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Arithmetic Sequence Problems with Solutions – Mastering Series Challenges

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Arithmetic Sequences Practice Problems and Solutions

Calculating terms in an arithmetic sequence, solving problems involving arithmetic sequences.

Feature Image How to Find the Sum of an Arithmetic Sequence Easy Steps with Examples

An arithmetic sequence is a series where each term increases by a constant amount, known as the common difference . I’ve always been fascinated by how this simple pattern appears in many mathematical problems and real-world situations alike.

Understanding this concept is fundamental for students as it not only enhances their problem-solving skills but also introduces them to the systematic approach of sequences in math .

The first term of an arithmetic sequence sets the stage, while the common difference dictates the incremental steps that each subsequent term will follow. This can be mathematically expressed as $a_n = a_1 + (n – 1)d$.

Whether I’m calculating the nth term or the sum of terms within a sequence , these formulas are the tools that uncover solutions to countless arithmetic sequence problems. Join me in unraveling the beauty and simplicity of arithmetic sequences ; together, we might just discover why they’re considered the building blocks in the world of mathematics .

When I work with arithmetic sequences , I always keep in mind that they have a unique feature: each term is derived by adding a constant value, known as the common difference , to the previous term. Let’s explore this concept through a few examples and problems.

Example 1: Finding a Term in the Sequence

Given the first term, $a_1$ of an arithmetic sequence is 5 and the common difference ( d ) is 3, what is the 10th term $a_{10}$?

Here’s how I determine it: $a_{10} = a_1 + (10 – 1)d ] [ a_{10} = 5 + 9 \times 3 ] [ a_{10} = 5 + 27 ] [ a_{10} = 32$

So, the 10th term is 32.

Sequence A: If $a_1 = 2 $and ( d = 4 ), find $a_5$.

Sequence B: For $a_3 = 7 $ and $a_7 = 19$, calculate the common difference ( d ).

I calculate $a_5$ by using the formula: $a_n = a_1 + (n – 1)d $ $ a_5 = 2 + (5 – 1) \times 4 $ $a_5 = 2 + 16 $ $a_5 = 18$

To find ( d ), I use the formula: $a_n = a_1 + (n – 1)d$ Solving for ( d ), I rearrange the terms from $a_3$ and $a_7$: $d = \frac{a_7 – a_3}{7 – 3}$ $d = \frac{19 – 7}{4}$ $d = \frac{12}{4}$ [ d = 3 ]

Here’s a quick reference table summarizing the properties of arithmetic sequences :

Property	Description
First Term	Denoted as $a_1$, where the sequence begins
Common Difference	Denoted as ( d ), the fixed amount between terms
( n )th Term	Given by $ a_n = a_1 + (n – 1)d $

Remember these properties to solve any arithmetic sequence problem effectively!

In an arithmetic sequence , each term after the first is found by adding a constant, known as the common difference ( d ), to the previous term. I find that a clear understanding of the formula helps immensely:

$a_n = a_1 + (n – 1)d$

Here, $a_n$ represents the $n^{th}$term, $a_1$ is the first term, and ( n ) is the term number.

Let’s say we need to calculate the fourth and fifth terms of a sequence where the first term $a_1 $ is 8 and the common difference ( d ) is 2. The explicit formula for this sequence would be $ a_n = 8 + (n – 1)(2) $.

To calculate the fourth term $a_4 $: $a_4 = 8 + (4 – 1)(2) = 8 + 6 = 14$

For the fifth term ( a_5 ), just add the common difference to the fourth term: $a_5 = a_4 + d = 14 + 2 = 16$

Here’s a table to illustrate these calculations:

Term Number (n)	Formula	Term Value ( a_n )
4	( 8 + (4 – 1)(2) )	14
5	( 8 + (5 – 1)(2) ) or ( 14 + 2 )	16

Remember, the formula provides a direct way to calculate any term in the sequence, known as the explicit or general term formula. Just insert the term number ( n ) and you’ll get the value for $a_n$. I find this methodical approach simplifies the process and avoids confusion.

When I approach arithmetic sequences , I find it helpful to remember that they’re essentially lists of numbers where each term is found by adding a constant to the previous term. This constant is called the common difference, denoted as ( d ). For example, in the sequence 3, 7, 11, 15, …, the common difference is ( d = 4 ).

To articulate the ( n )th term of an arithmetic sequence, $a_n $, I use the fundamental formula:

$a_n = a_1 + (n – 1)d $

In this expression, $a_1$ represents the first term of the sequence.

If I’m solving a specific problem—let’s call it Example 1—I might be given $a_1 = 5 $and ( d = 3 ), and asked to find $a_4 $. I’d calculate it as follows:

$a_4 = 5 + (4 – 1) \times 3 = 5 + 9 = 14$

In applications involving arithmetic series, such as financial planning or scheduling tasks over weeks, the sum of the first ( n ) terms often comes into play. To calculate this sum, ( S_n ), I rely on the formula:

$S_n = \frac{n}{2}(a_1 + a_n)$

Now, if I’m asked to work through Example 3, where I need the sum of the first 10 terms of the sequence starting with 2 and having a common difference of 5, the process looks like this:

$a_{10} = 2 + (10 – 1) \times 5 = 47$ $S_{10} = \frac{10}{2}(2 + 47) = 5 \times 49 = 245$

Linear functions and systems of equations sometimes bear a resemblance to arithmetic sequences, such as when I need to find the intersection of sequence A and sequence B. This would involve setting the nth terms equal to each other and solving the resulting linear equation.

