An arithmetic sequence goes from one term to the next by always adding **(or subtracting)** the same value. The number added (or subtracted) at each stage of the arithmetic sequence is called the common difference. Examples of arithmetic sequences occur when things change by the same amount each time.

Contents

- 1 How do you find the next term of an arithmetic sequence?
- 2 What is term in arithmetic sequence?
- 3 How do you find next terms?
- 4 What is sum of arithmetic sequence?
- 5 What is nth term?
- 6 What is the next 3 terms of the Fibonacci sequence?
- 7 Which of the following are examples of arithmetic sequences?
- 8 How will you find the next in an arithmetic sequence and geometric sequence?
- 9 What is the arithmetic between 10 and 24?
- 10 How to find the next term in an arithmetic sequence – ACT Math
- 11 How to find the next term in an arithmetic sequence – Algebra 1
- 12 Introduction to Arithmetic Progressions
- 13 How to Find Any Term of an Arithmetic Sequence
- 14 Video
- 15 About This Article
- 16 Did this article help you?
- 17 Key Concepts
- 18 Glossary
- 19 Contribute!
- 20 Arithmetic Sequences and Sums
- 21 Arithmetic Sequence
- 22 Advanced Topic: Summing an Arithmetic Series
- 23 Footnote: Why Does the Formula Work?
- 24 Arithmetic & Geometric Sequences
- 24.0.1 Find the common difference and the next term of the following sequence:
- 24.0.2 Find the common ratio and the seventh term of the following sequence:
- 24.0.3 Find the tenth term and then-th term of the following sequence:
- 24.0.4 Find then-th term and the first three terms of the arithmetic sequence havinga6= 5andd=
- 24.0.5 Find then-th term and the first three terms of the arithmetic sequence havinga4= 93anda8= 65.
- 24.0.6 Find then-th and the26 th terms of the geometric sequence withanda12= 160.

- 25 Arithmetic Sequences and Series
- 26 Find the next three terms of the arithmetic sequence. 3/4, 1/2, 1/4, 1, .
- 27 Arithmetic Sequence and It’s nth Term:
- 28 Answer and Explanation:1
- 29 Arithmetic Sequences and Series
- 30 Arithmetic Series

## How do you find the next term of an arithmetic sequence?

Correct answer: The difference between each term is constant, thus the sequence is an arithmetic sequence. Simply find the difference between each term, and add it to the last term to find the next term.

## What is term in arithmetic sequence?

A Sequence is a set of things (usually numbers) that are in order. Each number in the sequence is called a term (or sometimes ” element ” or “member”), read Sequences and Series for more details.

## How do you find next terms?

Correct answer: First, find the common difference for the sequence. Subtract the first term from the second term. Subtract the second term from the third term. To find the next value, add to the last given number.

## What is sum of arithmetic sequence?

The sum of an arithmetic sequence is “ the sum of the first n terms” of the sequence and it can found using one of the following formulas: Sn=n2(2a+(n−1)d)Sn=n2(a1+an) Here, a=a1 a = a 1 = the first term. d = the common difference.

## What is nth term?

The nth term is a formula that enables us to find any term in a sequence. The ‘n’ stands for the term number. To find the 10th term we would follow the formula for the sequence but substitute 10 instead of ‘n’; to find the 50th term we would substitute 50 instead of n.

## What is the next 3 terms of the Fibonacci sequence?

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811,

## Which of the following are examples of arithmetic sequences?

For example, the sequence 3, 5, 7, 9… is arithmetic because the difference between consecutive terms is always two. The sequence 21, 16, 11, 6 is arithmetic as well because the difference between consecutive terms is always minus five.

## How will you find the next in an arithmetic sequence and geometric sequence?

The common pattern in an arithmetic sequence is that the same number is added or subtracted to each number to produce the next number. The common pattern in a geometric sequence is that the same number is multiplied or divided to each number to produce the next number.

## What is the arithmetic between 10 and 24?

Using the average formula, get the arithmetic mean of 10 and 24. Thus, 10+24/2 =17 is the arithmetic mean.

