What Is The 15Th Term In The Arithmetic Sequence? (Perfect answer)

d=24−17=7. Now to write formula to find the fifteenth term: a15=a1+7(15−1) a15=17+98=115.

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How do you find the fifteenth term of a sequence?

To find the 15th term, follow these steps:

  1. Find the common ratio, r. In this sequence, each consecutive term is twice the previous term.
  2. Find the formula for the given sequence. In terms of the formula, a1 = 2 and r = 2.
  3. Find the term you’re looking for. If an = 2n, then a15 = 215 = 32,768.

What is the sum of the first 15 terms of the arithmetic sequence?

Thus, the sum of the first fifteen terms in the arithmetic sequence is 975.

What is the 25th term of arithmetic sequence?

3, 9, 15, 21, 27, … Solution: A sequence in which the difference between all pairs of consecutive numbers is equal is called an arithmetic progression. Therefore, the 25th term is 147.

How do you find the term of an arithmetic sequence?

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 15th term of the Series 2015 10?

∴ The 15th term is -50.

What is the term 15?

1: a number that is one more than fourteen — see Table of Numbers. 2: the first point scored by a side in a game of tennis. — called also five.

What is the arithmetic between 10 and 24?

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

What is the 5th term of the geometric sequence 15?

The fifth term of the geometric sequence 5, 15, 45 is 405.

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.

How do you find the 15th term in the arithmetic sequence class 9 maths CBSE

As an example, the general sequence of Arithmetic progression(A.P.) is $a, a+d, a+2d, a+3d, and so on. In this case, $a$ is the first phrase. The difference between any two successive words, often known as the common difference, is represented by the symbol $d$. The $n$ term of an A.P. may be calculated using the formula $a n= a+(n-1)d$. In order to identify the 15th term of the above arithmetic sequence, we must first compare the supplied numbers with the general sequence of A.P. and then with the given numbers again.

  1. The complete, step-by-step solution is as follows: The sequence in question is known as an Arithmetic Progression (A.P.).
  2. The common difference (d) may be found by subtracting any two successive terms from each other, yielding D = 4 + left(right) = 4 + 3 $, where d is the number of decimal places after the decimal point.
  3. is defined as$=a + (n – 1)d $where $a$ is the first term and $d$ is the common constant difference, as previously stated According to the question, we must determine the value of the 15th term, which means that $n = 15 $ is the answer.
  4. will be$ $Now, by inserting the values of$ n,a,and,d $ in the nth term of the A.P., we get$= a + (n – 1)d $ in the nth term of the A.P.
  5. Thus, the 15th term of the following arithmetic sequence is equivalent to $ 95 $ when written as $ left(right).
  6. or a random sequence.
  7. Step 2: Substitute $ n $ for $ n + 1 $ in$$ to get$ + 1 $ Step 3: Perform a $-$ calculation.
  8. If not, the sequence is not an A.P.
  9. Sequence: A sequence is a function whose domains are the set of all natural numbers greater than or equal to N.
  10. Real Sequence: A real sequence is a sequence whose range is a subset of R and whose length is a multiple of R.
  11. In mathematics, an arithmetic progression is a series in which the difference between each term and the preceding term is always equal to one hundred percent.

As a result, the constant$ (= d) $ is constant$ (= d) $ for all$ n in N $. The constant difference is commonly symbolized by the letter $d$, and it is sometimes referred to as the common difference.

Find the sum of an arithmetic progression: 2, 5, 8 . from the 15th term to the 50th term. A. 3775 B. 3474 C. 3210 D. 3154

Calculate the total of an arithmetic progression consisting of the numbers 2, 5, and 8 from the 15th term to the 50th term. _A. 3,775B. 3,474C. 3,210D. 3,154A. 3,775B. 3,474C. 3,210D.

Arithmetic Sequences

arithmetic progression or sequence A series of real numbers in which each successive term following the first is a product of adding or subtracting a non-zero constant from the preceding term Is known as an arithmetic progression or sequence The constant that is not zero is referred to as the common difference of the sequence. Consider the terms of an arithmetic series that are as follows: a 1,a 2,a 3,.,a ,a n The common difference of the sequence-$$d = a 2-a 1 = a 3-a 2=. = a – a $$Now, the sequence-$$d = a 2-a 1 = a 3-a 2=.

