problem solving involving arithmetic sequence

, and of the arithmetic sequence if you know :

, and of the arithmetic sequence if you know :

perimeter of the triangle is 72 cm
area of the triangle is 54 cm

the first consecutive odd numbers
the first consecutive even numbers

for digging the well on its depth in meters.

• Link Partners
• Privacy Policy

Copyright © 2015-2023 math-exercises.com - All rights reserved. Any use of website content without written permission is prohibited.

9.2 Arithmetic Sequences

Learning objectives.

In this section, you will:

• Find the common difference for an arithmetic sequence.
• Write terms of an arithmetic sequence.
• Use a recursive formula for an arithmetic sequence.
• Use an explicit formula for an arithmetic sequence.

Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year.

As an example, consider a woman who starts a small contracting business. She purchases a new truck for $25,000. After five years, she estimates that she will be able to sell the truck for $8,000. The loss in value of the truck will therefore be $17,000, which is $3,400 per year for five years. The truck will be worth $21,600 after the first year; $18,200 after two years; $14,800 after three years; $11,400 after four years; and $8,000 at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value.

Finding Common Differences

The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is –3,400.

The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence , and add 3 to find the subsequent term.

Arithmetic Sequence

An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference . If a 1 a 1 is the first term of an arithmetic sequence and d d is the common difference, the sequence will be:

Is each sequence arithmetic? If so, find the common difference.

• ⓐ { 1 , 2 , 4 , 8 , 16 , ... } { 1 , 2 , 4 , 8 , 16 , ... }
• ⓑ { − 3 , 1 , 5 , 9 , 13 , ... } { − 3 , 1 , 5 , 9 , 13 , ... }

Subtract each term from the subsequent term to determine whether a common difference exists.

The graph of each of these sequences is shown in Figure 1 . We can see from the graphs that, although both sequences show growth, a a is not linear whereas b b is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line.

If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference?

No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference.

Is the given sequence arithmetic? If so, find the common difference.

Writing Terms of Arithmetic Sequences

Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of n n and d d into formula below.

Given the first term and the common difference of an arithmetic sequence, find the first several terms.

• Add the common difference to the first term to find the second term.
• Add the common difference to the second term to find the third term.
• Continue until all of the desired terms are identified.
• Write the terms separated by commas within brackets.

Write the first five terms of the arithmetic sequence with a 1 = 17 a 1 = 17 and d = − 3 d = − 3 .

Adding − 3 − 3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.

The first five terms are { 17 , 14 , 11 , 8 , 5 } { 17 , 14 , 11 , 8 , 5 }

As expected, the graph of the sequence consists of points on a line as shown in Figure 2 .

List the first five terms of the arithmetic sequence with a 1 = 1 a 1 = 1 and d = 5 d = 5 .

Given any first term and any other term in an arithmetic sequence, find a given term.

• Substitute the values given for a 1 , a n , n a 1 , a n , n into the formula a n = a 1 + ( n − 1 ) d a n = a 1 + ( n − 1 ) d to solve for d . d .
• Find a given term by substituting the appropriate values for a 1 , n , a 1 , n , and d d into the formula a n = a 1 + ( n − 1 ) d . a n = a 1 + ( n − 1 ) d .

Given a 1 = 8 a 1 = 8 and a 4 = 14 a 4 = 14 , find a 5 a 5 .

The sequence can be written in terms of the initial term 8 and the common difference d d .

We know the fourth term equals 14; we know the fourth term has the form a 1 + 3 d = 8 + 3 d a 1 + 3 d = 8 + 3 d .

We can find the common difference d d .

Find the fifth term by adding the common difference to the fourth term.

Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation a n = a 1 + ( n − 1 ) d . a n = a 1 + ( n − 1 ) d .

Given a 3 = 7 a 3 = 7 and a 5 = 17 a 5 = 17 , find a 2 a 2 .

Using Recursive Formulas for Arithmetic Sequences

Some arithmetic sequences are defined in terms of the previous term using a recursive formula . The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given.