## How to find the next term in an arithmetic sequence – ACT Math

Given the following numerical sequence: 1, 5, 9, _, _, 21, What are the two missing terms in the arithmetic sequence that you’re looking for? Answers that might be given include: 14, 1714, 1613, 1613, 1712, and 18 are the numbers of the day. Reason: The sequence is defined bya n= 4 n–3 for suchn= 1,2,3,4 and a n= 3 for such. Who knows what the next word in the following sequence will be. The correct response is: An explanation of the question: What is the next word in the following sentence sequence?

The correct response is: Explanation: To begin, consider the following transitions from one number to another in this series: Add to the list of recipients Subtract from the beginning Add to the list of recipients Subtract from the beginning Based on what you’ve seen thus far, you may assume that the next stage will be to provide more information.

Which of the following is the next phrase in the following sequence: Correct response:Explanation:This sequence is a little difficult to follow.

Following that, the third element is equal to the sum of the first two elements, followed by the fourth element, which is equal to the total of the second and third elements, and so on for each element after that.

Who knows what the following word in the series is.

Simply calculate the difference between each term and multiply it by the last term to arrive at the next term in the series.

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## How to find the next term in an arithmetic sequence – Algebra 1

What is the first number in the following arithmetic sequence? Answers that might be given include: 652 None of the other options are acceptable. 7 Explanation: Given that the series is arithmetic, we may determine the next number in the sequence by adding (or removing) a constant term from the previous number in the sequence. We are aware of two of the values, which are separated by an unknown value. We know that is equal to half the distance between -1 and 13; as a result, is equal to half the distance between these two numbers The distance between them may be calculated by summing the absolute values of the two variables.

- We can then proceed forward or backward in order to figure out what happened.
- The correct response is: Explanation: The common difference between two consecutive numbers in the series may be calculated by taking the difference between two adjacent numbers in the sequence.
- Our next term will be compatible with the equation, which implies that the following term must be.
- Finally, following that, the following word will be, implying that the following term must be In order to answer the question, we must first add the three words in the following sentence to get the total sum.
- What is the monetary worth of?
- You will see that every time you travel from one number to the very next number, the total grows by seven numbers.
- Accordingly, we may multiply 37 by 7, resulting in the number 43.
- In order to proceed, find the next phrase in the sequence: 2, 7, 17, 37, and 77 are the numbers 2, 7, 17, and 37, respectively.
- The correct response is: Explanation: Determining the type of sequence you have, that is, determining whether the sequence varies by a constant difference or a constant ratio, is important.
- As a result, the series progresses by subtracting 16 from the previous number.
- The following arithmetic sequence has a next term that must be found: Explain why you got the correct answer:First, discover the common difference between the two sequences.

Subtract the first term from the second term to arrive at the answer. Subtract the second term from the third term to arrive at the answer. Subtract the third term from the fourth term to arrive at the answer. To determine the next value, add one to the last number that was provided.

The following arithmetic sequence has a next term that must be found: Explain why you got the correct answer:First, discover the common difference between the two sequences. Subtract the first term from the second term to arrive at the answer. Subtract the second term from the third term to arrive at the answer. To determine the next value, add one to the last number that was provided. Find the next word in the arithmetic sequence that has been provided: Explain why you got the correct answer:First, discover the common difference between the two sequences.

- Subtract the second term from the third term to arrive at the answer.
- The following arithmetic sequence has a next term that must be found: Explain why you got the correct answer:First, discover the common difference between the two sequences.
- Subtract the second term from the third term to arrive at the answer.
- The following arithmetic sequence has a next term that must be found: The correct response is: Explanation: To begin, identify the common difference between the sequences.
- Subtract the second term from the third term to arrive at the answer.
- To determine the next value, add one to the last number that was provided.

