What Makes A Sequence An Arithmetic Sequence? (Solved)

An arithmetic sequence is a sequence (list of numbers) that has a common difference (a positive or negative constant) between the consecutive terms.

Contents

What defines an arithmetic sequence?

An arithmetic sequence is a sequence where each term increases by adding/subtracting some constant k. This is in contrast to a geometric sequence where each term increases by dividing/multiplying some constant k.

How do you know if a sequence is arithmetic?

Definition and Basic Examples of Arithmetic Sequence. An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.

Which sequence is an arithmetic sequence?

An arithmetic progression, or arithmetic sequence, is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5,7,9,11,13,⋯ 5, 7, 9, 11, 13, ⋯ is an arithmetic sequence with common difference of 2.

How is an arithmetic sequence different from other sequences?

Sequences with such patterns are called arithmetic sequences. In an arithmetic sequence, the difference between consecutive terms is always the same. For example, the sequence 3, 5, 7, 9 is arithmetic because the difference between consecutive terms is always two.

What makes an arithmetic sequence different from a geometric sequence?

An arithmetic Sequence is a set of numbers in which each new phrase differs from the previous term by a fixed amount. Geometric Sequence is a series of integers in which each element after the first is obtained by multiplying the preceding number by a constant factor.

What is an arithmetic sequence? + Example

An arithmetic sequence is a series (list of numbers) in which there is a common difference (a positive or negative constant) between the items that are consecutively listed. For example, consider the following instances of arithmetic sequences: 1.) The numbers 7, 14, 21, and 28 are used because the common difference is 7. 2.) The numbers 48, 45, 42, and 39 are chosen because they have a common difference of – 3. The following are instances of arithmetic sequences that are not to be confused with them: It is not 2,4,8,16 since the difference between the first and second terms is 2, but the difference between the second and third terms is 4, and the difference between the third and fourth terms is 8 because the difference between the first and second terms is 2.

2.) The numbers 1, 4, 9, and 16 are incorrect because the difference between the first and second is 3, the difference between the second and third is 5, and the difference between the third and fourth is 7.

The reasons for this are that the difference between the first and second is three points, the difference between the second and third is two points, and the difference between third and fourth is five points.

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

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.

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. As a result, it would be advisable to express the sequence in such a way that the reader can grasp the overall pattern.

Finding Patterns

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

Example 1

An anarithmetic finite sequence is a sequence in which all pairs of succeeding terms share a common difference and is defined as follows: Figure out which of these arithmetic finite sequences has the most common difference: 2, 7, 12, 17,., 47 are the prime numbers. Because the common difference is 5, the first four terms of the series demonstrate that the common difference is 5. With another way of saying it, we may add 5 to any phrase in the series to get the following term in the sequence.

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 sum of all positive integers up to and including 1000 is 500 500.

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

Advanced Topic: Summing an Arithmetic Series

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

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

The values ofa,dandnare as follows:

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

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

Footnote: Why Does the Formula Work?

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

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

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

Arithmetic Sequences

In mathematics, an arithmetic sequence is a succession of integers in which the value of each number grows or decreases by a fixed amount each term. When an arithmetic sequence has n terms, we may construct a formula for each term in the form fn+c, where d is the common difference. Once you’ve determined the common difference, you can calculate the value ofcby substituting 1fornand the first term in the series fora1 into the equation. Example 1: The arithmetic sequence 1,5,9,13,17,21,25 is an arithmetic series with a common difference of four.

  1. For the thenthterm, we substituten=1,a1=1andd=4inan=dn+cto findc, which is the formula for thenthterm.
  2. As an example, the arithmetic sequence 12-9-6-3-0-3-6-0 is an arithmetic series with a common difference of three.
  3. It is important to note that, because the series is decreasing, the common difference is a negative number.) To determine the next3 terms, we just keep subtracting3: 6 3=9 9 3=12 12 3=15 6 3=9 9 3=12 12 3=15 As a result, the next three terms are 9, 12, and 15.
  4. As a result, the formula for the fifteenth term in this series isan=3n+15.