Recursive Formula for an Arithmetic Sequence

The recursive formula for an arithmetic sequence with common difference d d is:

Given an arithmetic sequence, write its recursive formula.

• Subtract any term from the subsequent term to find the common difference.
• State the initial term and substitute the common difference into the recursive formula for arithmetic sequences.

Writing a Recursive Formula for an Arithmetic Sequence

Write a recursive formula for the arithmetic sequence .

The first term is given as −18 −18 . The common difference can be found by subtracting the first term from the second term.

Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.

We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure 3 . The growth pattern of the sequence shows the constant difference of 11 units.

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

No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference.

Write a recursive formula for the arithmetic sequence.

Using Explicit Formulas for Arithmetic Sequences

We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.

To find the y -intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.

The common difference is − 50 − 50 , so the sequence represents a linear function with a slope of − 50 − 50 . To find the y y -intercept, we subtract − 50 − 50 from 200 : 200 − ( − 50 ) = 200 + 50 = 250 200 : 200 − ( − 50 ) = 200 + 50 = 250 . You can also find the y y -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure 4 .

Recall the slope-intercept form of a line is y = m x + b . y = m x + b . When dealing with sequences, we use a n a n in place of y y and n n in place of x . x . If we know the slope and vertical intercept of the function, we can substitute them for m m and b b in the slope-intercept form of a line. Substituting − 50 − 50 for the slope and 250 250 for the vertical intercept, we get the following equation:

We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is a n = 200 − 50 ( n − 1 ) a n = 200 − 50 ( n − 1 ) , which simplifies to a n = − 50 n + 250. a n = − 50 n + 250.

Explicit Formula for an Arithmetic Sequence

An explicit formula for the n th n th term of an arithmetic sequence is given by

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

• Find the common difference, a 2 − a 1 . a 2 − a 1 .
• Substitute the common difference and the first term into a n = a 1 + d ( n − 1 ) . a n = a 1 + d ( n − 1 ) .

Writing the n th Term Explicit Formula for an Arithmetic Sequence

Write an explicit formula for the arithmetic sequence.

The common difference can be found by subtracting the first term from the second term.

The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.

The graph of this sequence, represented in Figure 5 , shows a slope of 10 and a vertical intercept of − 8 − 8 .

Write an explicit formula for the following arithmetic sequence.

Finding the Number of Terms in a Finite Arithmetic Sequence

Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence.

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

• Find the common difference d . d .
• Substitute the common difference and the first term into a n = a 1 + d ( n – 1 ) . a n = a 1 + d ( n – 1 ) .
• Substitute the last term for a n a n and solve for n . n .

Find the number of terms in the finite arithmetic sequence .

The common difference is − 7 − 7 . Substitute the common difference and the initial term of the sequence into the n th n th term formula and simplify.

Substitute − 41 − 41 for a n a n and solve for n n

There are eight terms in the sequence.

Find the number of terms in the finite arithmetic sequence.

Solving Application Problems with Arithmetic Sequences

In many application problems, it often makes sense to use an initial term of a 0 a 0 instead of a 1 . a 1 . In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula:

A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week.

• ⓐ Write a formula for the child’s weekly allowance in a given year.
• ⓑ What will the child’s allowance be when he is 16 years old?

The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2.

Let A A be the amount of the allowance and n n be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get:

We can find the number of years since age 5 by subtracting.

We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16.

The child’s allowance at age 16 will be $23 per week.

A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today?

Access this online resource for additional instruction and practice with arithmetic sequences.

• Arithmetic Sequences

9.2 Section Exercises

What is an arithmetic sequence?

How is the common difference of an arithmetic sequence found?

How do we determine whether a sequence is arithmetic?

What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence?

Describe how linear functions and arithmetic sequences are similar. How are they different?

For the following exercises, find the common difference for the arithmetic sequence provided.

{ 5 , 11 , 17 , 23 , 29 , ... } { 5 , 11 , 17 , 23 , 29 , ... }

{ 0 , 1 2 , 1 , 3 2 , 2 , ... } { 0 , 1 2 , 1 , 3 2 , 2 , ... }

For the following exercises, determine whether the sequence is arithmetic. If so find the common difference.

