## Table of Contents

- What is an Arithmetic Sequence?
- Arithmetic Sequence Explicit Formula
- The Recursive Formula for Arithmetic Sequence
- Sum of Finite Terms of an Arithmetic Sequence
- Lesson Summary

Understand what an arithmetic sequence is and discover how to solve arithmetic sequence problems using the explicit and recursive formulas. Learn the formula that explains how to sum a finite number of terms of an arithmetic progression.
Updated: 09/23/2021

- What is an Arithmetic Sequence?
- Arithmetic Sequence Explicit Formula
- The Recursive Formula for Arithmetic Sequence
- Sum of Finite Terms of an Arithmetic Sequence
- Lesson Summary

A sequence is simply a succession of elements. In mathematics, a special type of sequence arises and, for that reason, receives a special name and is the object of study in this lesson. The elements of a sequence are called its terms. An arithmetic sequence, also known as arithmetic progression, is a sequence in which a term minus its predecessor is a constant, which we call the common difference, and denote by {eq}d {/eq}.

In order to verify if a given sequence has an arithmetic pattern, we need to check if the difference between all pairs of consecutive terms is the same number. Observe the following sequences and determine whether or not they are arithmetic progressions.

a. {eq}2, 4, 6, 8 {/eq}

This is the sequence of the first four even positive numbers. Since {eq}4 - 2 = 6 - 4 = 8 - 6 = 2 {/eq}, we conclude that this finite sequence is arithmetic with {eq}d = 2 {/eq}.

b. {eq}1, 3, 5, 7, 9, 11, \cdots {/eq}

This is an infinite sequence, but we obviously do not need to do an infinite number of subtractions. The ellipsis at the end of the list of numbers indicates that the procedure is the same for the subsequent terms. Since {eq}3 - 1 = 5 - 3 = 7 - 5 = 9 - 7 = 11 - 9 = 2 {/eq}, we conclude that the list of positive odd numbers is an arithmetic progression with {eq}d = 2 {/eq}.

c. {eq}3, 9, 27, 81, \cdots {/eq}

Since {eq}9 - 3 = 6 \neq 18 = 27 - 9 {/eq}, the difference between consecutive terms is not a constant, so the given sequence is not arithmetic. One could have noticed that, in this sequence, to go from one term to the next, we multiply by 3. When that happens, we have a geometric sequence (or progression), but that is not the object of study of this lesson.

Observe the following pattern in an arithmetic progression {eq}a_1, a_2, a_3, \cdots {/eq}

{eq}a_2 = a_1 + d\\ a_3 = a_2 + d = (a_1 + d) + d = a_1 +2 d\\ a_4 = a_3 + d = (a_1 + 2 d) + d {/eq}

So, if we want the n-th term of the sequence, it will be given by $$a_n = a_1 + (n - 1)d, $$ where {eq}a_1 {/eq} is the first term of the sequence, {eq}d {/eq} is the common difference and {eq}n {/eq}, the position of the term we want to evaluate.

Let {eq}a_1, a_2, \cdots {/eq} be an arithmetic sequence with {eq}a_1 = 5 {/eq} and {eq}d = 3 {/eq}. Find out the general term of the sequence.

We have {eq}a_n = 5 + (n - 1)3 = 5 + 3 n - 3 = 2 + 3 n {/eq}. Therefore, the general term is given by {eq}a_n = 2 + 3 n {/eq}

The first two terms of an arithmetic sequence are 6 and 2. Write an explicit formula for the general term.

Since {eq}a_1 = 6 {/eq} and {eq}a_2 = 2 {/eq}, it follows that {eq}d = 2 - 6 = -4 {/eq}. Plugging the values in the formula gives {eq}a_n = 6 + (n - 1) (- 4) = 6 - 4 n + 4 = 10 - 4 n {/eq}

Now that we saw the explicit formula, we are going to introduce the recursive formula. In an arithmetic progression, we have

{eq}a_1\\ a_2 = a_1 + d\\ a_3 = a_2 + d\\ \vdots {/eq}

So, in general, we have the relation a_n = a_{n-1} + d, which is called a recursive formula because it provides a way of knowing the general term knowing the previous one and the common difference.

In the fifteen-ball pool game, 15 balls are displayed in a triangle with the number of balls in each row forming an arithmetic sequence. The first row contains 5 balls, the last, 1 ball and, from one row to the next, the number of balls is decreased by 1 unity. Find the number of rows in the sequence.

The following is given: {eq}a_1 = 5 {/eq}, {eq}a_n = 1 {/eq} and {eq}d = - 1 {/eq}. We want to find {eq}n {/eq}. We are going to use the explicit formula to solve this problem.

{eq}1 = 5 + (n - 1)(- 1) \implies - 4 = - n + 1 \implies n = 5 {/eq} Therefore, there are 5 rows, as depicted below.

The first and eleventh terms of a sequence are equal to 4 and 19, respectively. Find the common difference and the twenty-eighth term of the arithmetic progression.