## 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:

- 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 Any Term of an Arithmetic Sequence

Solve the first problem, which is to find out how many terms are in the first 35 terms of the series 5,11,17,23. a = 5 in the given series, d = a 2–a in the given series. The number 1 equals 11 minus 5 equals 6, and the number n equals 35 In the case of S n= (n/2)(2a + (n – 1) x d), the formula is Eq. (35/2)(2 x 5 + (35 – 1) x 6; Sn= (35/2)(2 x 5 + (35 – 1)) [10 + 34x 6] S n= (35/2)(10 + 34x 6). S n= (35/2)(10 + 34 x 6). S n= (35/2)(10 + 34 x 6). (13/2)(10 + 204) S n= (13/2)(10 + 204) S n= (13/2)(10 + 204) S n= (13/2)(10 + 204) S n= (13/2)(10 + 204) S n= (13/2)(10 + 204) 2S n = 35 x 214/2S n = 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.

(31/2)(5 + 209) S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) 2S n = 35 x 214/2S n = 3745 3rd Problem: A sum of 21 rupees is split among three brothers, with each brother receiving three equal shares of money in AP and the sum of their squares being 155 rupees.

Because of the way the money is split in AP, let the three pieces of money be (a-d), (a), and (a+d).

two times fifteen hundred fifty-two (d) The answer is 2=155–148.

Three pieces of money are distributed as follows:a + d = 7 + 2 = 9a + d = 7 + 2 = 5a + 2 = 7a + d + 2 = 5a + 2 = 5a + 2 = 5a + 2 + 5a + 2 + 5a + 2 = 5a + 2 = 5a + 2 + 5a + 2 = 5a + 2 = 5a + 2 + 5a + 2 = 5a + 2 = 5a + 2 + 5a + 2 = 5a As a result, Rupees 9 constitutes the majority of the total amount of money.

- 1 Determine the common difference between the two sequences. A list of numbers may be given to you with the explanation that the list is an arithmetic sequence, or you may be required to figure it out for yourself. In each scenario, the initial step is the same as it is in the other. Choose the first two terms that appear consecutively in the list. Subtract the first term from the second term to arrive at the answer. It is the outcome of your sequence that is the common difference
- 2 Check to see if the common difference is constant across the board. Finding the common difference between the first two terms does not imply that your list is an arithmetic sequence in the traditional sense. You must ensure that the difference is continuous across the whole list. Subtract two separate consecutive terms from the list to see how much of a difference there is. If the result is consistent for one or two other pairs of words, then you have most likely discovered an arithmetic sequence of terms. Advertisement
- s3 Add the common difference to the last phrase that was supplied. Finding the next term in an arithmetic series is straightforward after you’ve determined the common difference. Simply add the common difference to the final phrase in the list, and you will arrive at the next number in the sequence. Advertisement

- 1 Double-check that you are starting with an arithmetic sequence before proceeding. Sometimes you will have a list of numbers with a missing phrase in the center, and this will be the case. As with the last step, begin by ensuring that your list is an arithmetic sequence. Make a choice between any two consecutive words and calculate the difference between them. Once you’ve done that, compare it to two additional consecutive terms in the list. You can proceed if the differences are the same, in which case you can assume you are working with an arithmetic series. 2 Before the space, add the common difference to the end of the word. This is analogous to appending a phrase to the end of a sequence of words. Locate the phrase in your sequence that comes directly before the gap in question. This is the “last” number that you are familiar with. By multiplying this term by your common difference, you may get the number that should be used to fill in the blank
- 3 To calculate the common difference, subtract it from the phrase that follows the space. Check your response from the other way to be certain that you have the proper answer. An arithmetic sequence should be consistent in both directions, regardless of the direction in which it is performed. If you travel from left to right and add 4, then you would proceed in the opposite direction, from right to left, and do the reverse and remove 4
- 4 is the sum of the two numbers. You should compare your results. The two outcomes that you obtain, whether you add up from the bottom or subtract down from the top, should be identical to one another. If they do, you have discovered the value for the word that was previously unknown. It is your responsibility to ensure that your work is error-free. The arithmetic sequence you have may or may not be correct. Advertisement