{ 11.4 , 9.3 , 7.2 , 5.1 , 3 , ... } { 11.4 , 9.3 , 7.2 , 5.1 , 3 , ... }

{ 4 , 16 , 64 , 256 , 1024 , ... } { 4 , 16 , 64 , 256 , 1024 , ... }

For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference.

a 1 = −25 a 1 = −25 , d = −9 d = −9

a 1 = 0 a 1 = 0 , d = 2 3 d = 2 3

For the following exercises, write the first five terms of the arithmetic series given two terms.

a 1 = 17 , a 7 = − 31 a 1 = 17 , a 7 = − 31

a 13 = − 60 , a 33 = − 160 a 13 = − 60 , a 33 = − 160

For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference.

First term is 3, common difference is 4, find the 5 th term.

First term is 4, common difference is 5, find the 4 th term.

First term is 5, common difference is 6, find the 8 th term.

First term is 6, common difference is 7, find the 6 th term.

First term is 7, common difference is 8, find the 7 th term.

For the following exercises, find the first term given two terms from an arithmetic sequence.

Find the first term or a 1 a 1 of an arithmetic sequence if a 6 = 12 a 6 = 12 and a 14 = 28. a 14 = 28.

Find the first term or a 1 a 1 of an arithmetic sequence if a 7 = 21 a 7 = 21 and a 15 = 42. a 15 = 42.

Find the first term or a 1 a 1 of an arithmetic sequence if a 8 = 40 a 8 = 40 and a 23 = 115. a 23 = 115.

Find the first term or a 1 a 1 of an arithmetic sequence if a 9 = 54 a 9 = 54 and a 17 = 102. a 17 = 102.

Find the first term or a 1 a 1 of an arithmetic sequence if a 11 = 11 a 11 = 11 and a 21 = 16. a 21 = 16.

For the following exercises, find the specified term given two terms from an arithmetic sequence.

a 1 = 33 a 1 = 33 and a 7 = − 15. a 7 = − 15. Find a 4 . a 4 .

a 3 = − 17.1 a 3 = − 17.1 and a 10 = − 15.7. a 10 = − 15.7. Find a 21 . a 21 .

For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence.

a 1 = 39 ; a n = a n − 1 − 3 a 1 = 39 ; a n = a n − 1 − 3

a 1 = − 19 ; a n = a n − 1 − 1.4 a 1 = − 19 ; a n = a n − 1 − 1.4

For the following exercises, write a recursive formula for each arithmetic sequence.

a = { 40 , 60 , 80 , ... } a = { 40 , 60 , 80 , ... }

a = { 17 , 26 , 35 , ... } a = { 17 , 26 , 35 , ... }

a = { − 1 , 2 , 5 , ... } a = { − 1 , 2 , 5 , ... }

a = { 12 , 17 , 22 , ... } a = { 12 , 17 , 22 , ... }

a = { − 15 , − 7 , 1 , ... } a = { − 15 , − 7 , 1 , ... }

a = { 8.9 , 10.3 , 11.7 , ... } a = { 8.9 , 10.3 , 11.7 , ... }

a = { − 0.52 , − 1.02 , − 1.52 , ... } a = { − 0.52 , − 1.02 , − 1.52 , ... }

a = { 1 5 , 9 20 , 7 10 , ... } a = { 1 5 , 9 20 , 7 10 , ... }

a = { − 1 2 , − 5 4 , − 2 , ... } a = { − 1 2 , − 5 4 , − 2 , ... }

a = { 1 6 , − 11 12 , − 2 , ... } a = { 1 6 , − 11 12 , − 2 , ... }

For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term.

a = { 7 ,  4 ,  1 ,  ... } ; a = { 7 ,  4 ,  1 ,  ... } ; Find the 17 th term.

a = { 4 ,  11 ,  18 ,  ... } ; a = { 4 ,  11 ,  18 ,  ... } ; Find the 14 th term.

a = { 2 ,  6 ,  10 ,  ... } ; a = { 2 ,  6 ,  10 ,  ... } ; Find the 12 th term.