Exemple No. 3: The number series 2,3,5,8,12,17,23,. is not an arithmetic sequence. Differencea2 is 1, but the following differencea3 is 2, and the differencea4 is 3. There is no way to write a formula in the form of forman=dn+c for this sequence. Geometric sequences are another type of sequence.

Arithmetic Sequence: Formula & Definition – Video & Lesson Transcript

Afterwards, the th term in a series will be denoted by the symbol (n). The first term of a series is a (1), and the 23rd term of a sequence is the letter a (1). (23). Parentheses will be used at several points in this course to indicate that the numbers next to thea are generally written as subscripts.

Finding the Terms

This will be represented by the letter A for the th term in the series (n). The first term of a series is a (1), and the 23rd term of a sequence is the letter a (3). (23). Parentheses will be used at several points in this course to indicate that the numbers adjacent to theaare normally written as subscripts.

Finding then th Term

Consider the identical sequence as in the preceding example, with the exception that we must now discover the 33rd word oracle (33). We may utilize the same strategy as previously, but it would take a long time to complete the project. We need to come up with a way that is both faster and more efficient. We are aware that we are starting with ata (1), which is a negative number. We multiply each phrase by 5 to get the next term. To go from a (1) to a (33), we’d have to add 32 consecutive terms (33 – 1 = 32) to the beginning of the sequence.

To put it another way, a (33) = -3 + (33 – 1)5.

a (33) = -3 + (33 – 1)5 = -3 + 160 = 157.

Then the relationship between the th term and the initial terma (1) and the common differencedis provided by:

Arithmetic Sequences and Series

An arithmetic sequence is a set of integers in which the difference between the words that follow is always the same as its predecessor.

Learning Objectives

Make a calculation for the nth term of an arithmetic sequence and then define the characteristics of arithmetic sequences.

Key Takeaways

  • When the common differenced is used, the behavior of the arithmetic sequence is determined. Arithmetic sequences may be either limited or infinite in length.

Key Terms

  • Arithmetic sequence: An ordered list of numbers in which the difference between the subsequent terms is constant
  • Endless: An ordered list of numbers in which the difference between the consecutive terms is infinite
  • Infinite, unending, without beginning or end
  • Limitless
  • Innumerable

For example, an arithmetic progression or arithmetic sequence is a succession of integers in which the difference between the following terms is always the same as the difference between the previous terms. A common difference of 2 may be found in the arithmetic sequence 5, 7, 9, 11, 13, cdots, which is an example of an arithmetic sequence.

  • 1: The initial term in the series
  • D: The difference between the common differences of consecutive terms
  • A 1: a n: Then the nth term in the series.

The behavior of the arithmetic sequence is determined by the common differenced arithmetic sequence. If the common difference,d, is the following:

  • Positively, the sequence will continue to develop towards infinity (+infty). If the sequence is negative, it will regress towards negative infinity (-infty)
  • If it is positive, it will regress towards positive infinity (-infty).

It should be noted that the first term in the series can be thought of asa 1+0cdot d, the second term can be thought of asa 1+1cdot d, and the third term can be thought of asa 1+2cdot d, and therefore the following equation givesa n:a n In the equation a n= a 1+(n1)cdot D Of course, one may always type down each term until one has the term desired—but if one need the 50th term, this can be time-consuming and inefficient.

6.2: Arithmetic and Geometric Sequences

Arithmetic sequences and geometric sequences are two forms of mathematical sequences that are commonly encountered. In an arithmetic sequence, there is a constant difference between each subsequent pair of words in the sequence. There are some parallels between this and linear functions of the type (y=m x+b). Among any pair of subsequent words in a geometric series, there is a constant ratio between them. This would have the effect of a constant multiplier being applied to the data. Examples The Arithmetic Sequence is as follows: Take note that the constant difference in this case is 6.

For the n-th term, one method is to use as the coefficient the constant difference between the two terms: (a_ =6n+?).

