Calculus, all content (2017 edition)
- Summation notation
- Converting explicit series terms to summation notation
- Converting explicit series terms to summation notation (n ≥ 2)
- Summation notation intro
- Arithmetic series formula
- Worked example: arithmetic series (sigma notation)
- Worked example: arithmetic series (sum expression)
- Worked example: arithmetic series (recursive formula)
- Arithmetic series
Arithmetic series formula
Learn how to write an arithmetic sequence in general terms, using a common difference d and a first term a. Next, we'll see that this formula is equivalent to multiplying the average of the first and last terms by the number of terms. Created by Sal Khan.
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- At2:00mins and after, I understand what you did, I don't understand why. Do we need to write every series backwards every time we want the equation for the sequence or is there a better way to go about this? And where did writing it backwards come from?(20 votes)
- No need to write all that out every time. The purpose of all that is to illustrate why the formula works. The fundamental insight that originally led to the creation of this formula probably started with the observation that the sum of the first term and last term in an arithmetic series is always the same as the sum of the 2nd and 2nd-to-last, 3rd and 3rd-to-last, etc. Try it in your head with a simple series, such as whole numbers from 1 to 10, to see what I mean. Writing a sequence forward then backward, one above the other, is just another way to look at that same phenomenon because it aligns the first term with the last term, 2nd term with 2nd-to-last, etc. It also leaves no confusion about what to do with the middle term when there are an odd number of terms.(38 votes)
- Why does Sal swap the order of the sequence and add it to itself ? I understand that it works in finding the series, but I don't understand why it works.(12 votes)
- Imagine the sequence of whole numbers from 1 to 10 written out. Then imagine the same sequence written in reverse order just below the first. When you add the vertical pairs of corresponding terms, you will get the same result each time, which in this example is 11 (1+10=11, 2+9=11, 3+8=11 ...). This is because as you move from one pair to the next, the upper term increases by the same amount that the lower term decreases. Sal is illustrating this principle in a general form, showing how it applies to ANY arithmetic sequence.(23 votes)
- What is the general formula for geometric series? If there is one.(8 votes)
- Geometric series:
finite geometric series- Sn= a sub 1(1-r^n) / 1-r
infinite geometric series- Sn= a sub 1 / 1-r(4 votes)
- Sorry about this question but, what is the difference between a arithmetic sequence and a arithmetic series. I can't seem to wrap my brain around this topic.(8 votes)
- series 1+2+3+4...
literally the only difference is being separated by commas and plus signs(6 votes)
- Just for example: What if we have sigma notation?Σ It has lower limit(a) as 1, upper limit as 4, then,Σ, a(a-1).(6 votes)
- What is the easiest way to see if a sequence is arithmetic or geometric and, similarly, to see if a series is one or the other?(4 votes)
- Consider the sequence of numbers we will represent as A, B, C, D, E .....
If B-A = C-B = D-C = E-D.... then it is arithmetic.
If A/B = B/C = C/D = E/D .... then it is geometric.
An arithmetic series is the sum of an arithmetic sequence
A geometric series is the sum of a geometric sequence.