For the following exercises, use the explicit formula to write the first five terms of the arithmetic sequence.

a n = 24 − 4 n a n = 24 − 4 n

a n = 1 2 n − 1 2 a n = 1 2 n − 1 2

For the following exercises, write an explicit formula for each arithmetic sequence.

a = { 3 , 5 , 7 , ... } a = { 3 , 5 , 7 , ... }

a = { 32 , 24 , 16 , ... } a = { 32 , 24 , 16 , ... }

a = { − 5 ,  95 ,  195 ,  ... } a = { − 5 ,  95 ,  195 ,  ... }

a = { −17 ,  −217 ,  −417 , ... } a = { −17 ,  −217 ,  −417 , ... }

a = { 1.8 ,  3.6 ,  5.4 ,  ... } a = { 1.8 ,  3.6 ,  5.4 ,  ... }

a = { −18.1 , −16.2 , −14.3 , ... } a = { −18.1 , −16.2 , −14.3 , ... }

a = { 15.8 , 18.5 , 21.2 , ... } a = { 15.8 , 18.5 , 21.2 , ... }

a = { 1 3 , − 4 3 , −3 ,  ... } a = { 1 3 , − 4 3 , −3 ,  ... }

a = { 0 , 1 3 , 2 3 , ... } a = { 0 , 1 3 , 2 3 , ... }

a = { − 5 , − 10 3 , − 5 3 , … } a = { − 5 , − 10 3 , − 5 3 , … }

For the following exercises, find the number of terms in the given finite arithmetic sequence.

a = { 3 , − 4 , − 11 ,  ... , − 60 } a = { 3 , − 4 , − 11 ,  ... , − 60 }

a = { 1.2 , 1.4 , 1.6 , ... , 3.8 } a = { 1.2 , 1.4 , 1.6 , ... , 3.8 }

a = { 1 2 , 2 , 7 2 , ... , 8 } a = { 1 2 , 2 , 7 2 , ... , 8 }

For the following exercises, determine whether the graph shown represents an arithmetic sequence.

For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence.

a 1 = 0 , d = 4 a 1 = 0 , d = 4

a 1 = 9 ; a n = a n − 1 − 10 a 1 = 9 ; a n = a n − 1 − 10

a n = − 12 + 5 n a n = − 12 + 5 n

For the following exercises, follow the steps to work with the arithmetic sequence a n = 3 n − 2 a n = 3 n − 2 using a graphing calculator:

• Select SEQ in the fourth line
• Select DOT in the fifth line
• Press [ENTER]
• n Min n Min is the first counting number for the sequence. Set n Min = 1 n Min = 1
• u ( n ) u ( n ) is the pattern for the sequence. Set u ( n ) = 3 n − 2 u ( n ) = 3 n − 2
• u ( n Min) u ( n Min) is the first number in the sequence. Set u ( n Min) = 1 u ( n Min) = 1
• Set TblStart = 1 TblStart = 1
• Set Δ Tbl = 1 Δ Tbl = 1
• Set Indpnt: Auto and Depend: Auto
• Press [2ND] then [GRAPH] to go to the TABLE

What are the first seven terms shown in the column with the heading u ( n ) ? u ( n ) ?

Use the scroll-down arrow to scroll to n = 50. n = 50. What value is given for u ( n ) ? u ( n ) ?

Press [WINDOW] . Set n Min = 1 n Min = 1 , n Max = 5 n Max = 5 , x Min = 0 x Min = 0 , x Max = 6 x Max = 6 , y Min = − 1 y Min = − 1 , and y Max = 14. y Max = 14. Then press [GRAPH] . Graph the sequence as it appears on the graphing calculator.

For the following exercises, follow the steps given above to work with the arithmetic sequence a n = 1 2 n + 5 a n = 1 2 n + 5 using a graphing calculator.

What are the first seven terms shown in the column with the heading u ( n ) u ( n ) in the TABLE feature?

Graph the sequence as it appears on the graphing calculator. Be sure to adjust the WINDOW settings as needed.

Give two examples of arithmetic sequences whose 4 th terms are 9. 9.