Occasionally, arithmetic sequences can be mistaken for geometric sequences , where each term is found by multiplying by a constant. It’s important to differentiate between them based on their definitions.

For exercises, it’s beneficial to practice finding nth terms, and sums , and even constructing sequences from given scenarios. This ensures a robust understanding when faced with a variety of problems involving arithmetic sequences .

In exploring the realm of arithmetic sequences , I’ve delved into numerous problems and their corresponding solutions. The patterns in these sequences—where the difference between consecutive terms remains constant—allow for straightforward and satisfying problem-solving experiences.

For a sequence with an initial term of $a_1 $ and a common difference of ( d ), the $n^{th}$term is given by $a_n = a_1 + (n – 1)d $.

I’ve found that this formula not only assists in identifying individual terms but also in predicting future ones. Whether calculating the $50^{th}$term or determining the sum of the first several terms, the process remains consistent and is rooted in this foundational equation.

In educational settings, arithmetic sequences serve as an excellent tool for reinforcing the core concepts of algebra and functions. Complexity varies from basic to advanced problems, catering to a range of skill levels. These sequences also reflect practical real-world applications, such as financial modeling and computer algorithms, highlighting the relevance beyond classroom walls.

Through practicing these problems, the elegance and power of arithmetic sequences in mathematical analysis become increasingly apparent. They exemplify the harmony of structure and progression in mathematics —a reminder of how simple rules can generate infinitely complex and fascinating patterns.

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How to Solve Arithmetic Sequences? (+FREE Worksheet!)

Do you want to know how to solve Arithmetic Sequences problems? you can do it in few simple and easy steps.

Step by step guide to solve Arithmetic Sequences problems

A sequence of numbers such that the difference between the consecutive terms is constant is called arithmetic sequence. For example, the sequence $6, 8, 10, 12, 14$, … is an arithmetic sequence with common difference of $2$.
To find any term in an arithmetic sequence use this formula: $\color{blue}{x_{n}=a+d(n-1)}$
$a =$ the first term ,$d =$ the common difference between terms , $n =$ number of items

Arithmetic Sequences – Example 1:

Find the first three terms of the sequence. $a_{17}=38,d=3$

First, we need to find $a_{1}$ or a. Use arithmetic sequence formula: $\color{blue}{x_{n}=a+d(n-1)}$ If $a_{8}=38$, then $n=8$. Rewrite the formula and put the values provided: $x_{n}=a+d(n-1)→38=a+3(3-1)=a+6$, now solve for $a$. $38=a+6→a=38-6=32$, First three terms: $32,35,38$

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Arithmetic sequences – example 2:.

Given the first term and the common difference of an arithmetic sequence find the first five terms. $a_{1}=18,d=2$

Use arithmetic sequence formula: $\color{blue}{x_{n}=a+d(n-1)}$ If $n=1$ then: $x_{1}=18+2(1)→x_{1}=18$ First five terms: $18,20,22,24,26$

Arithmetic Sequences – Example 3:

Given the first term and the common difference of an arithmetic sequence find the first five terms. $a_{1}=24,d=2$

Use arithmetic sequence formula: $\color{blue}{x_{n}=a+d(n-1)}$ If $n=1$ then: $x_{1}=22+2(1)→x_{1}=24$ First five terms: $24,26,28,30,32$

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Arithmetic sequences – example 4:.

Find the first five terms of the sequence. $a_{17}=152,d=4$

First, we need to find $a_{1}$ or $a$. Use arithmetic sequence formula: $\color{blue}{x_{n}=a+d(n-1)}$ If $a_{17}=152$, then $n=17$. Rewrite the formula and put the values provided: $x_{n}=a+d(n-1)→152=a+4(17-1)=a+64$, now solve for $a$. $152=a+64→a=152-64=88$, First five terms: $88,92,96,100,104$

Given the first term and the common difference of an arithmetic sequence find the first five terms and the explicit formula.

$\color{blue}{a_{1} = 24, d = 2}$
$\color{blue}{a_{1} = –15, d = – 5}$
$\color{blue}{a_{1} = 18, d = 10}$
$\color{blue}{a_{1 }= –38, d = –100}$

Download Arithmetic Sequences Worksheet

First Five Terms $\color{blue}{: 24, 26, 28, 30, 32, Explicit: a_{n} = 22 + 2n}$
First Five Terms $\color{blue}{: –15, –20, –25, –30, –35, Explicit: a_{n} = –10 – 5n}$
First Five Terms $\color{blue}{: 18, 28, 38, 48, 58, Explicit: a_{n} = 8 + 10n}$
First Five Terms $\color{blue}{: –38, –138, –238, –338, –438, Explicit: a_{n} = 62 – 100n}$

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Arithmetic sequence

In algebra , an arithmetic sequence , sometimes called an arithmetic progression , is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence.

$1, 2, 3, 4$

1 Properties
3.1 Introductory problems
3.2 Intermediate problems

$a_1$

An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value.

$\frac{n(a_1 + a_n)}{2}$

Here are some problems with solutions that utilize arithmetic sequences and series.