The nth term of an arithmetic sequence-

An arithmetic progression or sequence is a sequence of real numbers in which each successive term after the first is the result of the addition or subtraction of a non-zero constant into the preceding term. Common difference of the series refers to the non-zero constant that occurs more than once. Consider the terms of an arithmetic series that are shown below: a 1,a 2,a 3,.,a ,a n A – a_ is the common difference of the sequence-$$d = a 2 a 1 = a 3 a 2=. = a – a_ Now consider the coomon difference of the sequence-$$d

The sum of the first n terms of an arithmetic sequence-

TextS_ = fracleft $$displaystyle S n = fracleft $$textS_ = fracleft $$ $$

Answer and Explanation:1

The following arithmetic sequence is as follows: under-$$2,5,8,.$$ The first term and the common difference of the arithmetic are as follows: text (a) = 2 and, beginning text (d) = 5-2 = 3 end text (e) = 3 The formula for the nth term of an arithmetic sequence is derived as follows: $$a = a+(n-1)d$$ $$a = a+(n-1)d$$ Now, enter n=15 to find out what the 15th term is. $$begina = 2+ $$begina = 2+ $$ (15-1) 3 = 2+ = 3 (14) 3 = 2+42 = 44 end $$$ 3 = 2+42 = 44 In a similar vein, enter n=50 to discover the 50th word.

As a result, the total of the arithmetic series from the 15th term to the 50th term.

SOLUTION: the 15th term in an arithmetic sequence is 129 and the sum of the first fifteen terms is 1095. find the sum if the first ten terms.

byjosgarithmetic and greenestamps discovered 2 solutions: byjosgarithmetic(36812)(Show Source): The answer is: You are welcome to include this solution on YOUR website! A is the first letter of the alphabet. It is the fifteenth term. The total of the first 15 words is. In A and d, there are two linear equations. The following is a simplified system of equations: . What is the sum of the first 10 terms? Reduce complexity through computing. The following is the response bygreenestamps (10250)(Show Source): You are welcome to include this solution on YOUR website!

  1. (1) Any given collection of numbers has an average that is equal to the sum of all of them divided by the number of numbers in the collection.
  2. (3) Because the numbers in an arithmetic series are evenly spaced, the average of all the numbers is equal to the average of the first and last numbers in the sequence.
  3. Because we are given both the final (15th) phrase and the total of the 15 terms in this issue, we can apply the method described in (4) above to discover the first number in the series instantly.
  4. total = (how many there are) multiplied by (the average of the first and last numbers): The first term begins at the age of seventeen.
  5. The 15th term is the first term plus the common difference 14 times, which equals the following: The common difference is 8.

The tenth term is the first term plus the common difference nine times, for a total of 10 terms: The number 89 represents the tenth period. When you add up the first ten terms, you get the following: the number of terms multiplied by the average of the first and tenth terms.

Arithmetic Sequences and Series

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

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

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

How to find nth term of arithmetic sequence

What is the best way to determine the nth term in an arithmetic sequence? In addition, we may quickly locate any term in the sequence using the generic term formula. In this context, nth term refers to any term in the series, such as the 20th or 15th, or 28th. Finding the nth term may be calculated using the formula a = a + (n – 1)d Question 1: Calculate the common difference and 15th term of an A.P 125, 120, 115, 110,. Solution: The first term (a) is equal to 125.

  • A.P (an) is an abbreviation for A.P (an) (n – 1) 15 minus 1 (-25) =125 + 14 (-25) =125 + 14 (-25) =125–350a15 = -225 As a result, the 15th term of A.P is -225.
  • Is 3 in the following?
  • A2 – A1 =23 14 – 24 =(93/4) is the common difference between two numbers.
  • equals 33 + (n-1) (-3/4) = 24 + (n-1) 3 – 24 = (n-1)(-3/4)(-21 x 4)/(-3) = n -1 =84/3 = n -1 =28 = n – 1 =n=29 = (n-1)(-3/4)(-21 x 4)/(-3) = n -1 =84/3 = n – 1 =n=29 As a result, the number 3 represents the 29th term of A.P.
  • Solution: First term (a) = 2,Common difference = 3 2 – 2=2 n = 12, and second term (b) = 2.
  • As a result, the 12th of A.P.
  • 4: Identify the 17th word of the A.P 4, 9, 14,.
  • The common difference (d) is equal to 9 – 4 =5n = 17 points.
  • As a result, the 17th of A.P.
  • Now, consider the following example on “How to get the nth term of an arithmetic series.” In the case of an A.P., the 10th and 18th terms are 41 and 73, respectively.

