1st and 50th terms to compute the 50thpartial sum of the series: a1=4a50=5(50)−1=249 Then, using the formula, find the partial sum of the given arithmetic sequence that is 50th in length.

### Example 6

Evaluate:Σn=135(10−4n). This problem asks us to find the sum of the first 35 terms of an arithmetic series with a general terman=104n. The solution is as follows: This may be used to determine the 1 stand for the 35th period. a1=10−4(1)=6a35=10−4(35)=−130 Then, using the formula, find out what the 35th partial sum will be. Sn=n(a1+an)2S35=35⋅(a1+a35)2=352=35(−124)2=−2,170 2,170 is the answer.

### Example 7

In an outdoor amphitheater, the first row of seating comprises 26 seats, the second row contains 28 seats, the third row contains 30 seats, and so on and so forth. Is there a maximum capacity for seating in the theater if there are 18 rows of seats? The Roman Theater (Fig. 9.2) (Wikipedia) Solution: To begin, discover a formula that may be used to calculate the number of seats in each given row. In this case, the number of seats in each row is organized into a sequence: 26,28,30,… It is important to note that the difference between any two consecutive words is 2.

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where a1=26 and d=2.

As a result, the number of seats in each row may be calculated using the formulaan=2n+24.

In order to do this, we require the following 18 thterms: a1=26a18=2(18)+24=60 This may be used to calculate the 18th partial sum, which is calculated as follows: Sn=n(a1+an)2S18=18⋅(a1+a18)2=18(26+60) 2=9(86)=774 There are a total of 774 seats available.

Take a look at this! Calculate the sum of the first 60 terms of the following sequence of numbers: 5, 0, 5, 10, 15,. are all possible combinations. Answer_S60=−8,550

### Key Takeaways

• When the difference between successive terms is constant, a series is called an arithmetic sequence. According to the following formula, the general term of an arithmetic series may be represented as the sum of its initial term, common differenced term, and indexnumber, as follows: an=a1+(n−1)d
• An arithmetic series is the sum of the terms of an arithmetic sequence
• An arithmetic sequence is the sum of the terms of an arithmetic series
• As a result, the partial sum of an arithmetic series may be computed using the first and final terms in the following manner: Sn=n(a1+an)2

### Topic Exercises

1. Given the first term and common difference of an arithmetic series, write the first five terms of the sequence. Calculate the general term for the following numbers: a1=5
2. D=3
3. A1=12
4. D=2
5. A1=15
6. D=5
7. A1=7
8. D=4
9. D=1
10. A1=23
11. D=13
12. A 1=1
13. D=12
14. A1=54
15. D=14
16. A1=1.8
17. D=0.6
18. A1=4.3
19. D=2.1
1. Find a formula for the general term based on the arithmetic sequence and apply it to get the 100 th term based on the series. 0.8, 2, 3.2, 4.4, 5.6,.
2. 4.4, 7.5, 13.7, 16.8,.
3. 3, 8, 13, 18, 23,.
4. 3, 7, 11, 15, 19,.
5. 6, 14, 22, 30, 38,.
6. 5, 10, 15, 20, 25,.
7. 2, 4, 6, 8, 10,.
8. 12,52,92,132,.
9. 13, 23, 53,83,.
10. 14,12,54,2,114,. Find the positive odd integer that is 50th
11. Find the positive even integer that is 50th
12. Find the 40 th term in the sequence that consists of every other positive odd integer in the following format: 1, 5, 9, 13,.
13. Find the 40th term in the sequence that consists of every other positive even integer: 1, 5, 9, 13,.
14. Find the 40th term in the sequence that consists of every other positive even integer: 2, 6, 10, 14,.
15. 2, 6, 10, 14,. What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
16. What number is the phrase 172 in the arithmetic sequence 4, 4, 12, 20, 28,.
17. What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
18. Find an equation that yields the general term in terms of a1 and the common differenced given the arithmetic sequence described by the recurrence relationan=an1+5wherea1=2 andn1 and the common differenced
19. Find an equation that yields the general term in terms ofa1and the common differenced, given the arithmetic sequence described by the recurrence relationan=an1-9wherea1=4 andn1
20. This is the problem.
1. Calculate a formula for the general term based on the terms of an arithmetic sequence: a1=6anda7=42
2. A1=12anda12=6
3. A1=19anda26=56
4. A1=9anda31=141
5. A1=16anda10=376
6. A1=54anda11=654
7. A3=6anda26=40
8. A3=16andananda15=
1. Find all possible arithmetic means between the given terms: a1=3anda6=17
2. A1=5anda5=7
3. A2=4anda8=7
4. A5=12anda9=72
5. A5=15anda7=21
6. A6=4anda11=1
7. A7=4anda11=1