- 1 Determine which phrase is the first in the series. Not all sequences begin with the integers 0 or 1 as the first or second numbers. Take a look at the list of numbers you have and identify the first phrase on it. Your beginning position, which can be identified using variables such as a(1), is the following: 2Define your common difference as d in the following way: Find the common difference between the sequences, just like you did previously. The common difference in this working example is 5, which is the most significant. It is the same result if you check with any of the other words in the sequence. This is a common distinction between the algebraic variable d, which we shall observe. 3 Use the explicit formula to solve the problem. In algebra, an explicit formula is a mathematical equation that may be used to determine any term in an arithmetic series without having to write down the entire list of terms in the sequence. An algebraic series can be represented by the explicit formula

- It is possible to read the word a(n) as “the nth term of a,” where n denotes the number in the list that you are looking for and a(n) reflects the actual value of that number. The number n will be 100 if you are asked to locate the 100th item in an arithmetic series, for example. Notably, while n is 100 in this example, the value of the 100th term, rather than the number 100 itself, will be represented as a(n).

- 4 Fill in the blanks with your information to help us solve the problem. Make use of the explicit formula for your sequence to enter the information that you already know in order to locate the word that you want. Advertisement

- 1, rearrange the explicit formula such that it may be used to solve for additional variables. Several bits of information about an arithmetic sequence may be discovered by employing the explicit formula and some fundamental algebraic operations. As written in its original form, the explicit formula is intended to solve for an integer n and provide you with the nth term in a series of numbers. You may, however, modify this formula algebraically and solve for any of the variables in the equation. 2 Find the first phrase in a series by using the search function. For example, you may know that the 50th term of an arithmetic series is 300, and you may also know that the terms have been growing by 7 (the “common difference”), but you may wish to know what the sequence’s very first term was. To determine your solution, use the improved explicit formula that solves for a1 as previously stated

- Make use of the equation and fill in the blanks with the facts you already know. Because you know that the 50th term is 300, n=50, n-1=49, and a(n)=300 are the values of n. You are also informed that the common difference, denoted by the letter d, is seven. Therefore, the formula is as follows: This works out as well. The series that you have created began at 43 and increased by 7 each time. As a result, it appears as follows: 43,50,57,64,71,78.293,300

- 3 Determine the total length of a sequence. Consider the following scenario: you know the beginning and ending points of an arithmetic series, but you need to know how long it is. Make use of the updated formula

- Consider the following scenario: you know that a specific arithmetic sequence starts at 100 and grows by 13. In addition, you are informed that the ultimate term is 2,856. You can find out the length of the series by putting the terms a1=100, d=13, and a(n)=2856 together. Fill in the blanks with the terms from the formula to get the answer. If you do the math, you will come up with, which equals 212+1, which equals 213. 213 words are included inside a single sequence
- An example of this would be the following: 101-313-126-213-136-139.2843-2856.