Introductory problems

2005 AMC 10A Problem 17
2006 AMC 10A Problem 19
2012 AIME I Problems/Problem 2
2004 AMC 10B Problems/Problem 10
2006 AMC 10A, Problem 9
2006 AMC 12A, Problem 12

Intermediate problems

2003 AIME I, Problem 2

$x^5-5x^4-35x^3+ax^2+bx+c$

Geometric sequence
Harmonic sequence
Sequences and series

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Arithmetic Sequence

The arithmetic sequence is the sequence where the common difference remains constant between any two successive terms. Let us recall what is a sequence. A sequence is a collection of numbers that follow a pattern. For example, the sequence 1, 6, 11, 16, … is an arithmetic sequence because there is a pattern where each number is obtained by adding 5 to its previous term. We have two arithmetic sequence formulas.

The formula for finding n th term of an arithmetic sequence
The formula to find the sum of first n terms of an arithmetic sequence

If we want to find any term in the arithmetic sequence then we can use the arithmetic sequence formula. Let us learn the definition of an arithmetic sequence and arithmetic sequence formulas along with derivations and a lot more examples for a better understanding.

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6.

What is an Arithmetic Sequence?

An arithmetic sequence is defined in two ways. It is a "sequence where the differences between every two successive terms are the same" (or) In an arithmetic sequence, "every term is obtained by adding a fixed number (positive or negative or zero) to its previous term". The following is an arithmetic sequence as every term is obtained by adding a fixed number 4 to its previous term.

Arithmetic Sequence Example

Consider the sequence 3, 6, 9, 12, 15, .... is an arithmetic sequence because every term is obtained by adding a constant number (3) to its previous term.

The first term, a = 3
The common difference, d = 6 - 3 = 9 - 6 = 12 - 9 = 15 - 12 = ... = 3

Thus, an arithmetic sequence can be written as a, a + d, a + 2d, a + 3d, .... Let us verify this pattern for the above example.

a, a + d, a + 2d, a + 3d, a + 4d, ... = 3, 3 + 3, 3 + 2(3), 3 + 3(3), 3 + 4(3),... = 3, 6, 9, 12,15,....

A few more examples of an arithmetic sequence are:

5, 8, 11, 14, ...
80, 75, 70, 65, 60, ...
π/2, π, 3π/2, 2π, ....
-√2, -2√2, -3√2, -4√2, ...

Arithmetic Sequence Formula

The first term of an arithmetic sequence is a, its common difference is d, n is the number of terms. The general form of the AP is a, a+d, a+2d, a+3d,......up to n terms. We have different formulas associated with an arithmetic sequence used to calculate the n th term, the sum of n terms of an AP, or the common difference of a given arithmetic sequence.

The arithmetic sequence formula is given as,

N th Term: a n = a + (n-1)d
S n = (n/2) [2a + (n - 1)d]
d = a n - a n-1

Nth Term of Arithmetic Sequence

The n th term of an arithmetic sequence a 1 , a 2 , a 3 , ... is given by a n = a 1 + (n - 1) d . This is also known as the general term of the arithmetic sequence. This directly follows from the understanding that the arithmetic sequence a 1 , a 2 , a 3 , ... = a 1 , a 1 + d, a 1 + 2d, a 1 + 3d,... The following table shows some arithmetic sequences along with the first term, the common difference, and the n th term.

Arithmetic Sequence	First Term (a)	Common Difference (d)	n term = a
80, 75, 70, 65, 60, ...	80	-5	80 + (n - 1) (-5) = -5n + 85
π/2, π, 3π/2, 2π, ....	π/2	π/2	π/2 + (n - 1) (π/2) = nπ/2
-√2, -2√2, -3√2, -4√2, ...	-√2	-√2	-√2 + (n - 1) (-√2) = -√2 n

Arithmetic Sequence Recursive Formula

The above formula for finding the n t h term of an arithmetic sequence is used to find any term of the sequence when the values of 'a 1 ' and 'd' are known. There is another formula to find the n th term which is called the " recursive formula of an arithmetic sequence " and is used to find a term (a n ) of the sequence when its previous term (a n-1 ) and 'd' are known. It says

a n = a n-1 + d

This formula just follows the definition of the arithmetic sequence.

Example: Find a 21 of an arithmetic sequence if a 19 = -72 and d = 7.

By using the recursive formula,

a 20 = a 19 + d = -72 + 7 = -65

a 21 = a 20 + d = -65 + 7 = -58

Therefore, a 21 = -58.

Arithmetic Series

The sum of the arithmetic sequence formula is used to find the sum of its first n terms. Note that the sum of terms of an arithmetic sequence is known as arithmetic series. Consider an arithmetic series in which the first term is a 1 (or 'a') and the common difference is d. The sum of its first n terms is denoted by S n . Then

When the n th term is NOT known: S n = n/2 [2a 1 + (n-1) d]
When the n th term is known: S n = n/2 [a 1 + a n ]

Ms. Natalie earns $200,000 per annum and her salary increases by $25,000 per annum. Then how much does she earn at the end of the first 5 years?

The amount earned by Ms. Natalie for the first year is, a = 2,00,000. The increment per annum is, d = 25,000. We have to calculate her earnings in the first 5 years. Hence n = 5. Substituting these values in the sum sum of arithmetic sequence formula,

S n = n/2 [2a 1 + (n-1) d]

⇒ S n = 5/2(2(200000) + (5 - 1)(25000))

= 5/2 (400000 +100000)

= 5/2 (500000)

She earns $1,250,000 in 5 years. We can use this formula to be more helpful for larger values of 'n'.