### Part B: Arithmetic Series

1. Make a calculation for the provided total based on the formula for the general term an=3n+5
2. S100
3. An=5n11
4. An=12n
5. S70
6. An=132n
7. S120
8. An=12n34
9. S20
10. An=n35
11. S150
12. An=455n
13. S65
14. An=2n48
15. S95
16. An=4.41.6n
17. S75
18. An=6.5n3.3
19. S67
20. An=3n+5
1. Consider the following values: n=1160(3n)
2. N=1121(2n)
3. N=1250(4n3)
4. N=1120(2n+12)
5. N=170(198n)
6. N=1220(5n)
7. N=160(5212n)
8. N=151(38n+14)
9. N=1120(1.5n2.6)
10. N=1175(0.2n1.6)
11. The total of all 200 positive integers is found by counting them up. To solve this problem, find the sum of the first 400 positive integers.
1. The generic term for a sequence of positive odd integers is denoted byan=2n1 and is defined as follows: Furthermore, the generic phrase for a sequence of positive even integers is denoted by the number an=2n. Look for the following: The sum of the first 50 positive odd integers
2. The sum of the first 200 positive odd integers
3. The sum of the first 50 positive even integers
4. The sum of the first 200 positive even integers
5. The sum of the first 100 positive even integers
6. The sum of the firstk positive odd integers
7. The sum of the firstk positive odd integers the sum of the firstk positive even integers
8. The sum of the firstk positive odd integers
9. There are eight seats in the front row of a tiny theater, which is the standard configuration. Following that, each row contains three additional seats than the one before it. How many total seats are there in the theater if there are 12 rows of seats? In an outdoor amphitheater, the first row of seating comprises 42 seats, the second row contains 44 seats, the third row contains 46 seats, and so on and so forth. When there are 22 rows, how many people can fit in the theater’s entire seating capacity? The number of bricks in a triangle stack are as follows: 37 bricks on the bottom row, 34 bricks on the second row and so on, ending with one brick on the top row. What is the total number of bricks in the stack
10. Each succeeding row of a triangle stack of bricks contains one fewer brick, until there is just one brick remaining on the top of the stack. Given a total of 210 bricks in the stack, how many rows does the stack have? A salary contract with a 10-year term pays \$65,000 in the first year, with a \$3,200 increase for each additional year after. Calculate the total salary obligation over a ten-year period (see Figure 1). In accordance with the hour, a clock tower strikes its bell a specified number of times. The clock strikes once at one o’clock, twice at two o’clock, and so on until twelve o’clock. A day’s worth of time is represented by the number of times the clock tower’s bell rings.

### Part C: Discussion Board

1. Is the Fibonacci sequence an arithmetic series or a geometric sequence? How to explain: Using the formula for the then th partial sum of an arithmetic sequenceSn=n(a1+an)2and the formula for the general terman=a1+(n1)dto derive a new formula for the then th partial sum of an arithmetic sequenceSn=n2, we can derive the formula for the then th partial sum of an arithmetic sequenceSn=n2. How would this formula be beneficial in the given situation? Explain with the use of an example of your own creation
2. Discuss strategies for computing sums in situations when the index does not begin with one. For example, n=1535(3n+4)=1,659
3. N=1535(3n+4)=1,659
4. Carl Friedrich Gauss is the subject of a well-known tale about his misbehaving in school. As a punishment, his instructor assigned him the chore of adding the first 100 integers to his list of disciplinary actions. According to folklore, young Gauss replied accurately within seconds of being asked. The question is, what is the solution, and how do you believe he was able to come up with the figure so quickly?

1. 5, 8, 11, 14, 17
2. An=3n+2
3. 15, 10, 5, 0, 0
4. An=205n
5. 12,32,52,72,92
6. An=n12
7. 1,12, 0,12, 1
8. An=3212n
9. 1.8, 2.4, 3, 3.6, 4.2
10. An=0.6n+1.2
11. An=6n3
12. A100=597
13. An=14n
14. A100=399
15. An=5n
16. A100=500
17. An=2n32
1. 2,450, 90, 7,800, 4,230, 38,640, 124,750, 18,550, 765, 10,000, 20,100, 2,500, 2,550, K2, 294 seats, 247 bricks, \$794,000, and

## How and Why to Use the General Term of an Arithmetic Sequence – Video & Lesson Transcript

Any number in an arithmetic sequence may be calculated using this broad word, which refers to the mathematical formula. To identify a specific number in our sequence,xsubn, we would start with our starting number, a, and multiply it by the common difference (d), multiplied by nminus 1, which is the position of our desired number minus one, according to the formula. For example, if we are seeking for the 30th number, we will use the notation nis 30, and our formula will begin withxsub 30 andn- 1 = 29.

## Using the General Term

To further understand how we utilize this generic phrase, let’s look at an example. Here’s a fresh sequence to try out: We’re looking for the value of the number that occupies the 25th position in our series of numbers. In this instance, ourn is 25. It is necessary to first determine whether or not this sequence is an arithmetic sequence before proceeding forward. To do so, we examine the difference between each succeeding pair of numbers to see if the difference is the same for each pair of numbers.

If it isn’t, we won’t be able to apply our general term method to come up with an answer.

They both add up to 2; therefore, they are equivalent.