Create a new question

- Question How can I determine the first three terms if I only have the tenth and fifteenth terms? Subtract the tenth term from the fifteenth term and divide by five to get D, which is the difference between any two consecutive terms in the series of terms. Calculate the first term by multiplying D by 9 and subtracting that amount from the tenth term
- This is the first term. Question What is the mathematical formula for the numbers 8, 16, 32, 64, and ? This is not an arithmetic sequence in the traditional sense. Research geometric sequences for any formula you’re interested in learning about. Question How do I compute the 5 terms of an arithmetic sequence if the first term is 8 and the final term is 100, and the first term is 8 and the last term is 100? Take 8 away from 100 to get 92. 92 divided by 4 equals (because with five terms there will be four intervals between the first and last term). This gives you the number 23, which is the length of each interval. As a result, the sequence starts with 8 and has a common difference of 23
- Question How can I find out which term in the arithmetic sequence has the value of -38 in it? The common difference (d) is equal to 4 minus 7 = -3. The first term (a) equals 7. The given period (t) equals -38. (n-1)d = t + (a + (n-1)d, or, -38 = 7 + (n-1)-3, is the formula for time. As a result, n=16, which means that -38 is the sixteenth term
- Question The first three terms of 4n+3 are as follows: The first three terms, starting with n = 1, are 7, 11, and 15
- Question In the sequence 1/2, 1, 2, 4, 8, what is the formula for determining the nth term in the sequence? Alexandre Lima’s full name is Alexandre Lima. Community Answer This is a geometric progression in which each phrase is computed by multiplying the previous term by a predetermined constant before proceeding to the next. When using the example, the constant (q) is two since 2 * (1/2) = one, 2 * one = two, and 2 * two equals four. The formula is: a = a1 x q(n-1)
- For example, a = 1/2 x 2 in the example (n-1). For example, the tenth term is written as a(10) = 1/2 x 2(9) = 256. Question What is the best way to discover the 100th term if I only have the first five terms available? Take a look at Method 3 above, particularly Step 3. Question What if you have the common difference and the first term, but you need to know the a specific number is in relation to what nth number? For example, d=-4, a1=35, and 377 is a term number, correct? The formula for the nth term, denoted by the letter a(n), is provided in Method 3 above. Fill in the blanks with your numbers and solve for n
- Question What is the proper way to use the formula? If you want to discover the “nth” term in an arithmetic series, begin with the first term, which is a. (1). In addition, the product of “n-1” and “d” should be considered (the difference between any two consecutive terms). Consider the arithmetic sequences 3, 9, 15, 21, and 27 as an example. Because the difference between successive terms is always six, a(1) = three, and d = six. Consider the following scenario: you wish to locate the seventh word in the series (n = 7). Then a(7) = a(1) + (n-1)(d) = 3 + (6)(6) = 39, and a(7) = a(1) + (n-1)(d) = 39. In this sequence, number 39 corresponds to the seventh word
- Question What is the best way to locate the first three terms? Suppose you have the fourth, fifth, and sixth terms in the series, for example, 6, 8, and ten, respectively. The formula for finding any term in the series is Un (or Ur) = the first term + the term you are attempting to find minus one (for example, if you were trying to find the fifth term, the formula would be 5 -1) x d, where d is the length of the sequence (the common difference). Because you already know some of the terms in the sequence, you can put in the terms you already know into the formula and solve for the first term to get the answer: U(4) = 6 = U(1) + U(2) = U(4) (4-1) 2. The value of the fourth term, U(4), was provided as 6, and the common difference was found to be 2. After being simplified, the formula looks somewhat like this: 6 is equal to U(1) plus 6. The result of removing 6 from both sides is that U(1) equals 0, and you can use this to get any other term in the series using this formula.

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- There are several distinct types of number sequences to choose from. Do not make the mistake of assuming that a list of integers is an arithmetic series. Make sure to verify at least two pairings of words, and ideally three or four, in order to identify the common difference between them.

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## About This Article

Summary of the Article When looking for a term in an arithmetic series, locate the common difference between the first and second numbers by subtracting the first from the second. Verify that the difference is consistent between each number in the series by re-running the preceding equation with the second and third numbers, third and fourth numbers, and so on until the difference is no longer consistent. Once you’ve determined the common difference, all that’s left to do to locate the missing number is to multiply the common difference by the term that came before it in the series.

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## Did this article help you?

recursive formula for nth term of an arithmetic sequence | _ = _ +d textnge 2 |

explicit formula for nth term of an arithmetic sequence | _ = _ +dleft(n – 1right) |

## Key Concepts

- An arithmetic sequence is a series in which the difference between any two successive terms is a constant
- An example would be The common difference is defined as the constant that exists between two successive terms. It is the number added to any one phrase in an arithmetic sequence that creates the succeeding term that is known as the common difference. The terms of an arithmetic series can be discovered by starting with the first term and repeatedly adding the common difference
- A recursive formula for an arithmetic sequence with common differencedis provided by = +d,nge 2
- A recursive formula for an arithmetic sequence with common differencedis given by = +d,nge 2
- As with any recursive formula, the first term in the series must be specified
- Otherwise, the formula will fail. An explicit formula for an arithmetic sequence with common differenced is provided by = +dleft(n – 1right)
- An example of this formula is = +dleft(n – 1right)
- When determining the number of words in a sequence, it is possible to apply an explicit formula. In application situations, we may modify the explicit formula to = +dn, which is a somewhat different formula.