Sum of Arithmetic Sequence

Let us take an arithmetic sequence that has its first term to be a 1 and the common difference to be d. Then the sum of the first 'n' terms of the sequence is given by

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

Let us write the same sum from right to left (i.e., from the n th term to the first term).

S n = a n + (a n – d) + (a n – 2d) + … + a 1 ... (2)

Adding (1) and (2), all terms with 'd' get canceled.

2S n = (a 1 + a n ) + (a 1 + a n ) + (a 1 + a n ) + … + (a 1 + a n )

2S n = n (a 1 + a n )

S n = [n(a 1 + a n )]/2

By substituting a n = a 1 + (n – 1)d into the last formula, we have

S n = n/2 [a 1 + a 1 + (n – 1)d] (or)

S n = n/2 [2a 1 + (n – 1)d]

Thus, we have derived both formulas for the sum of the arithmetic sequence.

Difference Between Arithmetic Sequence and Geometric Sequence

Here are the differences between arithmetic and geometric sequence :


In this, the differences between every two consecutive numbers are the same.	In this, the of every two consecutive numbers are the same.
It is identified by the first term (a) and the common difference (d).	It is identified by the first term (a) and the (r).
There is a linear relationship between the terms.	There is an between the terms.

Important Notes on Arithmetic Sequence:

In arithmetic sequences, the difference between every two successive numbers is the same.
The common difference of an arithmetic sequence a 1 , a 2 , a 3 , ... is, d = a 2 - a 1 = a 3 - a 2 = ...
The n th term of an arithmetic sequence is a n = a 1 + (n−1)d.
The sum of the first n terms of an arithmetic sequence is S n = n/2[2a 1 + (n − 1)d].
The common difference between arithmetic sequences can be either positive or negative or zero.

☛ Related Topics:

Sequence Calculator
Series Calculator
Arithmetic Sequence Calculator
Geometric Sequence Calculator

Arithmetic Sequence Examples

Example 1: Find the n th term of the arithmetic sequence -5, -7/2, -2, ....

The given sequence is -5, -7/2, -2, ...

Here, the first term is a = -5, and the common difference is, d = -(7/2) - (-5) = -2 - (-7/2) = ... = 3/2.

The n th term of an arithmetic sequence is given by

a n = a 1 + (n−1)d

a n = -5 +(n - 1) (3/2)

= -5+ (3/2)n - 3/2

= 3n/2 - 13/2

Answer: The n th term of the given arithmetic sequence is, a n = 3n/2 - 13/2.

Example 2: Which term of the arithmetic sequence -3, -8, -13, -18,... is -248?

The given arithmetic sequence is -3, -8, -13, -18,...

The first term is, a = -3

The common difference is, d = -8 - (-3) = -13 - (-8) = ... = -5.

It is given that the n th term is, a n = -248.

Substitute all these values in the n th term of an arithmetic sequence formula,

a n = a 1 + (n−1)d ⇒ -248 = -3 + (-5)(n - 1) ⇒ -248 = -3 -5n + 5 ⇒ -248 = 2 - 5n ⇒ -250 = -5n ⇒ n = 50

Answer: -248 is the 50 th term of the given sequence.

Example 3: Find the sum of the arithmetic sequence -3, -8, -13, -18,.., -248.

This sequence is the same as the one that is given in Example 2 .

There we found that a = -3, d = -5, and n = 50.

So we have to find the sum of the 50 terms of the given arithmetic series.

S n = n/2[a 1 + a n ]

S 50 = [50 (-3 - 248)]/2 = -6275

Answer: The sum of the given arithmetic sequence is -6275.

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Arithmetic Sequence Questions

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FAQs on Arithmetic Sequence

What is an arithmetic sequence in algebra.

An arithmetic sequence in algebra is a sequence of numbers where the difference between every two consecutive terms is the same. Generally, the arithmetic sequence is written as a, a+d, a+2d, a+3d, ..., where a is the first term and d is the common difference.

What are Arithmetic Sequence Formulas?

Here are the formulas related to an arithmetic sequence where a₁ (or a) is the first term and d is a common difference:

The common difference, d = a n - a n-1 .
n th term of sequence is, a n = a + (n - 1)d
Sum of n terms of sequence is , S n = [n(a 1 + a n )]/2 (or) n/2 (2a + (n - 1)d)

What is the Definition of an Arithmetic Sequence?

A sequence of numbers in which every term (except the first term) is obtained by adding a constant number to the previous term is called an arithmetic sequence . For example, 1, 3, 5, 7, ... is an arithmetic sequence as every term is obtained by adding 2 (a constant number) to its previous term.

How to Identify An Arithmetic Sequence?

If the difference between every two consecutive terms of a sequence is the same then it is an arithmetic sequence. For example, 3, 8, 13, 18 ... is arithmetic because the consecutive terms have a fixed difference.

18-13 = 5 and so on.

What is the n th term of an Arithmetic Sequence?

The n th term of arithmetic sequences is given by a n = a + (n – 1) × d. Here 'a' represents the first term and 'd' represents the common difference.

What is an Arithmetic Series?

An arithmetic series is a sum of an arithmetic sequence where each term is obtained by adding a fixed number to each previous term.

