# How To Find A Specific Term In An Arithmetic Sequence? (Correct answer)

sequence determined by a = 2 and d = 3. Solution: To find a specific term of an arithmetic sequence, we use the formula for finding the nth term. Step 1: The nth term of an arithmetic sequence is given by an = a + (n – 1)d. So, to find the nth term, substitute the given values a = 2 and d = 3 into the formula.

## What is the formula for finding terms in an arithmetic sequence?

Finding the number of terms in an arithmetic sequence might sound like a complex task, but it’s actually pretty straightforward. All you need to do is plug the given values into the formula tn = a + (n – 1) d and solve for n, which is the number of terms.

## 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.

## How do you find the terms of a quadratic sequence?

Sequence – 2n² If we combine an2 with bn + c, we get the nth term rule of our quadratic sequence 2n2 + 3n + 1. You might be wondering why, to find the nth term of a quadratic sequence, we divide the second difference by 2 to find the value of a, when in a linear sequence we can just use the difference itself.

## 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.

## Finite Sequence: Definition & Examples – Video & Lesson Transcript

When there is a finite series, the first term is followed by a second term, and so on until the last term. In a finite sequence, the letternoften reflects the total number of phrases in the sequence. In a finite series, the first term may be represented by a (1), the second term can be represented by a (2), etc. Parentheses are often used to separate numbers adjacent to thea, but parentheses will be used at other points in this course to distinguish them from subscripts. This terminology is illustrated in the following graphic.

## Examples

Despite the fact that there are many other forms of finite sequences, we shall confine ourselves to the field of mathematics for the time being. The prime numbers smaller than 40, for example, are an example of a finite sequence, as seen in the table below: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, Another example is the range of natural numbers between zero and one hundred. Due to the fact that it would be tedious to write down all of the terms in this finite sequence, we will demonstrate it as follows: 1, 2, 3, 4, 5,., 100 are the numbers from 1 to 100.

Alternatively, there might be alternative sequences that begin in the same way and end in 100, but which do not contain the natural numbers less than or equal to 100.

## Finding Patterns

Let’s take a look at some finite sequences to see if there are any patterns.

#### Example 1

Examine a few finite sequences to see if any patterns can be discovered.

## Finding Sums of Finite Arithmetic Series – Sequences and Series (Algebra 2)

The sum of all terms for a finitearithmetic sequencegiven bywherea1 is the first term,dis the common difference, andnis the number of terms may be determined using the following formula: wherea1 is the first term,dis the common difference, andnis the number of terms Consider the arithmetic sum as an example of how the total is computed in this manner. If you choose not to add all of the words at once, keep in mind that the first and last terms may be recast as2fives. This method may be used to rewrite the second and second-to-last terms as well.

There are 9fives in all, and the aggregate is 9 x 5 = 9.

expand more Due to the fact that the difference between each term is constant, this sequence from 1 through 1000 is arithmetic.

The first phrase, a1, is one and the last term, is one thousand thousand. The total number of terms is less than 1000. The sum of all positive integers up to and including 1000 is 500 500. }}}!}}}Test}}} arrow back} arrow forwardarrow leftarrow right

## How to Find Any Term of an Arithmetic Sequence

Documentation Download Documentation Download Documentation An arithmetic sequence is a collection of integers that differ from one another by a fixed amount from one to the next. Consider the following example: the list of even integers. This is an arithmetic sequence since the difference between one number in the list and the next is always 2. It is possible to be requested to discover the very next phrase from a list of terms if you are aware that you are working with an arithmetic sequence.

Finally, you could be interested in knowing, for example, the 100th phrase without having to write down all 100 words one by one.

1. 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. 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
3. 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. 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
2. 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
3. 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. 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).
1. 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. 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
• 3 Determine the length of a series of characters or numbers. Allow us to imagine that we already know the beginning and ending points of some sort of mathematical series, but we want to know how long it is. To make use of the new formula,
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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.

• 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.

## Video

• Remember that depending on whether it is being added or removed, the result might be either positive or negative.

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.

Did you find this overview to be helpful?

• 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.

1. Considering that the average difference is 50, the series represents a linear function with an associated slope of 50.
2. 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.
3. When working with sequences, we substitute _instead of y and ninstead of n.
4. 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.

For this sequence, there is another explicit formula, which is_ =200 – 50left(n – 1right), which may be simplified to_ =-50n+250.

### 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

What’s the most significant distinction between you and everyone else? To solve for = +dleft(n – 1right), substitute the common difference and the first term.

### 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.

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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

### 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?

## Number Sequence Calculator

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

## Geometric Sequence Calculator

For example, the numbers 1, 2, 4, 8, 16, 32, 64, and 128 are all represented by the symbol a n.

## Fibonacci Sequence Calculator

For example, the numbers 1, 2, 4, 8, 16, 32, 64, and 128 are all represented as a n.