We may state the following about the sequence: (a_ =6 n-1); (a_ =6 n-1); (a_ =6 n-1); The following is an example of a formula that you can memorize: Any integer sequence with a constant difference (d) is stated as follows: (a_ =a_ +(n-1) d) = (a_ =a_ +(n-1) d) = (a_ =a_ +(n-1) d) It’s important to note that if we use the values from our example, we receive the same result as we did before: (a_ =a_ +(n-1) d)(a_ =5, d=6)(a_ =5, d=6)(a_ =5, d=6) As a result, (a_ +(n-1) d=5+(n-1) * 6=5+6 n-6=6 n-1), or (a_ =6 n-1), or (a_ =6 n-1) A negative integer represents the constant difference when the terms of an arithmetic sequence are growing smaller as time goes on.

  • (a_ =-5 n+29) (a_ =-5 n+29) (a_ =-5 n+29) Sequence of Geometric Shapes With geometric sequences, the constant multiplier remains constant throughout the whole series.
  • Unless the multiplier is less than (1,) then the terms will get more tiny.
  • Similarly, if the terms are becoming smaller, the multiplier would be in the denominator.
  • The exercises are as follows: (a_ =frac) or (a_ =frac) or (a_ =50 *left(fracright)) and so on.
  • If the problem involves arithmetic, find out what the constant difference is.

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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

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

The sum of the first 50 terms of the following sequence: 4, 9, 14, 19, 24,. is to be determined. The solution is to determine whether or not there is a common difference between the concepts that have been presented. d=9−4=5 Keep in mind that the difference between any two consecutive phrases is a factor of 5 The sequence is, in fact, an arithmetic progression, and we may writean=a1+(n1)d=4+(n1)5=4+5n5=5n1 as an anagram of an=a1+(n1)d. The generic phrase isan=5n1, as a result of which For this sequence, we require the 1st and 50th terms to compute the 50 thpartial sum of the series.

Sn=n(a1+an)2S50=50.(a1+a50)2=50(4+249) 2=25(253)=6,325 Answer_S50=6,325

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,.
  5. Answer:S60=−8,550
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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 all possible arithmetic means between the given terms: a1=3anda6=17
  2. A1=5anda5=7
  3. A2=4anda8=7
  4. A5=12anda9=72
  5. A5=15anda7=21
  6. A6=4anda11=1
  7. A7=4anda11=1

Part B: Arithmetic Series

  1. Make a calculation for the provided total based on the formula for the general term an=3n+5
  2. S100
  3. An=5n11
  4. An=12n
  5. S70
  6. An=132n
  7. S120
  8. An=12n34
  9. S20
  10. An=n35
  11. S150
  12. An=455n
  13. S65
  14. An=2n48
  15. S95
  16. An=4.41.6n
  17. S75
  18. An=6.5n3.3
  19. S67
  20. An=3n+5
  1. Consider the following values: n=1160(3n)
  2. N=1121(2n)
  3. N=1250(4n3)
  4. N=1120(2n+12)
  5. N=170(198n)
  6. N=1220(5n)
  7. N=160(5212n)
  8. N=151(38n+14)
  9. N=1120(1.5n2.6)
  10. N=1175(0.2n1.6)
  11. The total of all 200 positive integers is found by counting them up. To solve this problem, find the sum of the first 400 positive integers.
  1. The generic term for a sequence of positive odd integers is denoted byan=2n1 and is defined as follows: Furthermore, the generic phrase for a sequence of positive even integers is denoted by the number an=2n. Look for the following: The sum of the first 50 positive odd integers
  2. The sum of the first 200 positive odd integers
  3. The sum of the first 50 positive even integers
  4. The sum of the first 200 positive even integers
  5. The sum of the first 100 positive even integers
  6. The sum of the firstk positive odd integers
  7. The sum of the firstk positive odd integers the sum of the firstk positive even integers
  8. The sum of the firstk positive odd integers
  9. There are eight seats in the front row of a tiny theater, which is the standard configuration. Following that, each row contains three additional seats than the one before it. How many total seats are there in the theater if there are 12 rows of seats? In an outdoor amphitheater, the first row of seating comprises 42 seats, the second row contains 44 seats, the third row contains 46 seats, and so on and so forth. When there are 22 rows, how many people can fit in the theater’s entire seating capacity? The number of bricks in a triangle stack are as follows: 37 bricks on the bottom row, 34 bricks on the second row and so on, ending with one brick on the top row. What is the total number of bricks in the stack
  10. Each succeeding row of a triangle stack of bricks contains one fewer brick, until there is just one brick remaining on the top of the stack. Given a total of 210 bricks in the stack, how many rows does the stack have? A 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. Is the Fibonacci sequence an arithmetic series or a geometric sequence? How to explain: Using the formula for the then th partial sum of an arithmetic sequenceSn=n(a1+an)2and the formula for the general terman=a1+(n1)dto derive a new formula for the then th partial sum of an arithmetic sequenceSn=n2, we can derive the formula for the then th partial sum of an arithmetic sequenceSn=n2. How would this formula be beneficial in the given situation? Explain with the use of an example of your own creation
  2. Discuss strategies for computing sums in situations when the index does not begin with one. For example, n=1535(3n+4)=1,659
  3. N=1535(3n+4)=1,659
  4. Carl Friedrich Gauss is the subject of a well-known tale about his misbehaving in school. As a punishment, his instructor assigned him the chore of adding the first 100 integers to his list of disciplinary actions. According to folklore, young Gauss replied accurately within seconds of being asked. The question is, what is the solution, and how do you believe he was able to come up with the figure so quickly?