What is the Arithmetic Series Formula?

The sum of the first n terms of an arithmetic sequence (arithmetic series ) with the first term 'a' and common difference 'd' is denoted by Sₙ and we have two formulas to find it.

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

What is the Formula to Find the Common Difference in Arithmetic sequence?

The common difference of an arithmetic sequence, as its name suggests, is the difference between every two of its successive (or consecutive) terms. The formula for finding the common difference of an arithmetic sequence is, d = a n - a n-1 .

How to Find n in Arithmetic Sequence?

When we have to find the number of terms (n) in arithmetic sequences, some of the information about a, d, a n or S n might have been given in the problem. We will just substitute the given values in the formulas of a n or S n and solve it for n.

How To Find the First Term in Arithmetic sequence?

The first term of an arithmetic sequence is the number that occurs in the first position from the left. It is denoted by 'a'. If 'a' is NOT given in the problem, then some information about d (or) a n (or) S n might be given in the problem. We will just substitute the given values in the formulas of a n or S n and solve it for 'a'.

What is the Difference Between Arithmetic Sequence and Arithmetic Series?

An arithmetic sequence is a collection of numbers in which all the differences between every two consecutive numbers are equal to a constant whereas an arithmetic series is the sum of a few or more terms of an arithmetic sequence.

What are the Types of Sequences?

There are mainly 3 types of sequences in math. They are:

Arithmetic sequence
Geometric sequence
Harmonic sequence

What are the Applications of Arithmetic Sequence?

Here are some applications: the salary of a person which is increased by a constant amount by each year, the rent of a taxi which charges per mile, the number of fishes in a pond that increase by a constant number each month, etc.

How to Find the n th Term in Arithmetic Sequence?

Here are the steps for finding the n th term of arithmetic sequences:

Identify its first term, a
Common difference , d
Identify which term you want. i.e., n
Substitute all these into the formula a n = a + (n – 1) × d.

How to Find the Sum of n Terms of Arithmetic Sequence?

To find the sum of the first n terms of arithmetic sequences,

Identify its first term (a)
Common difference (d)
Identify which term you want (n)
Substitute all these into the formula S n = (n/2)(2a + (n - 1)d)

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Arithmetic Sequence Problems

There are many problems we can solve if we keep in mind that the n th term of an arithmetic sequence can be written in the following way: a n = a 1 +(n - 1)d Where a 1 is the first term, and d is the common difference. For example, if we are told that the first two terms add up to the fifth term, and that the common difference is 8 less than the first term we can take this equation: a 1 + a 2 = a 5 and rewrite it as follows: a 1 + [a 1 + d] = [a 1 + 4d] This leads to a 1 = 3d. Combine this with d = a 1 - 8, and we have: a 1 = 3(a 1 - 8) or a 1 = 12. This leads to d = 4, and from this information, we can find any other term of the sequence.

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Arithmetic Sequence

Arithmetic sequence formula.

If you wish to find any term (also known as the [latex]{{nth}}[/latex] term) in the arithmetic sequence, the arithmetic sequence formula should help you to do so. The critical step is to be able to identify or extract known values from the problem that will eventually be substituted into the formula itself.

When you’re done with this lesson, you may check out my other lesson about the Arithmetic Series Formula .

Let’s start by examining the essential parts of the arithmetic sequence formula:

Parts of the Arithmetic Sequence Formula

[latex]\large{a_n}[/latex] = the term that you want to find

[latex]\large{a_1}[/latex] = first term in the sequence

[latex]\large{n}[/latex] = the term position (ex: for 5th term, n = 5 )

[latex]\large{d}[/latex] = common difference of any pair of consecutive or adjacent numbers

Let’s put this formula in action!

Examples of How to Apply the Arithmetic Sequence Formula

Example 1: Find the 35 th term in the arithmetic sequence 3, 9, 15, 21, …

There are three things needed in order to find the 35 th term using the formula:

the first term ( [latex]{a_1}[/latex] )
the common difference between consecutive terms ([latex]d[/latex] )
and the term position ([latex]n[/latex] )

From the given sequence, we can easily read off the first term and common difference. In a nutshell, the common difference is the constant difference between any pair of adjacent terms.The term position is just the [latex]n[/latex] value in the [latex]{n^{th}}[/latex] term, thus in the [latex]{35^{th}}[/latex] term, [latex]n=35[/latex].

Therefore, the known values that we will substitute in the arithmetic formula are

So the solution to finding the missing term is,

Therefore, the 35th term of the arithmetic sequence is [latex]207[/latex].

Example 2: Find the 125 th term in the arithmetic sequence 4, −1, −6, −11, …

This arithmetic sequence has the first term [latex]{a_1} = 4[/latex], and a common difference of −5.

the sequence 4, -1, -6, -11, ... has a common difference of -5, and its first term is 4

Since we want to find the 125 th term, the [latex]n[/latex] value would be [latex]n=125[/latex]. The following are the known values we will plug into the formula:

The missing term in the sequence is calculated as,

Therefore, the 125th term of the arithmetic sequence is [latex]-616[/latex].

Example 3: If one term in the arithmetic sequence is [latex]{a_{21}} = – 17[/latex] and the common difference is [latex]d = – 3[/latex]. Find the following:

a) Write a rule that can find any term in the sequence.

b) Find the twelfth term ( [latex]{a_{12}}[/latex] ) and eighty-second term ( [latex]{a_{82}}[/latex] ) term.