### Arithmetic Sequence

When it comes to number sequences, an arithmetic series is one in which the difference between each consecutive term remains constant. Because of this disparity, the arithmetic series will either go towards positive or negative infinity, depending on the sign of the difference between the two values. The generic form of an arithmetic sequence is represented by the notation:

 a n= a 1+ f × (n-1)or more generally wherea nrefers to then thterm in the sequence a n= a m+ f × (n-m) a 1is the first term i.e. a 1, a 1+ f, a 1+ 2f,. fis the common difference EX: 1, 3, 5, 7, 9, 11, 13,.

It is evident from the preceding sequence that the common differencef is number two. Using the equation above, we can compute the fifth term as follows:

 EX: a 5= a 1+ f × (n-1)a 5= 1 + 2 × (5-1)a 5= 1 + 8 = 9

When the sequence is compared to the equation, it can be observed that the 5th term, a 5, which was discovered using the equation, fits the sequence exactly as predicted. The sum of an arithmetic series may also be computed in a straightforward and straightforward manner using the following formula, in conjunction with the prior approach to finda n: Determine how many terms are in the arithmetic series through the 5 thterm, using the same numerical sequence as in the preceding example:

 EX: 1 + 3 + 5 + 7 + 9 = 25(5 × (1 + 9))/2 = 50/2 = 25

### Geometric Sequence

When a number series begins with a single number, it is called a geometric sequence. Each succeeding number following the initial number is the product of the preceding number multiplied by a fixed, non-zero integer (common ratio). The following is an example of the general form of a geometric sequence:

 a n= a × r n-1 wherea nrefers to then thterm in the sequence i.e. a, ar, ar 2, ar 3,. ais the scale factor andris the common ratio EX: 1, 2, 4, 8, 16, 32, 64, 128,.

According to the preceding example, the common ratioris 2 and the scale factorais 1. Calculate the eighth term based on the equation provided above: When the value obtained from the equation is compared to the geometric sequence shown above, it is confirmed that they are identical. The following is the mathematical equation for computing the sum of a geometric sequence: Determine how much the total of the geometric sequence through the 3 rdterm is using the same geometric sequence as previously.

### Fibonacci Sequence

An example of this is a Fibonacci sequence, in which every number following the initial two numbers is equal to the sum of the two numbers that came before it. Based on where you begin, the first two numbers in a Fibonacci sequence are defined as either 1 and 1, or 0 and 1, depending on the starting point you choose. Fibonacci numbers appear frequently and unpredictably in mathematics, and they have been the topic of a plethora of investigations. These equations have applications in computer systems (such as Euclid’s method to compute the greatest common factor), economics, and biological situations, such as the branching of trees and the blossoming of an artichoke, among many other areas of study.

The Fibonacci sequence is stated as follows in mathematical notation:

 a n= a n-1+ a n-2 wherea nrefers to then thterm in the sequence EX: 0, 1, 1, 2, 3, 5, 8, 13, 21,. a 0= 0; a 1= 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.

The terms of an arithmetic sequence that occur between two supplied terms.

### Example 3

Locate 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: Find the common difference, d=4+6+2 to get started. Afterwards, determine the general term’s formula, which isa1=6andd=2. an=a1+(n−1)d=6+(n−1)⋅(−2)=6−2n+2=8−2n The 75 thterm may be computed as follows: an=8n and the 75 thterm are equal to an=8n. a75=8−2(75)=8−150=−142 Answer_an=8−2n;a100=−142 Arithmetic means are the terms that appear between two supplied words in an arithmetic sequence.

### 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!

## 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.

1. where a1=26 and d=2.
2. As a result, the number of seats in each row may be calculated using the formulaan=2n+24.
3. 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.
4. Calculate the sum of the first 60 terms of the following sequence of numbers: 5, 0, 5, 10, 15,.
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### 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 a formula for the general term given 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=16anda15=76
9. A4

### 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=5n−11
4. An=12−n
5. S70
6. An=1−32n
7. S120
8. An=12n−34
9. S20
10. An=n−35
11. S150
12. An=45−5n
13. S65
14. An=2n−48
15. S95
16. An=4.4−1.6n
17. S75
18. An=6.5n−3.3
19. S67
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 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

1. It may be expressed asan=2n1 where an=2n1 is the generic term for the sequence of positive odd numbers. in this case, byan=2n is used to denote a generic phrase for a succession of positive even numbers. The following are to be found
2. There are four different types of sums: the sum of the first 50 positive odd integers
3. The sum of the first 200 positive odd integers
4. The sum of the first 50 positive even integers
5. And the sum of the first 200 positive even integers. The sum of the firstk positive odd integers
6. The sum of the firstk positive odd integers (a) The sum of the firstkpositive even integers
7. (b) The sum of the firstknegative even integers
8. In a tiny theater, the first row of seating has a total of 8 places. Following that, each row has three additional seats than the one before it. How many total seats are there in the theater if there are 12 rows? If you’re looking for seats in an outdoor amphitheater, the first row has 42 seats, the second row has 44 seats, the third row has 46 seats, and so on. What is the overall seating capacity of the theater if there are 22 rows of seats
9. And 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 pile? Each succeeding row of a triangle stack of bricks contains one fewer brick, until there is just one brick remaining on the summit of the pile. If there are 210 bricks in all, how many rows does the stack have? a wage contract with a 10-year term that pays \$65,000 in the first year and increases by \$3,200 every year afterwards Calculate the entire salary obligation over the course of ten years. It is customary for a clock tower to chime its bell a certain number of times every hour. The clock strikes once at one o’clock, twice at two o’clock, and so on throughout the day. I’m curious how many times the clock tower’s bell rings in a day.