## Glossary

Arithmetic sequencea sequence in which the difference between any two consecutive terms is a constantcommon difference is a series in which the difference between any two consecutive terms is a constant an arithmetic series is the difference between any two consecutive words in the sequence

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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. Here’s how to make advantage of it:

### 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 & Geometric Sequences

The arithmetic and geometric sequences are the two most straightforward types of sequences to work with. An arithmetic sequence progresses from one term to the next by adding (or removing) the same value on each successive term. For example, the numbers 2, 5, 8, 11, 14,.are arithmetic because each step adds three; while the numbers 7, 3, –1, –5,.are arithmetic because each step subtracts four. The number that is added (or subtracted) at each stage of an arithmetic sequence is referred to as the “common difference”d because if you subtract (that is, if you determine the difference of) subsequent terms, you will always receive this common value as a result of the process.

Below In a geometric sequence, the terms are connected to one another by always multiplying (or dividing) by the same value.

Each step of a geometric sequence is represented by a number that has been multiplied (or divided), which is referred to as the “common ratio.” If you divide (that is, if you determine the ratio of) subsequent terms, you’ll always receive this common value.

#### Find the common difference and the next term of the following sequence:

3, 11, 19, 27, and 35 are the numbers. In order to get the common difference, I must remove each succeeding pair of terms from the total. There’s no point in choosing which couple I want to sit with as long as they’re right next to one other. To be thorough, I’ll go over each and every subtraction: 819 – 11 = 827 – 19 = 835 – 27 = 819 – 11 = 827 – 19 = 835 – 27 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 = 819 – 11 Due to the fact that the difference is always 8, the common difference isd=8.

By adding the common difference to the fifth phrase, I can come up with the next word: 35 plus 8 equals 43 Then here’s my response: “common difference: six-hundred-and-fortieth-term

#### Find the common ratio and the seventh term of the following sequence:

To get the common ratio, I must divide each succeeding pair of terms by the number of terms in the series. There’s no point in choosing which couple I want to sit with as long as they’re right next to one other. I’ll go over all of the divisions to be thorough: The ratio is always three, hence sor= three. As a result, I have five terms remaining; the sixth term will be the next term, and the seventh will be the term after that. The value of the seventh term will be determined by multiplying the fifth term by the common ratio two times.

When it comes to arithmetic sequences, the common difference isd, and the first terma1is commonly referred to as “a “.

As a result of this pattern, the then-th terma n will take the form: n=a+ (n– 1)d When it comes to geometric sequences, the typical ratio isr, and the first terma1 is commonly referred to as “a “.

This pattern will be followed by a phrase with the following form: a n=ar(n– 1) is equal to a n.

#### Find the tenth term and then-th term of the following sequence:

, 1, 2, 4, 8, and so forth. Identifying whether sort of sequence this is (arithmetic or geometric) is the first step in solving the problem. As soon as I look at the differences, I see that they are not equal; for example, the difference between the second and first terms is 2 – 1 = 1, while the difference between the third and second terms is 4 – 2 = 2. As a result, this isn’t a logical sequence. As an alternative, the ratios of succeeding terms remain constant. For example, Two plus one equals twenty-four plus two equals twenty-eight plus four equals two.

The division, on the other hand, would have produced the exact same result.) The series has a common ratio of 2 and the first term is a.