Thus, with the series you just see if the relationship between the terms is arithmetic (each term increases or decreases by adding a constant to the previous term ) or geometric (each term is found by multiplying the previous term by a constant).(7 votes)
- So basically sigma = (First term + last term)/2 * no. of terms(5 votes)
- For Arithmetic series specifically. Other things will likely have other formula.(4 votes)
- What exactly is "a"? Like, when in the equation a + d(n-1) ? What number in the sequence is "a"??(4 votes)
- A is the first term of a sequence, you may also see it written as t1 (t sub 1).(6 votes)
- In an infinite arithmetic series, how can you do the average of the terms ?(4 votes)
- there is no average, it just keeps getting bigger and bigger(4 votes)
- at3:40how did
1d + (n-2)d become (n - 2 + 1)d?(3 votes)
- It's an arithmetic factoring, you can see it more clearly if you you first expand the terms in the parenthesis and then take our the common
1d + (n - 2)d
1d + nd - 2d
(1 + n - 2) d
(n - 2 + 1) d(5 votes)
Let's write an arithmetic sequence in general terms. So we can start with some number a. And then we can keep adding d to it. And that number that we keep adding, which could be a positive or a negative number, we call our common difference. So the second term in our sequence will be a plus d. The third term in our sequence will be a plus 2d. So we keep adding d all the way to the n-th term in our sequence. And you already see here that in our first term, we added d zero times. Our second term, we added d once. In our third term, we added d twice. So you see, whatever the index of the term is, we're adding d one less than that many times. So if we go all the way to the n-th term, we're going to add d one less than n times. So it's going to be n minus 1 times d. Fair enough. And let me write that. This right over here is our n-th term. Now what I want to do is think about what the sum of this arithmetic sequence would be. And the sum of an arithmetic sequence we call an arithmetic series. So let me write that in yellow. Color changing is sometimes difficult. So the arithmetic series is just the sum of an arithmetic sequence. So let's call my arithmetic series s sub n. And let's say it's going to be the sum of these terms, so it's going to be a plus d, plus a plus 2d, plus all the way to adding the n-th term, which is a plus n minus 1 times d. Now I'm going to do the same trick that I did when I did the most basic arithmetic sequence. I'm going to add this to itself, but I'm going to swap the order in which I write this sum. So s sub n I can write as this, but I'm going to write it in reverse order. I'm going to write the last term first. So the n-th term is a plus n minus 1 times d. Then the second to last term is going to be a plus n minus 2 times d. The third to last is going to be a plus n minus 3 times d. And we're going to go all the way down to the first term, which is just a. Now let's add these two equations. We are going to get, on the left hand side, s sub n plus s sub n. You're going to get 2 times s sub n. Well, what's the sum of these two first terms right over here? I'm going to have a plus a plus n minus 1 times d. So it's going to be 2a plus n minus 1 times d. Now let's add both of these second terms. So if I were to add both of these second terms, what do I get it? I'm going to get 2a plus 2a. And what's d plus n minus 2 times d? So you could view it several ways. Let me write this over here. What is d plus n minus 2 times d? Well, this is just the same thing as 1d plus n minus 2 times d. And so you could just add the coefficients. So this is going to be n minus 2 plus 1 times d, which is equal to n minus 1 times d. So the second term also becomes 2a plus n minus 1 times d. Now let's add the third term. I'll do it in green. The third terms, I should say. And I think you're going to see a pattern here. It's 2a plus 2a. And if I have 2 plus n minus 3 of something and then I add 2, I'm going to have n minus one of that something. So plus n minus 1 times d. And you're going to keep doing that all the way until your n-th pair of terms, all the way until you add these two characters over here, which is just 2a plus n minus 1 times d. So you have this 2a plus n minus 1 d being added over and over again. And how many times are you doing that? Well, you had n pairs of terms when you were adding these two equations. In each of them, you had n terms. This is the first term, this is the second term, this is the third term, all the way to the n-th term. So I can rewrite 2 times the sum 2 times s sub n is going to be n times this quantity. It's going to be n times 2a plus n minus 1 times d. And then if we want to solve for s sub n, you just divide both sides by 2. And you get s sub n is equal to, and we get ourselves a little bit of a drum roll here, n times 2a plus n minus 1 times d. All of that over 2. Now, we've come up with a general formula, just a function of what our first term is, what our common difference is, and how many terms we're adding up. And so this is the generalized sum of an arithmetic sequence, which we call an arithmetic series. But now, let's ask ourselves this question. This is hard to remember. The n times 2a plus n minus 1 times d over 2. But in the last video, when I did a more concrete example, I said well, it looks like the sum of an arithmetic sequence could be written as perhaps the average of the first term a1 plus an. The average of the first term and the last term times the number of terms that you have. So is this actually the case? Do these two things gel? Because this is very easy to remember-- the average of the first and the last terms multiplied by the number of terms you had and actually makes intuitive sense, because you're just increasing by the same amount every time. So let's just average the first and the last term and then multiply times the number of terms we have. Well, all we have to do is rewrite this a little bit to see that it is indeed the exact same thing as this over here. So all we have to do is break out the a. So let me rewrite it. So, this could be rewritten as s sub n is equal to n times a plus a plus n minus 1 times d. I just broke up this 2a into an a plus a. All of that over 2. And you see, based on how we defined this thing, our first term a1 is a. And then our last term, a sub n, is a plus n minus 1 times d. So this whole business right over here really is the average of the first and last terms. I got my first term, adding it to my last term, dividing it by 2. And then I'm multiplying by the number of terms we have. And that's true for any arithmetic sequence, as we've just shown here.