Since {eq}a_1 = 4 {/eq} and {eq}a_{1 1} = 19 {/eq}, we have {eq}1 9 = 4 + 1 0 d \implies d = \displaystyle \frac{3}{2} {/eq}. With the common difference we can find the twenty-eighth term: {eq}a_{28} = 4 + 27 \displaystyle \frac{3}{2} = \displaystyle \frac{89}{2} {/eq}.

Suppose you are given an arithmetic sequence (finite or infinite) for which you want to determine the sum of a finite number of consecutive terms. Because we are interested in a finite number of terms of the sequence, we can relabel them to {eq}a_1, a_2, a_3, \cdots, a_{n} {/eq} in case the terms we are interested in are not the first terms of the sequence. Assume that, in the given sequence, the common difference is {eq}d {/eq}. We can rewrite the terms as follows:

$$a_1\\ a_2 = a_1 + d\\ a_3 = a_1 + 2 d\\ \vdots\\ a_{n-2} = a_1 + (n-3)d\\ a_{n-1} = a_1 + (n - 2)d\\ a_n = a_1 + (n-1)d\\ $$

If we add the first and last terms, second and second to last, and so on, we are going to notice a pattern. We are going to do the first three of those sums.

{eq}a_1 + a_n = a_1 + a_1 + (n - 1)d = 2 a_1 + (n - 1)d\\ a_2 + a_{n-1} = a_1 + d + a_1 + (n - 2)d = 2 a_1 + (n - 1)d\\ a_3 + a_{n-2} = a_1 + 2 d + a_1 + (n - 3)d = 2 a_1 + (n - 1)d {/eq}

With the first three sums, we can observe that each one of them is equal to {eq}a_1 + a_n {/eq}. From {eq}a_1 {/eq} to {eq}a_n {/eq} there are {eq}n {/eq} terms and because we are grouping these terms in pairs, it turns out we have {eq}\displaystyle \frac{n}{2} {/eq} summands. Therefore,

$$a_1 + a_2 + \cdots + a_n = (a_1 + a_n) \displaystyle \frac{n}{2} $$

Let's use the above formula to evaluate the sum of the first 30 terms of a sequence with {eq}a_1 = 3 {/eq} and {eq}d = - 2 {/eq}.

We need to find {eq}a_{3 0} {/eq} before we can plug the values in the sum formula. The general term formula is {eq}a_n = a_1 + (n - 1)d {/eq}. So, we have {eq}a_{3 0} = 3 + 29 (- 2) = - 5 5 {/eq}. Now, substituting the values in the sum formula gives us {eq}a_1 + a_2 + \cdots + a_{3 0} = \displaystyle \frac{(3 - 5 5) 3 0}{2} = - 7 8 0 {/eq}

We saw that an arithmetic sequence (or progression) is a sequence where the difference between two consecutive terms is a constant. The explicit formula for the general term is {eq}a_n = a_1 + (n - 1) d {/eq}, and the recursive formula is {eq}a_n = a_{n - 1} + d {/eq}, but, in this case, we also need to be given the term {eq}a_1 {/eq}. We finished the lesson presenting the formula that gives the sum of a finite number of terms of an arithmetic sequence: {eq}a_1 + a_2 + \cdots + a_n = (a_1 + a_n) \displaystyle \frac{n}{2} {/eq}.

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- Activities
- FAQs

- The explicit formula for the
*n*th term of an arithmetic sequence is*a*n =*a*1 + d(n - 1), where*a*n is the*n*th term of the sequence,*a*1 is the first term of the sequence, and*d*is the common difference of the sequence.

- Nancy is saving money for a bike that costs $275. She starts with $50, and adds $15 at the end of each week. How many weeks will it take her to get enough money to buy the bike?
- As a New Year's Resolution, on January 1, Bob has taken up running, and has a goal of running for 1 hour, or 60 minutes, straight. He starts by jogging for 5 minutes on the first day, and he adds 2 minutes to his jogging time each successive day after that. Will Bob have reached his goal by the end of the month (31 days)?

- It will take 16 weeks to save $275. If she starts with $50, and adds $15 each week, then the amount of money she has saved at the end of each week follows the arithmetic sequence of 50, 65, 90, 105, . . ., or the arithmetic sequence with
*a*1 = 50 and*d*= 15. Since the bike costs $275, we want to know what term will be 275, or for what value*n*will*a*n = 275. Plugging these into our explicit formula gives the equation 275 = 50 + 15(n - 1). Solving this for*n*gives*n*= 16. - Yes. To show this is the case, we verify that after 31 days, Bob is jogging 60 minutes or more following this pattern. He starts at 5 minutes and adds 2 minutes each day, so this can be modeled with an arithmetic sequence with a first term of 5 and a common difference of 2 (or 5, 7, 9, 11, . . .). We want to know if the 31st term will be greater than or equal to 60, so we plug
*a*1 = 5,*d*= 2, and n = 31 into our explicit formula to get*a*(31) = 5 + 2(31 - 1). Simplifying gives*a*(31) = 65, so at the end of the month, Bob can run for 65 minutes straight, which is greater than 60 minutes.

When we are given the first term a_1 of an arithmetic sequence, the recursive formula is given by a_n = a_(n-1) + d.

An arithmetic sequence with the first term being a_1 and common difference d has general term a_n = a_1 + (n - 1)d.

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