Answers

  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

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:

The arithmetic and geometric sequences are the two basic types of sequences to work with. It is possible to go from one term to another by adding (or deleting) the same value on each successive term. 2. 5. 8. 11. 14.is arithmetic because each step adds three; and 7. 3. –1, 5.is arithmetic because each step subtracts four; for example, 2, 5, 8, 11, 14. When a number is added (or subtracted) at each stage of an arithmetic sequence, it is referred to as the “common difference”d. This is because if you subtract (that is, if you discover the difference of) subsequent terms, you will always receive the same common value.

The numbers 1, 2, 4, 8, 16, and 81, 27, 9, 3, 1,.

are geometric because each step divides by three.

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. Before the next test, make a note of the formulae for the tenth term.

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 – Precalculus

Sequences, probability, and counting theory are all topics covered in this course.

Learning Objectives

You will learn the following things in this section:

  • Calculate the common difference between two arithmetic sequences
  • Make a list of the terms in an arithmetic sequence
  • When dealing with an arithmetic series, use a recursive formula. When dealing with an arithmetic series, use an explicit formula.

Large purchases, such as computers and automobiles, are frequently made by businesses for their own use. For taxation reasons, the book-value of these supplies diminishes with each passing year. Depreciation is the term used to describe this decline in value. Depreciation may be calculated in several ways, one of which is straight-line depreciation, which means that the value of the asset drops by the same amount each year. Consider the case of a lady who decides to start her own modest contracting firm.

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She expects to be able to sell the truck for $8,000 after five years, according to her estimations.

After one year, the vehicle will be worth?21,600; after two years, it will be worth?18,200; after three years, it will be worth?14,800; after four years, it will be worth?11,400; and after five years, it will be worth?8,000.

Finding Common Differences

The values of the vehicle in the example are said to constitute an anarithmetic sequence since they vary by a consistent amount each year, according to the definition. Every term grows or decreases by the same constant amount, which is referred to as the common difference of the sequence. –3,400 is the common difference between the two sequences in this case. Another example of an arithmetic series may be seen in the sequence below. In this situation, the constant difference is three times more than one.

Sequence of Arithmetic Operations When two successive words are added together, the difference between them is a constant.

Given that and is the initial term of an arithmetic sequence, and is the common difference, then the sequence will be as follows: Identifying Commonalities and Dissimilarities Is each of the sequences mathematical in nature?

To establish whether or not there is a common difference between two terms, subtract each phrase from the succeeding term.

  1. The series is not arithmetic because there is no common difference between the elements
  2. The sequence is arithmetic because there is a common difference between the elements. The most often encountered difference is 4

Analysis Each of these sequences is represented by a graph, which is depicted in (Figure). We can observe from the graphs that, despite the fact that both sequences exhibit increase, is not linear whereas is linear, as we previously said. Given that arithmetic sequences have an invariant rate of change, their graphs will always consist of points on a straight line. If we are informed that a series is arithmetic, do we have to subtract every term from the term after it in order to identify the common difference between the terms?