I can simply insert the following into the formulaa n=ar(n– 1) to obtain the then-th term: So, for example, I may plugn= 10 into the then-th term formula and simplify it as follows_n= 10 Then here’s what I’d say: n-th term: tenth term: 256 n-th term

#### Find then-th term and the first three terms of the arithmetic sequence havinga6= 5andd=

The n-th term in an arithmetic series has the form n=a+ (n– 1) d, which stands for n=a+ (n– 1) d. In this particular instance, that formula results in me. When I solve this formula for the value of the first term in the sequence, I obtain the resulta= Then:I have the first three terms in the series as a result of this. Because I know the value of the first term and the common difference, I can also develop the expression for the then-th term, which will be easier to remember: In such case, my response is as follows:n-th word, first three terms:

#### Find then-th term and the first three terms of the arithmetic sequence havinga4= 93anda8= 65.

Due to the fact thata4 anda8 are four places apart, I can determine from the definition of an arithmetic sequence that I can go from the fourth term to the eighth term by multiplying the common difference by four times the fourth term; in other words, the definition informs me that a8=a4 + 4 d. I can then use this information to solve for the common differenced: 65 = 93 + 4 d –28 = 4 d –7 = 65 = 93 + 4 d Also, I know that the fourth term is related to the first term by the formulaa4=a+ (4 – 1) d, so I can get the value of the first terma by using the value I just obtained ford and the value I just discovered fora: 93 =a+ 3(–7) =a+ 3(–7) =a+ 3(–7) =a+ 3(–7) =a+ 3(–7) =a+ 3(–7) 93 plus 21 equals 114.

As soon as I know what the first term’s value is and what the value of the common difference is, I can use the plug-and-chug method to figure out what the first three terms’ values are, as well as the general form of the fourth term: The numbers are as follows: a1= 114, a2= 114– 7, a3= 107– 7, and an= 114 + (n – 1)(–7)= 114 – 7, n+ 7, and an= 121–7, respectively.

#### Find then-th and the26 th terms of the geometric sequence withanda12= 160.

Given that the two words for which they’ve provided numerical values are separated by 12 – 5 = 7 places, I know that I can go from the fifth term to the twelfth term by multiplying the fifth term by the common ratio seven times; that is, a12= (a5) (r7). I can use this to figure out what the value of the common ratior should be: I also know that the fifth component is related to the first by the formulaa5=ar4, so I can use that knowledge to solve for the value of the first term, which is as follows: Now that I know the value of the first term as well as the value of the common ratio, I can put both into the formula for the then-th term to obtain the following result: I can assess the twenty-sixth term using this formula, and it is as follows, simplified: Then here’s my response:n-th term: 2,621,440 for the 26th term Once we have mastered the art of working with sequences of arithmetic and geometric expressions, we may move on to the concerns of combining these sequences together.

## 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. |

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.

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 Termthe 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. In this case, issequence B.

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. |

## Find the next three terms of the arithmetic sequence. 3/4, 1/2, 1/4, 1, .

Find the arithmetic sequence’s following three terms by rearranging them. frac, frac, frac, 1, frac, frac, 1.

## Arithmetic Sequence and It’s nth Term:

The arithmetic sequence is a sequence of numbers in mathematics that consists of a constant difference between any two continuous terms of the sequence and is composed of a constant difference between any two continuous terms of the sequence. This constant difference is referred to as the common difference of the series, and it is symbolized by the letter ‘d’ in the notation. General representation of an arithmetic series -$$a,(a+d),(a+2d),.a+(n-1)d$$, where an is the first term and a+(n-1)d$$ is the total number of terms in the arithmetic sequencen is the common difference of all the terms in the sequence

#### nth term

Formula-a =a+ is a mathematical formula (n-1) Arithmetic sequence term with the value of n is represented by the symbol dherea_

## Answer and Explanation:1

The arithmetic sequence that has been provided is-$$frac,frac,frac,. In the given sequence-$$begintext(a)= fractext(d)= frac-$$begintext(d)= frac-$$begintext(d)= frac-$$begintext(d)= frac-$$begintext(d)= frac-$$begintext(d)= frac-$$begina = frac+(4-1)fraca

## 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.

Take a moment to confirm that this equation accurately reflects the sequence you’ve been given. 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!

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. There are two variables, a1 and a100. 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.

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.

Calculate the sum of the first 60 terms of the following sequence of numbers: 5, 0, 5, 10, 15,.