Give two examples of arithmetic sequences whose 10 th terms are 206. 206.

Find the 5 th term of the arithmetic sequence { 9 b , 5 b , b , … } . { 9 b , 5 b , b , … } .

Find the 11 th term of the arithmetic sequence { 3 a − 2 b , a + 2 b , − a + 6 b … } . { 3 a − 2 b , a + 2 b , − a + 6 b … } .

At which term does the sequence { 5.4 , 14.5 , 23.6 , ... } { 5.4 , 14.5 , 23.6 , ... } exceed 151?

At which term does the sequence { 17 3 , 31 6 , 14 3 , ... } { 17 3 , 31 6 , 14 3 , ... } begin to have negative values?

For which terms does the finite arithmetic sequence { 5 2 , 19 8 , 9 4 , ... , 1 8 } { 5 2 , 19 8 , 9 4 , ... , 1 8 } have integer values?

Write an arithmetic sequence using a recursive formula. Show the first 4 terms, and then find the 31 st term.

Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28 th term.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/9-2-arithmetic-sequences

© Jun 28, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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.

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 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 :

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.

go to slide go to slide go to slide

Book a Free Trial Class

Arithmetic Sequence Questions

go to slide go to slide

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)

Arithmetic Sequence: Definition and Basic Examples

Definition and basic examples of arithmetic sequence.

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

The constant difference in all pairs of consecutive or successive numbers in a sequence is called the common difference , denoted by the letter [latex]d[/latex]. We use the common difference to go from one term to another. How? Take the current term and add the common difference to get to the next term, and so on. That is how the terms in the sequence are generated.

• If the common difference between consecutive terms is positive , we say that the sequence is increasing .
• On the other hand, when the difference is negative we say that the sequence is decreasing .

Illustrative Examples of Increasing and Decreasing Arithmetic Sequences

Here are two examples of arithmetic sequences. Observe their common differences.

With this basic idea in mind, you can now solve basic arithmetic sequence problems.

Examples of How to Apply the Concept of Arithmetic Sequence

Example 1: Find the next term in the sequence below.

First, find the common difference of each pair of consecutive numbers.

• [latex]15−7 = 8[/latex]
• [latex]23−15 = 8[/latex]
• [latex]31−23 = 8[/latex]

Since the common difference is [latex]8[/latex] or written as [latex]d=8[/latex], we can find the next term after [latex]31[/latex] by adding [latex]8[/latex] to it. Therefore, we have [latex]31 + 8 = 39[/latex].

Example 2: Find the next term in the sequence below.

Observe that the sequence is decreasing. We expect to have a common difference that is negative in value.

• [latex]24−31 = −7[/latex]
• [latex]17−24 = −7[/latex]
• [latex]10−17 = −7[/latex]

To get to the next term, we will add this common difference of [latex]d=-7[/latex] to the last term in the sequence. Therefore, [latex]10 + \left( { – 7} \right) = 3[/latex].

Example 3: Find the next three terms in the sequence below.

Be careful here. Don’t assume that if the terms in the sequence are all negative numbers, it is a decreasing sequence. Remember, it is decreasing whenever the common difference is negative. So let’s find the common difference by taking each term and subtracting it by the term that comes before it.

The common difference here is positive four [latex]\left( { + \,4} \right)[/latex] which makes this an increasing arithmetic sequence. We can obtain the next three terms by adding the last term by this common difference. Whatever is the result, add again by [latex]4[/latex], and do it one more time.

Here’s the calculation:

The next three terms in the sequence are shown in red.

Example 4: Find the seventh term (7 th ) in the sequence below.

Sometimes you may encounter a problem in an arithmetic sequence that involves fractions. So be ready to use your previous knowledge on how to add or subtract fractions .

Also, always make sure that you understand what the question is asking so that you can have the correct strategy to approach the problem.

In this example, we are asked to find the seventh term, not simply the next term. It is a good practice to write all the terms in the sequence and label them, if possible.

Now we have a clear understanding of how to work this out. Find the common difference, and use this to find the seventh term.