As long as we know that the sequence is arithmetic, we may take any one term from it and subtract it from the following term to determine the common difference.

If this is the case, identify the common difference. The sequence follows a mathematical pattern. The main distinction is whether or not the provided sequence is arithmetic. If this is the case, identify the common difference. Because of this, the sequence is not arithmetic.

Writing Terms of Arithmetic Sequences

After recognizing an arithmetic sequence, we can determine the terms if we are provided the first term as well as the common difference between the two terms. The terms may be discovered by starting with the first term and repeatedly adding the common difference to the end of the list. In addition, any term may be obtained by inserting the values of and into the formula below, which is shown below. Find the first many terms of an arithmetic series based on the first term and the common difference of the sequence.

  1. To determine the second term, add the common difference to the first term
  2. And so on. To determine the third term, add the common difference to the second term
  3. This will give you the third term. Make sure to keep going until you’ve found all of the needed keywords
  4. Create a list of words separated by commas and enclosed inside brackets

Creating Arithmetic Sequences in the Form of Terms Fill in the blanks with the first five terms of the arithmetic sequencewithand. Adding three equals the same as deleting three. Starting with the first phrase, remove 3 from each word to arrive at the next term in the sequence. The first five terms are as follows: Analysis As predicted, the graph of the series is made up of points on a line, as seen in the figure (Figure). List the first five terms of the arithmetic sequence beginning with and ending with and.

There are two ways to write the sequence: in terms of the first word, 8, and the common difference.

We can identify the commonalities between the two situations.

Analysis Observe how each term’s common difference is multiplied by one in order to identify the following terms: once to find the second term, twice to get the third term, and so on.

Using Recursive Formulas for Arithmetic Sequences

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. Suppose the common difference is 5, and each phrase is the preceding term multiplied by five.

For an Arithmetic Sequence, a Recursive Formula can be used.

  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

Making a Recursive Formula for an Arithmetic Sequence is a difficult task. Write a recursive formula for the arithmetic series in the following format: The first term is defined as follows. It is possible to calculate the common difference by subtracting the first term from the second term. In the recursive formula for arithmetic sequences, substitute the initial term and the common difference in place of the first term. Analysis We can observe that the common difference is the slope of the line generated when the terms in the sequence are graphed, as illustrated in the figure below (Figure).

Is it necessary to deduct the first term from the second term in order to obtain the common difference between the two?

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

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. 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.

The common difference is that the series indicates a linear function with a slope of, whereas the difference is that We subtract from in order to determine the-intercept.

The graph is displayed in the following: (Figure).

The following equation is obtained by substituting for the slope as well as for the vertical intercept: To create an explicit formula for an arithmetic series, we do not need to identify the vertical intercept of the sequence.

  1. Identify the commonalities and differences
  2. Replace the common difference and the first word with the following:

After that, I’ll write the term paper. An Arithmetic Sequence with a Clearly Defined Formula Create an explicit formula for the arithmetic series using the following syntax: It is possible to calculate the common difference by subtracting the first term from the second term. The most often encountered difference is ten. To simplify the formula, substitute the common difference and the first term in the series into it. Analysis It can be shown in (Figure) that the slope of this sequence is 10 and that the vertical intercept is 10.

Finding the Number of Terms in a Finite 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. The Number of Terms in a Finite Arithmetic Sequence can be determined by the following method: The number of terms in the infinite arithmetic sequence is to be determined.

The most noticeable change is.

substitute for and find a solution for There are a total of eight terms in the series. The number of terms in the finite arithmetic sequence has to be determined. There are a total of 11 terms included in the sequence.