Solution to part a)

Because we know a term in the sequence which is [latex]{a_{21}} = – 17[/latex] and the common difference [latex]d = – 3[/latex], the only missing value in the formula which we can easily solve is the first term, [latex]{a_1}[/latex].

Since we found [latex]{a_1} = 43[/latex] and we know [latex]d = – 3[/latex], the rule to find any term in the sequence is

How do we really know if the rule is correct? What I would do is verify it with the given information in the problem that [latex]{a_{21}} = – 17[/latex].

So we ask ourselves, what is [latex]{a_{21}} = ?[/latex]

We already know the answer though but we want to see if the rule would give us −17.

Since [latex]{a_1} = 43[/latex], [latex]n=21[/latex] and [latex]d = – 3[/latex], we substitute these values into the formula then simplify.

Which it does! Great.

Solution to part b)

To answer the second part of the problem, use the rule that we found in part a) which is

Here are the calculations side-by-side.

a sub 10 equals 10 and a sub 82 equals negative 200

Example 4: Given two terms in the arithmetic sequence, [latex]{a_5} = – 8[/latex] and [latex]{a_{25}} = 72[/latex];

b) Find the 100 th term ( [latex]{a_{100}}[/latex] ).

The problem tells us that there is an arithmetic sequence with two known terms which are [latex]{a_5} = – 8[/latex] and [latex]{a_{25}} = 72[/latex]. The first step is to use the information of each term and substitute its value in the arithmetic formula. We have two terms so we will do it twice.

This is wonderful because we have two equations and two unknown variables. We can solve this system of linear equations either by the Substitution Method or Elimination Method . You should agree that the Elimination Method is the better choice for this.

Place the two equations on top of each other while aligning the similar terms.

We can eliminate the term [latex]{a_1}[/latex] by multiplying Equation # 1 by the number −1 and adding them together.

Since we already know the value of one of the two missing unknowns which is [latex]d = 4[/latex], it is now easy to find the other value. We can find the value of [latex]{a_1}[/latex] by substituting the value of [latex]d[/latex] on any of the two equations. For this, let’s use Equation #1.

After knowing the values of both the first term ( [latex]{a_1}[/latex] ) and the common difference ( [latex]d[/latex] ), we can finally write the general formula of the sequence.

To find the 100 th term ( [latex]{a_{100}}[/latex] ) of the sequence, use the formula found in part a)

Arithmetic sequence

An arithmetic sequence is a type of sequence in which the difference between each consecutive term in the sequence is constant. For example, the difference between each term in the following sequence is 3:

2, 5, 8, 11, 14, 17, 20...

To expand the above arithmetic sequence, starting at the first term, 2, add 3 to determine each consecutive term. This is simple for the first few terms, but using this method to determine terms further along in the sequence gets tedious very quickly. Fortunately, the n th term of an arithmetic sequence can be determined using

a n = a 1 + (n - 1)d

where a n is the n th term, a 1 is the initial term, and d is the constant difference between each term. Using the above sequence, the formula becomes:

a n = 2 + 3n - 3 = 3n - 1

Therefore, the 100th term of this sequence is:

a 100 = 3(100) - 1 = 299

This formula allows us to determine the n th term of any arithmetic sequence.

Arithmetic sequence vs arithmetic series

An arithmetic series is the sum of a finite part of an arithmetic sequence. For example, 2 + 5 + 8 = 15 is an arithmetic series of the first three terms in the sequence above. The sum of a finite arithmetic sequence can be found using the following formula,

where n is the number of terms in the sequence, a 1 is the first term in the sequence, and a n is the n th term, and d is the constant difference between each term.

Find the sum of the first 7 terms in the arithmetic sequence 1, 7, 13, 19, 25, ...

First determine the rest of the first 7 terms. The difference between each consecutive term in the sequence is 6. Since we only need to determine 2 more terms, it is simpler to just add 6 to the last given term and the term after that. The first 7 terms are:

1, 7, 13, 19, 25, 31, 37

The sum of these terms is:

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Arithmetic & Geometric Sequences

Intro Examples Arith. & Geo. Seq. Arith. Series Geo. Series

The two simplest sequences to work with are arithmetic and geometric sequences.

An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value. For instance, 2, 5, 8, 11, 14,... is arithmetic, because each step adds three; and 7, 3, −1, −5,... is arithmetic, because each step subtracts 4 .

The number added (or subtracted) at each stage of an arithmetic sequence is called the "common difference" d , because if you subtract (that is, if you find the difference of) successive terms, you'll always get this common value.

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The number multiplied (or divided) at each stage of a geometric sequence is called the "common ratio" r , because if you divide (that is, if you find the ratio of) successive terms, you'll always get this common value.

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

3, 11, 19, 27, 35, ...

To find the common difference, I have to subtract a successive pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other. To be thorough, I'll do all the subtractions:

11 − 3 = 8

19 − 11 = 8

27 − 19 = 8

35 − 27 = 8

The difference is always 8 , so the common difference is d = 8 .

They gave me five terms, so the sixth term of the sequence is going to be the very next term. I find the next term by adding the common difference to the fifth term:

35 + 8 = 43

Then my answer is:

common difference: d = 8

sixth term: 43

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

To find the common ratio, I have to divide a successive pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other. To be thorough, I'll do all the divisions:

The ratio is always 3 , so r = 3 .