Finding the common difference,

Then we find the 7th term by adding the common difference starting with the 4th term, and so on. Here’s the complete calculation.

Therefore, the seventh term of the sequence is zero (0). We can write the final answer as,

Example 5: Find the [latex]\color{red}{35^{th}}[/latex] term in the arithmetic sequence [latex]3[/latex], [latex]9[/latex], [latex]15[/latex], [latex]21[/latex], …

You can solve this problem by listing the successive terms using the common difference. This method is tedious because you will have to keep adding the common difference (which is [latex]6[/latex]) thirty-five times starting with the last term in the sequence.

You don’t have to do this because it is cumbersome. And not only that, it is easy to commit a careless error during the repetitive addition process.

If you decide to find the [latex]\color{red}{35^{th}}[/latex] term of the sequence using this “ successive addition ” method, your solution will look similar below. The “dot dot dot” means that there are calculations there but not shown as it can easily occupy the entire page.

You might also like these tutorials:

• Arithmetic Sequence Formula
• Arithmetic Series Formula
• More Practice Problems with the Arithmetic Sequence Formula

Reset password New user? Sign up

Existing user? Log in

Arithmetic and Geometric Progressions Problem Solving

Already have an account? Log in here.

Recommended Course

Algebra through puzzles.

Supercharge your algebraic intuition and problem solving skills!

• Vilakshan Gupta
• Andres Gonzalez

To solve problems on this page, you should be familiar with

• arithmetic progressions
• geometric progressions
• arithmetic-geometric progressions .

You can boost up your problem solving on arithmetic and geometric progressions through this wiki. Make sure you hit all the problems listed in this page.

Problem Solving - Basic

Problem solving - intermediate, problem solving - advanced.

This section contains basic problems based on the notions of arithmetic and geometric progressions. Starting with an example, we will head into the problems to solve.

I have an arithmetic progression such that the initial term is 5 and the common difference is 10. What is minimum value of \(n\) such that the \(n^\text{th}\) term is larger than 100? We can just start by listing out the numbers: \[ 5,15,25,35,45,55,65,75,85, 95,105. \] We can clearly see that the \(11^\text{th}\) number is larger than 100, and thus \(n=11.\) However, note that this will become impractical if the common difference becomes smaller and/or the number we are looking for becomes larger. A practical way to solve it is via applying the \(n^\text{th}\) term formula. With \(a = 5, d = 10\), we have \( T_n = a + (n-1) d > 100 \). Then \(5 + (n-1) \cdot 10 > 100 \). Solving for \(n\) yields \(n > 10.5\). So the \(11^\text{th}\) term is the smallest term that satisfies the condition. \(_\square\)

Here comes the problems for you to solve.

The average of the first 100 positive integers is \(\text{__________}.\)

If \(A, B, C, D \) are consecutive terms in an arithmetic progression, what is the value of

\[ \frac{ D^2 - A^2 } { C^2 - B^2} ? \]

Assume \( C^2 - B^2 \neq 0.\)

\[ 54+51+48+45+ \cdots\]

You are given the sum of an arithmetic progression of a finite number of terms, as shown above.

What is the minimum number of terms used to make a total value of 513?

One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely.

Find the sum of the perimeters of all these triangles that are defined above.

Once a man did a favor to a king that made the king very happy. Out of joy the king told the man to wish for anything and he would be granted. The man wanted to ask for the whole kingdom which was worth 1500 trillion dollars, but obviously that would make the king mad and he would never be granted that wish.

The man who happened to be a mathematician thought a little bit and said the following:

"Bring in a big piece of rug with an \(8\times 8\) grid in it. Starting from the top left square, put one dollar in that square. Put two dollars in the square next to it and then double of that, four dollars, in the next square and so on. When you reach the end of the first row, continue on to the next row, doubling the amount every time as you move to the next square, all the way until the \(64^\text{th}\) square at the bottom right."

The king thought for a second. The first square will take one dollar, the second two dollars, the third, four dollars, and next 8 dollars, and then 16 dollars, and then 32 dollars, 64 dollars, 128 dollars, 256 dollars, and so on. That's not too bad. I can do it.