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

- Given the first term and common difference of an arithmetic series, write the first five terms of the sequence. Find a formula that describes the generic term. The values of a1 are 5 and 3
- 12 and 2
- 15 and 5
- 7 and 4 respectively
- 12 and 1
- A1=23 and 13 respectively
- 1 and 12 respectively
- A1=54 and 14. The values of a1 are 1.8 and 0.6
- 4.3 and 2.1
- And a1=5.4 and 2.1 respectively.

- Locate a formula for the general term and apply it to get the 100 thterm, given the arithmetic series given the sequence 0.8, 2, 3.2, 4.4, 5.6,.
- 4.4, 7.5, 13.7, 16.8,.
- 3, 8, 13, 18, 23,.
- 3, 7, 11, 15, 19,.
- 6, 14, 22, 30, 38,.
- 5, 10, 15, 20, 25,.
- 2, 4, 6, 8, 10,.
- 12,52,92,132,.
- 13, 23, 53,83,.
- 14,12,54,2,114,. Find the positive odd integer that is 50th
- Find the positive even integer that is 50th
- Find the 40 th term in the sequence that consists of every other positive odd integer in the following format: 1, 5, 9, 13,.
- Find the 40th term in the sequence that consists of every other positive even integer: 1, 5, 9, 13,.
- Find the 40th term in the sequence that consists of every other positive even integer: 2, 6, 10, 14,.
- 2, 6, 10, 14,. What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
- What number is the phrase 172 in the arithmetic sequence 4, 4, 12, 20, 28,.
- What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
- 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
- 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
- This is the problem.

- Find a formula for the general term from the terms of an arithmetic sequence given the terms of the series. 1 = 6 and 7 = 42
- 1 = 12 and 12= 6
- 1 = 19 and 26 = 56
- 1 = 9 and 31 = 141
- 1 = 16 and 10 = 376
- 1 = 54 and 11 = 654. 1 = 6 and 7 = 42
- 1= 9 and 31 = 141
- 1 = 6 and 7

- Find all of the arithmetic means that exist between the two supplied terms. a1=3anda6=17
- A1=5anda5=7
- A2=4anda8=7
- A5=12anda9=72
- A5=15anda7=21
- A6=4anda11=1

### Part B: Arithmetic Series

- In light of the general term’s formula, figure out how much the suggested total is. an=3n+5
- S100
- An=5n11
- An=12n
- S70
- An=132n
- S120
- An=12n34
- S20
- An=n35
- S150
- An=455n
- S65
- An=2n48
- S95
- An=4.41.6n
- S75
- An=6.5n3.3
- S67
- An=3n+5

- Evaluate. 1160(3n)
- 1121(2n)
- 1250(4n-3)
- 1120(2n+12)
- 170 (198n)
- 1220(5n)
- 160(5212n)
- 151(38+14
- 1120(1.5n+2.6)
- 1175(0.2N1.6)
- 1170 (19 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.

- 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 numbers
- The sum of the first 200 positive odd integers
- The sum of the first 500 positive odd integers
- The sum of the first 50 positive even numbers
- The sum of the first 200 positive even integers
- The sum of the first 500 positive even integers
- The sum of the firstk positive odd integers
- The sum of the firstk positive odd integers the sum of the firstk positive even integers
- The sum of the firstk positive odd integers
- 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
- 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 wage contract with a 10-year term pays $65,000 in the first year, with a $3,200 raise for each consecutive year after. Calculate the entire salary obligation over a ten-year period (see Figure 1). In accordance with the hour, a clock tower knocks 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

- 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
- Discuss strategies for computing sums in situations when the index does not begin with one. For example, n=1535(3n+4)=1,659
- N=1535(3n+4)=1,659
- 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?

### Answers

- An=3n+2
- An=5n+3
- An=6n
- An=3n+2
- An=6n+3
- An=6n+2

- 1,565,450, 2,500,450, k2,
- 90,800, k4,230,
- 38640, 124,750,
- 18,550, k765
- 10,578
- 20,100,
- 2,500,550, k2,
- 294 seats, 247 bricks, $794,000, and so on.