Solving Application Problems with Arithmetic Sequences

Using an initial term ofinstead of in many application difficulties makes logical sense in many situations In order to account for the variation in beginning terms in both cases, we make a little modification to the explicit formula. The following is the formula that we use: Problem-Solving using Arithmetic Sequences in Practical Situations Week after week, a youngster of five years old receives an allowance of one dollar. His parents have 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
  1. In this case, an arithmetic sequence with a starting term of 1 and a common difference of 2 may be used to represent what happened. Let be the amount of the allowance, and let be the number of years after reaching the age of five years. Using the modified explicit formula for an arithmetic series, we get the following results: By subtracting, we may find out how many years have passed since we were five. We’re asking for the child’s allowance after 11 years of being without one. In order to calculate the child’s allowance at the age of 16, substitute 11 into the calculation. What will the child’s allowance be when he or she reaches the age of sixteen? 23 hours each week

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? The formula is, and it will take her 42 minutes to complete the task.

Section Exercises

What is an arithmetic sequence, and how does it work? A series in which each subsequent term grows (or lowers) by a constant value is known as a constant value sequence. What is the procedure for determining the common difference of an arithmetic sequence? What is the best way to tell if a sequence is arithmetic or not? If the difference between all successive words is the same, we may say that they are sequential. This is the same as saying that the sequence has a common difference between it and the rest of the series.

What are the primary distinctions between the two methods?

What is the difference between them?

These two types of functions are distinct because their domains are not the same; linear functions are defined for all real numbers, whereas arithmetic sequences are specified for natural numbers or a subset of the natural numbers, respectively.

Algebraic

Work on finding the common difference for the arithmetic sequence that is supplied in the following problems. The most noticeable distinction is Determine if the sequence in the following exercises is mathematical or not in the following exercises. If this is the case, identify the common difference. Because of this, the sequence is not arithmetic. Write the first five terms of the arithmetic sequence given the first term and the common difference in the following tasks. Fill in the blanks with the first five terms of the arithmetic series given two terms in the following problems.

  1. The first term is 3, the most common difference is 4, and the fifth term is 3.
  2. The first term is 5, the most common difference is 6, and the eighth term is The first term is 6, the most common difference is 7, and the sixth term is 6.
  3. To complete the following problems, you must discover the first term from an arithmetic series given two terms.
  4. Make use of the recursive formula in order to write the first five terms of the arithmetic sequence in each of the following tasks.
  5. The following activities require you to build a recursive formula for the provided arithmetic sequence, followed by a search for the required term in the formula.
  6. Find the fourteenth term.
  7. For the following problems, write the first five terms of the arithmetic sequence using the explicit formula that was provided.

The number of terms in the finite arithmetic sequence presented in the following task is to be determined for the following exercises: The series has a total of ten terms. Six words are contained within the sequence of letters.

Graphical

Determine whether or not the graph given reflects an arithmetic sequence in the following tasks. The graph does not reflect an arithmetic sequence in the traditional sense. To complete the following activities, use the information supplied to graph the first 5 terms of the arithmetic sequence using the information provided.

Technology

In order to complete the following activities with an arithmetic sequence and a graphing calculator, follow the methods listed below:

  • Select SEQ in the fourth line
  • Select DOT in the fifth line
  • Then press Enter or Return.

Press To go to the TBLSET, press the button. To go to the TABLE, press the button. In the column with the header “First Seven Terms,” what are the first seven terms listed? Use the scroll-down arrow to move to the next page. What is the monetary value assigned to Press. Set the parameters and then press. The sequence should be graphed exactly as it appears on the graphing calculator. Follow the techniques outlined above to work with the arithmetic sequence using a graphing calculator for the following problems.

Make careful to modify the WINDOW settings as necessary.

Extensions

Examples of arithmetic sequences with four terms that are the same are shown. Give two instances of arithmetic sequences whose tenth terms are as follows: There will be a range of responses. Examples: and The fifth term of the arithmetic series must be discovered Determine the eleventh term in the arithmetic series. At what point does the sequence reach the number 151? When does the series begin to contain negative values and at what point does it stop? For which terms does the finite arithmetic sequence have integer values?

Create an arithmetic series using a recursive formula to demonstrate your understanding.

There will be a range of responses.

Example: Formula for recursion: The first four terms are as follows: Create an arithmetic sequence using an explicit formula to demonstrate your understanding.

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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