They gave me five terms, so the sixth term is the very next term; the seventh will be the term after that. To find the value of the seventh term, I'll multiply the fifth term by the common ratio twice:

a 6 = (18)(3) = 54

a 7 = (54)(3) = 162

common ratio: r = 3

seventh term: 162

Since arithmetic and geometric sequences are so nice and regular, they have formulas.

For arithmetic sequences, the common difference is d , and the first term a 1 is often referred to simply as " a " . Since we get the next term by adding the common difference, the value of a 2 is just:

a 2 = a + d

Continuing, the third term is:

a 3 = ( a + d ) + d = a + 2 d

The fourth term is:

a 4 = ( a + 2 d ) + d = a + 3 d

At each stage, the common difference was multiplied by a value that was one less than the index. Following this pattern, the n -th term a n will have the form:

a n = a + ( n − 1) d

For geometric sequences, the common ratio is r , and the first term a 1 is often referred to simply as " a " . Since we get the next term by multiplying by the common ratio, the value of a 2 is just:

a 3 = r ( ar ) = ar 2

a 4 = r ( ar 2 ) = ar 3

At each stage, the common ratio was raised to a power that was one less than the index. Following this pattern, the n -th term a n will have the form:

a n = ar ( n − 1)

Memorize these n -th-term formulas before the next test.

Find the tenth term and the n -th term of the following sequence:

The first thing I have to do is figure out which type of sequence this is: arithmetic or geometric. I quickly see that the differences don't match; for instance, the difference of the second and first term is 2 − 1 = 1 , but the difference of the third and second terms is 4 − 2 = 2 . So this isn't an arithmetic sequence.

On the other hand, the ratios of successive terms are the same:

2 ÷ 1 = 2

4 ÷ 2 = 2

8 ÷ 4 = 2

(I didn't do the division with the first term, because that involved fractions and I'm lazy. The division would have given the exact same result, though.)

a n = (1/2) 2 n −1 = (2 -1 )(2 n −1 )

=2 (−1) + ( n − 1) = 2 n − 2

To find the value of the tenth term, I can plug n = 10 into the n -th term formula and simplify:

a 10 = 2 10−2 = 2 8 = 256

tenth term: 256

Find the n -th term and the first three terms of the arithmetic sequence having a 6 = 5 and d = katex.render("\\boldsymbol{\\color{green}{\\frac{3}{2}}}", typed07); 3/2

This gives me the first three terms in the sequence. Since I have the value of the first term and the common difference, I can also create the expression for the n -th term, and simplify:

−5/2 + ( n − 1)(3/2)

= −5/2 + (3/2) n − 3/2

= −8/2 + (3/2) n = (3/2) n − 4

Find the n -th term and the first three terms of the arithmetic sequence having a 4 = 93 and a 8 = 65 .

Since a 4 and a 8 are four places apart, then I know from the definition of an arithmetic sequence that I'd get from the fourth term to the eighth term by adding the common difference four times to the fourth term; in other words, the definition tells me that a 8 = a 4 + 4 d . Using this, I can then solve for the common difference d :

65 = 93 + 4 d

−28 = 4 d

−7 = d

Also, I know that the fourth term relates to the first term by the formula a 4 = a + (4 − 1) d , so, using the value I just found for d , I can find the value of the first term a :

93 = a + 3(−7)

93 + 21 = a

Now that I have the value of the first term and the value of the common difference, I can plug-n-chug to find the values of the first three terms and the general form of the n -th term:

a 2 = 114 − 7 = 107

a 3 = 107 − 7 = 100

a n = 114 + ( n − 1)(−7)

= 114 − 7 n + 7 = 121 − 7 n

n -th term: 121 − 7 n

first three terms: 114, 107, 100

Find the n -th and the 26 th terms of the geometric sequence with katex.render("\\boldsymbol{\\color{green}{ \\mathit{a}_5 = \\frac{5}{4} }}", typed16); a 5 = 5/4 and a 12 = 160 .

The two terms for which they've given me numerical values are 12 − 5 = 7 places apart, so, from the definition of a geometric sequence, I know that I'd get from the fifth term to the twelfth term by multiplying the fifth term by the common ratio seven times; that is, a 12 = ( a 5 )( r 7 ) . I can use this to solve for the value of the common ratio r :

160 = (5/4)( r 7 )

Also, I know that the fifth term relates to the first by the formula a 5 = ar 4 , so I can solve for the value of the first term a :

5/4 = a (2 4 ) = 16 a

Now that I have the value of the first term and the value of the common ratio, I can plug each into the formula for the n -th term to get:

a n = (5/64)2 ( n − 1)

= (5/2 6 )(2 n −1 )

= (5)(2 −6 )(2 n −1 )

= 5(2 n −7 )

With this formula, I can evaluate the twenty-sixth term, and simplify:

a 26 = 5(2 19 )

= 2,621,440

26 th term: 2,621,440

Once we know how to work with sequences of arithmetic and geometric terms, we can turn to considerations of adding these sequences.

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Sequences - Finding a Rule

To find a missing number in a Sequence, first we must have a Rule

A Sequence is a set of things (usually numbers) that are in order.

Each number in the sequence is called a term (or sometimes "element" or "member"), read Sequences and Series for a more in-depth discussion.