The king agreed. What happened next?

\[1+2 \cdot 2+ 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots+ 100 \cdot 2^{99}= \, ?\]

This section contains a bit harder problems than the previous section. But all these can be solved using arithmetic and geometric problems. Here we go:

Real numbers \(a_1,a_2,\ldots,a_{99}\) form an arithmetic progression.

Suppose that \[ a_2+a_5+a_8+\cdots+a_{98}=205.\] Find the value of \( \displaystyle \sum_{k=1}^{99} a_k\).

The value of \(\displaystyle \sum_{n=1}^ \infty \frac{ 2n}{ 3^n } \) can be expressed in the form \( \frac{a}{b} \), where \(a\) and \(b\) are coprime positive integers. Find \( a - b \).

Let \(a,b,c\) be positive integers such that \(\frac{b}{a}\) is an integer. If \(a,b,c\) are in geometric progression and the arithmetic mean of \(a,b,c\) is \(b+2,\) find the value of

\[\dfrac{a^2+a-14}{a+1}.\]

If an infinite GP of real numbers has second term \(x\) and sum \(4,\) where does \(x\) belong?

4 positive integers form an arithmetic progression.

If we subtract \(2,6,7\) and \(2,\) respectively, from the 4 numbers, it forms a geometric progression.

What is the sum of these 4 numbers?

Let \(a, b \in R^{+}.\)

\(a, A_{1}, A_{2}, b\) is an arithmetic progression. \(a, G_{1}, G_{2}, b\) is a geometric progression.

Which of the following must be true?

We have three numbers in an arithmetic progression, and another three numbers in a geometric progression. Adding the corresponding terms of the two series, we get \( 120 , 116 , 130 \). If the sum of all the terms in the geometric progression is \( 342 \), what is the largest term in the geometric progression?

\(\) Details and Assumptions:

• If the terms of the AP are A, B, C, and the terms of the GP are X, Y, Z, then adding the corresponding terms will give us A+X, B+Y, C+Z.

This section has problems which need advanced understanding of the notions and generally get solved on using multiple notions at a time. Let's give these problems an attempt.

\[(1-x)(1-2x)(1-4x)\cdots \left(1-2^{101}x\right)\]

What is the coefficient of \(x^{101} \) in the expansion of the above?

\[ \frac {2+6}{4^{100}} + \frac {2+2 \times 6}{4^{99}} + \frac {2+ 3 \times 6}{4^{98}} + \cdots + \frac {2+ 100 \times 6}{4} \]

Evaluate the above expression.

In JEE examination the paper consists of 90 questions. The marks are awarded in such a way that if a person gets a question correct, he gets \(+4\) marks; if he does it wrong, he gets \(-2\) marks; if he leaves the question unanswered, he gets \(0\) marks (as per 2015). Find the sum of all possible marks that a student can get in JEE.

Let \(A=\{a_1, a_2, \ldots, a_n\}\) be a set of the first \(n\) terms of an arithmetic progression. Similarly, let \(B=\{b_1, b_2, \ldots, b_n\}\) be a set of the first \(n\) terms of a geometric progression.

If a new set \(C=A+B=\{a_1+b_1, a_2+b_2, \ldots, a_n+b_n\}\) and the first four terms of \(C\) are \(\{0, 0, 1, 0\},\) what is the \(11^\text{th}\) term of \(C?\)

A linear function \(f(x)=bx+a\) has the property that \(f\big(f(x)\big)=dx+c\) is another linear function such that \(a,b,c,d\) are integers that are consecutive terms in an arithmetic sequence. Find the last three digits of the sum of all possible values of \(f(2013)\).

\[ \begin{eqnarray} &\displaystyle\sum_{n=0}^{7}\log_{3}(x_{n}) &= 308 \\ 56 \leq & \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right )& \leq 57 \\ \end{eqnarray} \]

The increasing geometric sequence \(x_{0},x_{1},x_{2},\ldots\) consists entirely of integral powers of 3. If they satisfy the two conditions above, find \(\log_{3}(x_{14}).\)

Suppose \(2015\) people of different heights are arranged in a straight line from shortest to tallest such that

(i) the tops of their heads are collinear, and

(ii) for any two successive people, the horizontal distance between them is equal to the height of the shorter of the two people.

