Main content

## Precalculus

### Course: Precalculus > Unit 9

Lesson 4: Arithmetic series- Arithmetic series intro
- Arithmetic series formula
- Arithmetic series
- Worked example: arithmetic series (sigma notation)
- Worked example: arithmetic series (sum expression)
- Worked example: arithmetic series (recursive formula)
- Arithmetic series worksheet
- Arithmetic series
- Proof of finite arithmetic series formula
- Series: FAQ

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# Arithmetic series intro

Sal explains the formula for the sum of a finite arithmetic series. Created by Sal Khan.

## Want to join the conversation?

- How do you figure out the sum of an alternating sequence like the sum of the first 39 terms of

1 + (-1)^n?(8 votes)- You can look at it as a sum of two sequences--the first is arithmetic, with initial term a=1 and term difference of d=0. The first term is 1, the 39th ("last") term is 1+0*39=1. Sum of an arithmetic sequence is (first + last)*(#terms/2) = (1+1)*(39/2) = (2)*(39/2) = 39. The second sequence is geometric, with initial term a=-1 and term ratio r=-1. Sum of a geometric series, from another video, is a*(1-r^n)/(1-r)

= (-1)*(1-(-1)^39)/(1-(-1))

= (-1)*(1-(-1))/(1-(-1))

= (-1)*(1+1)/(1+1)

= (-1)*2/2

= -1

You can verify this intuitively by considering that the first term is -1, and then the next 38 terms cancel each other out in pairs (2nd and 3rd cancel, 4th and 5th cancel...38th and 39th cancel).

Sum the two results to get 39 + (-1) = 38.

Not every sequence can be so simply broken down into a sum of other (simple) sequences, but this one can.(13 votes)

- At1:22how is 2+(n-1)=(n+1)? That doesn't make any sense to me.(3 votes)
- 2 + 𝑛 – 1

We can combine the 2 and the -1 to get 2 – 1 = 1. Thus:

2 + 𝑛 – 1 = 𝑛 + 1

Proficiency in this type of algebra is expected for studying arithmetic series. I would recommend looking at the "solving one step equations" playlist and the "combining like terms" video.(8 votes)

- If this is true, why is the sum of all positive integers -1/12?(4 votes)
- It's important to keep in mind that the people who make that series add up to -1/12 are willfully breaking some of the rules of standard mathematics (and usually not mentioning that fact). For the purposes of what they're doing (unless they're just doing it to freak out the newbies), they're not wrong, but it just makes things confusing to people learning the rules of standard math. For what you're learning now, the intuition that the sum of all positive integers should be undefinable because it is infinitely large is the correct attitude.(6 votes)

- how do you write the sigma notation for the sequence 1+5+9+13+17+21 what is n? and what would the general form be?(2 votes)
- nth term: 1 + 4 * ( n - 1 ) = 1 + 4n - 4 = 4n - 3

6

Σ ( 4n - 3 ) = 1 + 5 + 9 + 13 + 17 + 21

n = 1(6 votes)

- So how would you find the sum of an infinite geometric series?(4 votes)
- You cannot add up an infinite number of numbers, but you can take the limit of the sum as n approaches infinity. Sometimes this gives you a number, sometimes it gives you infinity, sometimes it isn't helpful at all. Much of the time, you must be content to know if the sum "converges" or "diverges".

You'll learn about this in calculus. Unfortunately, Sal only has a couple of videos on the basics of this broad subject. Try a Google search for "Infinite Series".(2 votes)

- What does the n stand for?(3 votes)
- "n" is used to represent the term in the sequence. If you need the 3rd term, then n=3. If you need the 10th term, the n=10.

Hope this helps.(3 votes)

- Isn't that called the "little Gauss" after the German mathematician Carl Friedrich Gauss, who discovered that formula in elementary school? Maybe the story is apocryphal...(3 votes)
- Hi I just wanted to ask what does n-1 and n-2 mean(2 votes)
- Hi! Consider series 1,2,3,4,5 for example.

There are 5 terms in this Arithmetic series.

The fifth term is 5 and it is n

The fourth term is 4 and it is n-1

The third term is 3 and it is n-2

The second term is 2 and it is n-3

The first term is 1 and it is n-4. We know that n = 5 so n-4 is 1, n-3=2....(2 votes)

- What does "arithmetic series" have to do with the real world? How could one apply this concept in a job or something?(1 vote)
- Many jobs have regular salary increases. That means you get hired at an initial yearly amount, and then it increases steadily some fixed amount each year. (If it increases a fixed percent each year, you would use a geometric sequence.) So you can calculate: how long until you get certain amount of money or how much you can get in specific year and many more.. I hope that helps(4 votes)

- I think before exploring the next video that this is true. The average of the first and last term tells the average value of each term , and when we multiply it by the number of terms we get the overall increment.(2 votes)
- Essentially, if you wanted to add all of the terms of a finite arithmetic series together, you would add the first and the last terms together and multiply by the total number of terms divided by two (you have halved the number of adding operations, remember?). This works because of 1+100=101, the the next pair of terms gives you 2+99=101; and you will do this 50 times (there are 100/2 pairs) So ([first]+[last])*([num.of.terms]/2)=(100+1)*(50)= 5050

You can get this from the series formula:

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

The n/2 will give you the *50, n+1=101

I hope this helps!(1 vote)

## Video transcript

Let's say we have the simplest of arithmetic sequences and probably the simplest of sequences one, two... we're going to start at one and just increment by one one, two, three and we're going to go all the way to n And what I want to think about is what is the sum of this sequence going to be? And the sum of a sequence, we already know we call a series so what is the sum and I will just call it Sn what is this going to be equal to one plus two plus three plus going all the way to n well, we're going to do a neat little trick here where I'm going to rewrite this sum so I will write it again as Sn but now I'm just going to write it in reverse order I'm going to write it as n plus n minus one plus n minus two all the way to one and now I'm going to add these two equations So we know that Sn is equal to this so we are adding the same thing to both sides of this equation up here So on the left hand side we're going to have Sn plus Sn is just two times Sn and on the right hand side and this is where we start to see something kind of cool you have a one plus an n which is just going to be n plus one you have a two plus n minus one which is... well two plus n minus one is going to be n plus one again plus n plus one you have a three plus n minus two well that's going to be n plus one again I think you see what is going on here And we're going to go all the way to this last pair I guess you could say, or call it these last two terms and we have an n plus one again plus n plus one So how many of these n plus one's do we have? well we have n of them there were n of these terms in each of these equations one two three... all the way to n So, we can rewrite this thing as two times Sn is equal to, and you have n (n+1) terms so we could write it as n times n plus one n times n plus one and now to solve for Sn to solve for our sum we can just divide both sides by two And so we are going to be left with the sum from one to n this arithmetic sequence where we're just incrementing by one starting at one, is going to be equal to n times n plus one over two And this is neat because now you can quickly find the sum let's say from one to a hundred it will be 100 times 101 over 2 So very quickly you can find these sums And what I'm curious about and what we will explore in the next video is can we generalize this for any arithmetic sequence we started with a very simple one we started at one, we just incremented by one And it looks like so if I were to write it this way this is n times (n + 1) over 2 So this right over here, this n this is the nth term in our sequence and this right over here, this one was the first term in our sequence So at least in this case it looks like I took the average of the first term and the nth term so this right over here this is the average this right over here is the average of a1 and an and then I'm multiplying that times n and what I'm curious about is whether this is going to be true for any arithmetic sequence that the sum of it is going to be the average of the first and the last term times the number of terms