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

## 7.6.2: Finding the nth Term Given Two Terms for an Arithmetic Sequence

You’re making monthly payments toward the repayment of a student debt. Your loan debt will be \$17,500 once you have made your fifth payment on the loan. You will have a residual amount of \$12,000 after making your 16th payment. The n-th term rule for the circumstance described by this situation is: Previously, we were either given the common difference explicitly or were given two successive words from which we might deduce the common difference (see below). In this notion, given any two terms in a series, we will determine the common difference and write the n thterm rule for each term.

We will begin by creating two equations in two variables using the (n) term rule for an arithmetic sequence, as shown in the following example: Because ((a_ =17)), we can write ((a_ +(7-1) d=17) or more simply: ((a_ +6 d=17)(((a_ =82)), we can write ((a_ +(20-1) d=82) or more simply: ((((a_ +19 d=82)) Solve the following system that has resulted: Then, instead of starting with a 1 +6d =17, we start with a 1 +6d =17.

underlinequadquadRightarrow quadquadunderline -19d=-82-13d=-65d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d=5d= By substituting the number 5 for the letter d in one of the equations, we obtain the following: (begin a_ +6(5)=7 a_ +30=17 a =-13 end a_) Using these numbers, we can calculate the n thterm rule as follows: ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( (5) The expression (a_ =-13+5 n-5 a_ =5 n-18 a_ =13+5 n-5 a_ =5 n-18 end ) is equivalent to In this section, we’ll look at the common difference rule, first term rule, and (n) term rule for an arithmetic sequence in which (a_ =-13) and (a_ =-71) are both true.

• Despite the fact that this is precisely the same sort of problem as the last problem above, we will use a different strategy to solving it.
• We’re going to utilize that concept to figure out what the common denominator is.
• The difference between the term values is 71(13), which is -58 in this case.
• We may calculate the common difference by subtracting the terms and dividing the result by the difference in term number.
• ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( The following results from applying the (n) term rule: (a_ =7+(n-1)(-22)=7-2 n+2=-2 n+9).
• Have you ever noticed that the simplified n-term rule, (a_ =p n+q), in which p and q represent constants, looks a bit like (y=m x+b), the slope-intercept form of the equation of a line, in which p and q represent constants?
• to see what we can learn.
• The points are as follows: (1, 1), (2, 4), (3, 7), and (4).
• This is due to the fact that for every one rise in the term number (x), the term value (y) increases by three times.

As an example of how we may use the slope, 3, and the point (1, 1) in the equation (y=m x+b) to obtain the equation of this line, consider the following: ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( It is as follows: (a_ =3 n-1), where n is the number of terms in the sequence.

• This time, in order to formulate an equation, we will make use of the idea that the terms in an arithmetic sequence are really points on a line.
• We can determine the slope and the equation by following the steps outlined above.
• (Beginning at -50=-6 (10), going to 50=-60, and ending at 10 = B.) As a result, (y=-6 x+10) and (a_ =-6 n+10) are true.
• 1 The sequence represented by the student loan problem was given to you before, and you were required to discover the (n) term rule for it.
• We can determine the slope and the equation by following the steps outlined above.
• Exemple No.
• Solution We may deduce the equation (a_ =-13) from the expression (a_ +(6-1) d=a_ +5 d=-13).

Use the following two equations to find the values of (a_ ) and (d_ ): ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( Rightarrow a 1-15=-13-9d=17a 1=2 a 1-15=-13-9d=17a 1=2 a 1-15=-13-9d=17a 1=2 a 1-15=-13-9d=17a 1=2 a 1-15=-13-9d=17a 1=2 (d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d=-3, d= To solve this problem, use the following rule: (a_ =2+(n-1)(-3)=2-3 n+3=-3 n+5); (b_ =2+(n-1)(-3)=2-3 n+3=3 n+5; Example 3Find the n-term rule for the arithmetic sequence with the values (a_ =13) and (a_ =77) as the first and second terms.

1. Solution The most often encountered difference is (frac=frac=4).
2. Exemple No.
3. Solution The point (7,-75) is obtained by dividing (a_ =-75) by (a_ =-75).
4. The slope between these two places is (frac=frac=-11) (frac=frac=-11).

Make use of the two terms provided to come up with a (n) term rule for the sequence.

1. The values of (a_ =-17) and (a_ =-71)
2. (a_ =23) and (a_ =85)
3. (a_ =-6)
4. (a_ =-3)
5. (a_ =24)
6. (a_ =9)
7. (a_ =20)
8. And (a_ =-54)
9. (A) and (B)
10. (a_ =23)
11. (a_ =14)
12. (a_ =3 So, which way do you favor the most? Why

To examine the solutions to the Review questions, open this PDF file and go down to section 11.6.