## Writing Rules for Arithmetic Sequences – Video & Lesson Transcript

To further understand how we utilize this generic phrase, let’s look at an illustration. a new sequence is presented as follows: This problem requires us to determine the value of a number that occurs at the 25th position in our series of numbers. In this instance, ourn is 25 years old. We must first determine whether or not this sequence is an arithmetic sequence before proceeding any farther with the investigation. To do so, we examine the difference between each subsequent pair of numbers to check if the difference is the same as the difference between the previous two numbers.

This means that we can’t discover our solution using our general term formula if it isn’t there.

Because they both add up to 2, they are equivalent.

## Arithmetic Sequences

All sorts of arithmetic sequences may be found all throughout the world. Some of these may be used on a daily basis. A series of terms in which the difference between each subsequent pair of terms equals one is known as anarithmetic sequence, by definition. In the case of counting by 5s, you will receive an arithmetic sequence since the difference between each pair of terms is 5: For example, if you count by 5s, you will get the following: These sequences may not appear to be utilized for anything significant, yet they are in fact useful in a variety of situations and situations.

If the park charges an entrance fee of \$10.00 as well as a per ride fee of \$2.00, your total cost will be calculated as follows based on the number of rides you wish to ride:

For no rides, the arithmetic sequence begins with \$10, then proceeds on to \$12 for one ride, then \$14 for two rides, and so on.

## The Rule

For the reason that all arithmetic sequences follow the same pattern, you may apply a generic formula to obtain the formula for any particular sequence. The formula is as follows: Thea nrefers to the terms of the sequence, and thenrefers to the position of the term in the series. This word refers to the first term in the sequence if nis is equal to 1. The difference between all of the consecutive integers in your series is represented by the letter d. The explicit formula for an arithmetic series is referred to as the explicit formula.

## Given Two Terms

Arithmetic sequences contain the same difference between succeeding pairs of terms in the sequence; as a result, you only need to know the first two terms of the series to construct the formula; the further terms of the sequence are not required. Let’s have a look at this. Take a look at this situation. Create a formula for the arithmetic series that begins with the numbers 4, 7, and so on. Only the first two words are provided to you. Because you already know the explicit formula rule, all you need to know is the first term and the difference between each succeeding pair of terms in the following formula.

You also know that the difference between the first term and the second term is 7.

• An=a1+d(n-1)
• An= 4 + 3 (n-1)
• An= 4 + 3 n- 3
• An= 3 n+ 1
• An= 4 n+ 1

You should leave the then s alone because they will always be a variable. Thesens are what allow you to utilize this formula to locate the remaining terms in your sequence using the information in this formula. They correspond to the term you are looking for in the arithmetic sequence.By applying the rule to determine the formula for an arithmetic sequence, you can see that your arithmetic sequence follows the explicit rule 3 n+ 1 for all of the terms in the series. In other words, you can discover any term in the sequence only by entering in the term’s location.

## Real World Applications of Arithmetic Sequences:

• In an arithmetic sequence, the explicit formula for the then th term is defined asa n =a 1 + d(n – 1), wherea n is the then th term of the sequence, a 1 is the first term of the series, and dis the common difference of the sequence.
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### Applications:

1. Nancy is putting money aside to purchase a bike that will cost \$275. She begins with \$50 and continues to add \$15 at the end of each week until she reaches her goal. It will take her about how many weeks to save up enough money to purchase the bike. Bob decided to start running as a New Year’s Resolution on January 1st, with the objective of running for one hour, or 60 minutes, straight. He begins by running for 5 minutes on the first day, and he increases his jogging time by 2 minutes on each subsequent day after that, until he reaches his goal. Bob’s objective is to attain it before the end of the month (which is 31 days from now).

1. It will take 16 weeks to save \$275 in this situation. For example, if she starts with \$50 and adds \$15 each week, the amount of money she has saved at the end of each week follows the mathematical sequence of 50 (first week), 65 (second week), 90 (third week), 105 (fourth week), and so on. In light of the fact that the bike costs \$275, we are interested in knowing what term will be 275, or for what valuenwilla n = 275. The equation 275 = 50 + 15 is obtained by plugging these values into our explicit formula (n – 1). Solving the forngivesn=16 equation Yes. To demonstrate that this is the case, we check to see if Bob is still running for 60 minutes or more every day after 31 days of following this routine. He starts with 5 minutes and adds 2 minutes each day, therefore this may be represented mathematically by an arithmetic sequence with a beginning term of 5 and a common difference of 2 as shown in the diagram (or 5, 7, 9, 11,.). To determine if the 31st term will be more than or equal to 60, we pluga 1 = 5,d= 2, and n = 31 into our explicit formula to obtaina (31) = 5 + 2 (as in the explicit formula) (31 – 1). Bob can run for 65 minutes straight by the end of the month, which is more than 60 minutes, according to the simplified formula (31) = 65.

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## 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:

## 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: There are no variations!

As well as the fact that there are “n” of them The number 2S equals the sum of the numbers (2a + (n1)d) and the number 2S equals two times the number two.

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.

+ 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.