If the shortest person is \(49\) inches tall and the tallest is \(81\) inches tall, then how tall is the person at the middle of the line, (in inches)?

Given that \(a_1,a_2,a_3\) is an arithmetic progression in that order so that \(a_1+a_2+a_3=15\) and \(b_1,b_2,b_3\) is a geometric progression in that order so that \(b_1b_2b_3=27\).

If \(a_1+b_1, a_2+b_2, a_3+b_3\) are positive integers and form a geometric progression in that order, determine the maximum possible value of \(a_3\).

The answer is of the form \(\dfrac{a+b\sqrt{c}}{d}\), where \(a\), \(b\), \(c\), and \(d\) are positive integers and the fraction is in its simplest form and \(c\) is square free. Submit the value of \( a + b + c + d \).

Telescoping Series - Sum and Telescoping Series - Product

Limits of Sequences

Master concepts like these

Learn more in our Algebra through Puzzles course, built by experts for you.

Problem Loading...

Note Loading...

Set Loading...

SOLVING WORD PROBLEMS INVOLVING ARITHMETIC SEQUENCE

Problem 1 :

The 17 th term of an AP exceeds its 10 th term by 7. Find the common difference.

a 17  =   a 10 + 7

a + 16d  =  a + 9d  + 7

a - a + 16d - 9d  =  7

7d  =  7

d  =  7/7

d  =  1

So, the common difference is 1.

Problem 2 :

Which term of the AP :

3, 15, 27, 39,.......... will be 132 more than its 54th term?

a n = 132 + a 54

a + (n - 1) d = 132 + a + 53d

 a = 3    d = 15 - 3   =  12

3 + (n - 1) 12  =  132 + 3 + 53 (12)

3 + 12 n - 12  =  135 + 636

-9 + 12 n  =  771

12 n  =  771 + 9

12 n  =  780

n  =  780/12

n  =  65

Problem 3 :

Two APs have the same common difference. The difference between their 100th term is 100, What is the difference between their 1000th terms?

Let the first two terms of two APs a₁  and a ₂  respectively and the common difference of these two A.Ps be d

for first A.P

a₁₀₀ = a₁ + (100 - 1) d

  =  a₁ + 99 d

a₁₀₀ ₀ = a₁ + (1000 - 1) d

  =  a₁ + 999 d

For second AP

a₁₀₀  =  a ₂ + (100 - 1) d

  =  a ₂ + 99 d

a₁₀₀ ₀  =  a ₂ + (1000 - 1) d

=  a ₂ + 999 d

given that,

difference between two  100 th term of two APs = 100

( a₁ + 99 d) - ( a ₂ + 99 d) = 100

a₁ + 99 d -  a ₂ - 99 d = 100

a₁  -  a ₂ = 100

Difference between 1000th term of two APs

   = ( a₁ + 999 d) - ( a ₂ + 999 d)

   =  a₁ + 999 d -  a ₂ - 999 d

   =  a₁  -  a ₂

   =  100

Kindly mail your feedback to   [email protected]

We always appreciate your feedback.

© All rights reserved. onlinemath4all.com

• Sat Math Practice
• SAT Math Worksheets
• PEMDAS Rule
• BODMAS rule
• GEMDAS Order of Operations
• Math Calculators
• Transformations of Functions
• Order of rotational symmetry
• Lines of symmetry
• Compound Angles
• Quantitative Aptitude Tricks
• Trigonometric ratio table
• Word Problems
• Times Table Shortcuts
• 10th CBSE solution
• PSAT Math Preparation
• Privacy Policy
• Laws of Exponents

Recent Articles

Digital sat math problems and solutions (part - 9).

Aug 29, 24 07:03 PM

Digital SAT Math Problems and Solutions (Part - 8)

Aug 29, 24 06:57 PM

Digital SAT Math Problems and Solutions (Part - 7)

Aug 28, 24 08:09 PM