Solution: The tenth term is 41 =a + 9 d = 41- (1) The 18th term is equal to 73 =a + 17 d = 73- (2) Taking the second equation and subtracting it from the first equation + 17 d – (a + 9 d) = 73 -41a + 17d – (a + 9 d) = 32 =8d = 32 =d =4 + 17d – (a + 9 d) = 73 -41a + 17d – (a + 9 d) = 73 -41a In the first equationa + 9, replace d = 4 with d = 4.

=a + is an abbreviation for (n – 1) here n = 27=5 + dhere (27-1) 4 =5 + 26 (4) =5 + 104 =109 4 =5 + 26 (4) =5 + 104 =109 As a result, the 27th term in the series is 109.

and 100, 95, 90,.

Solution: a = a plus (n – 1) dnth term of the initial sequencea = 1d = t2-t1 =7-1 =d = 6a = 1d = t2-t1 =7-1 =d = 6an = 1 + (n-1) 6-n =1 + 6-n – 6-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n =6 n – 5-n = (1) nth term of the second sequencea = 100d = a2-a1 =95 – 100 =-5an = 100 + (n-1) (-5) =100 – 5 n + 5 =105 – 5 n -(2)(1) =100 – 5 n -(2)(1) =100 – 5 n -(2)(1) =100 – 5 n -(2)(1) =100 – 5 n -(2)(1) =100 – (2) The equation is 6n – 5=105 – 5 n6 n + 5 = 105 + 511 = 110 =110/11 =11 6n + 5 = 105 =105 + 511 n6 n + 5 = 105 =105 + 5 n6 n + 5 = 110 =105 + 511 n6 n + 5 = 110 =105 =110/11 =11 As a result, the 11th and 12th terms in the above sequence are equivalent.

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Formulas for Arithmetic Sequences

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

Using Explicit Formulas for Arithmetic Sequences

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

  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.

A General Note: Explicit Formula for an Arithmetic Sequence

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

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

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

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

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

Try It

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

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

A General Note: Recursive Formula for an Arithmetic Sequence

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

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

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

Example: Writing a Recursive Formula for an Arithmetic Sequence

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

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

No.

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

Try It

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

Find the Number of Terms in an Arithmetic Sequence

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

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

  1. Find the common differences between the two
  2. To solve for = +dleft(n – 1right), substitute the common difference and the first term into the equation Fill in the blanks with the final word from and solve forn

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?

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Arithmetic Sequences and Series

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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

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,.
  5. Answer:S60=−8,550

Key Takeaways

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

Topic Exercises

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

Part B: Arithmetic Series

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

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.

You can also be asked to fill in a blank if a phrase has been left out. Finally, you could be interested in knowing, for example, the 100th phrase without having to write down all 100 words one by one. You may do any of these tasks with the aid of a few easy steps.

  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. Promotional material
  3. 3 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 Identify the first phrase in the series by looking at the initial letter of the word. 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. For an algebraic series, the explicit formula is written as
  • It is possible to read the word a(n) as “the nth term of a,” where n denotes the number in the list that you are looking for and a(n) reflects the actual value of that number. The number n will be 100 if you are asked to locate the 100th item in an arithmetic series, for example. Notably, while n is 100 in this example, the value of the 100th term, rather than the number 100 itself, will be represented as a(n).

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

  1. 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 new explicit formula that solves for a1. This formula is available online.
  • Make use of the equation and fill in the blanks with the facts you already know. Because you know that the 50th term is 300, n=50, n-1=49, and a(n)=300 are the values of n. You are also informed that the common difference, denoted by the letter d, is seven. Therefore, the formula is as follows: This works out as well. The series that you have created began at 43 and increased by 7 each time. As a result, it appears as follows: 43,50,57,64,71,78.293,300

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

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

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