Finding Missing Numbers

To find a missing number, first find a Rule behind the Sequence.

Sometimes we can just look at the numbers and see a pattern:

Example: 1, 4, 9, 16, ?

Answer: they are Squares (1 2 =1, 2 2 =4, 3 2 =9, 4 2 =16, ...)

Rule: x n = n 2

Sequence: 1, 4, 9, 16, 25, 36, 49, ...

Did you see how we wrote that rule using "x" and "n" ?

x n means "term number n", so term 3 is written x 3

And we can calculate term 3 using:

x 3 = 3 2 = 9

We can use a Rule to find any term. For example, the 25th term can be found by "plugging in" 25 wherever n is.

x 25 = 25 2 = 625

How about another example:

Example: 3, 5, 8, 13, 21, ?

After 3 and 5 all the rest are the sum of the two numbers before ,

That is 3 + 5 = 8, 5 + 8 = 13 etc, which is part of the Fibonacci Sequence :

3, 5, 8, 13, 21, 34, 55, 89, ...

Which has this Rule:

Rule: x n = x n-1 + x n-2

Now what does x n-1 mean? It means "the previous term" as term number n-1 is 1 less than term number n .

And x n-2 means the term before that one .

Let's try that Rule for the 6th term:

x 6 = x 6-1 + x 6-2

x 6 = x 5 + x 4

So term 6 equals term 5 plus term 4. We already know term 5 is 21 and term 4 is 13, so:

x 6 = 21 + 13 = 34

One of the troubles with finding "the next number" in a sequence is that mathematics is so powerful we can find more than one Rule that works.

What is the next number in the sequence 1, 2, 4, 7, ?

Here are three solutions (there can be more!):

Solution 1: Add 1, then add 2, 3, 4, ...

So, 1+ 1 =2, 2+ 2 =4, 4+ 3 =7, 7+ 4 =11, etc...

Rule: x n = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, ...

(That rule looks a bit complicated, but it works)

Solution 2: After 1 and 2, add the two previous numbers, plus 1:

Rule: x n = x n-1 + x n-2 + 1

Sequence: 1, 2, 4, 7, 12, 20, 33, ...

Solution 3: After 1, 2 and 4, add the three previous numbers

Rule: x n = x n-1 + x n-2 + x n-3

Sequence: 1, 2, 4, 7, 13, 24, 44, ...

So, we have three perfectly reasonable solutions, and they create totally different sequences.

Which is right? They are all right.

... it may be a list of the winners' numbers ... so the next number could be ... anything!

Simplest Rule

When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.

Finding Differences

Sometimes it helps to find the differences between each pair of numbers ... this can often reveal an underlying pattern.

Here is a simple case:

The differences are always 2, so we can guess that "2n" is part of the answer.

Let us try 2n :

n:	1	2	3	4	5
Terms (x ):	7	9	11	13	15
2n:	2	4	6	8	10
Wrong by:	5	5	5	5	5

The last row shows that we are always wrong by 5, so just add 5 and we are done:

Rule: x n = 2n + 5

OK, we could have worked out "2n+5" by just playing around with the numbers a bit, but we want a systematic way to do it, for when the sequences get more complicated.

Second Differences

In the sequence {1, 2, 4, 7, 11, 16, 22, ...} we need to find the differences ...

... and then find the differences of those (called second differences ), like this:

The second differences in this case are 1.

With second differences we multiply by n 2 2

In our case the difference is 1, so let us try just n 2 2 :

n:	1	2	3	4	5
):

:
Wrong by:	0.5	0	-0.5	-1	-1.5

We are close, but seem to be drifting by 0.5, so let us try: n 2 2 − n 2

−
Wrong by:	1	1	1	1	1

Wrong by 1 now, so let us add 1:

− + 1
Wrong by:	0	0	0	0	0

The formula n 2 2 − n 2 + 1 can be simplified to n(n-1)/2 + 1

So by "trial-and-error" we discovered a rule that works:

Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ...

Other Types of Sequences

Read Sequences and Series to learn about:

Arithmetic Sequences
Geometric Sequences
Fibonacci Sequence
Triangular Sequence

And there are also:

Prime Numbers
Factorial Numbers

And many more!

Visit the On-Line Encyclopedia of Integer Sequences to be amazed.

If there is a special sequence you would like covered here let me know .

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Arithmetic sequence

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What is the Formula to Find the Common Difference in Arithmetic sequence?

How to Find n in Arithmetic Sequence?

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

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

Find the tenth term and the n -th term of the following sequence:

Find the n -th term and the first three terms of the arithmetic sequence having a 6 = 5 and d = katex.render("\\boldsymbol{\\color{green}{\\frac{3}{2}}}", typed07); 3/2

Find the n -th term and the first three terms of the arithmetic sequence having a 4 = 93 and a 8 = 65 .

Find the n -th and the 26 th terms of the geometric sequence with katex.render("\\boldsymbol{\\color{green}{ \\mathit{a}_5 = \\frac{5}{4} }}", typed16); a 5 = 5/4 and a 12 = 160 .

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Sequences - Finding a Rule

Finding Missing Numbers

Example: 1, 4, 9, 16, ?

Example: 3, 5, 8, 13, 21, ?

What is the next number in the sequence 1, 2, 4, 7, ?

Simplest Rule

Finding Differences

Second Differences

Other Types of Sequences

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