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## 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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# Worked example: arithmetic series (recursive formula)

Sal evaluates the sum of the first 650 terms in the sequence defined recursively as {aᵢ=aᵢ₋₁+11, a₁=4}. He does that by finding the 650th term and using the arithmetic series formula (a₁+aₙ)*n/2.

## Want to join the conversation?

- What is the difference between a series and a sequence?(5 votes)
- In a series, the individual terms are being added to each other.

A sequence just lists each individual term and uses commas to separate them.(12 votes)

- Isn't it easier (in evaluating this) to simply convert the recursive expression to an explicit one? Then you can just plug the values you want in, and solve for the answer. (With a bit of algebraic manipulation, of course.)(6 votes)
- i jus need to know how to write the equation(4 votes)
- When finding n (number in a series) when x is the rate of change, is this a correct formula:

n=|(an-a1)/x|+1(2 votes)- You are correct but, in this example, solving for n was not necessary since we were provided with it at the beginning. Sal told us that there are 650 terms.(4 votes)

- why does sal add 4 twice into the calculation where the formula only asks for it once.(3 votes)
- they first time, he added 4 as part of the formua for a(n) ie the 650th term

he second time it was as part of the formula for S(n) ie sum of 650 terms

look through the video once again im sure u'll get it

hope it helps OwO(2 votes)

- When Sal writes a shouldn't it be, if a = 4, 3+11

I-1(2 votes)- a1 or the first term is 4 and the difference between consecutive numbers is 11.(4 votes)

- I don't understand how we got 649, can someone explain?(2 votes)
- Ok, so think about it this way. I'm adding 650 terms in a sequence, right? Ok, so I'm given the first term, right? Yes, I am. If I'm given the first term, then I don't do anything to it, because it's already defined. This means that I have to change all but the first term, which is given. So, you take that first one from 650. 650-1, obviously, is 649, and that's how he got that number. Hope this helps.(3 votes)

- If the last number of ai= is minus, do you add 1 to the a650?(2 votes)
- Can we find the 650th term by using an= a + (n-1)d, where a is the first term and d is the common difference?(2 votes)
- In this video Sal said the AP is given by ai=ai-1+11 could it be possible to write ai=a1(1 vote)
- You could have a arithmetic series defined by ai = a1 (which has a difference of 0). But it isn't the same as ai=ai-1+11.(1 vote)

## Video transcript

- [Voiceover] So I'm going
to recursively define an arithmetic sequence. So we're going to say that
they ith term of the sequence is equal to the i minus oneth term of the sequence plus 11. So each term is going to be 11 more than the term before it. Now we have to establish a base case here, and so we're going to
say that the first term of our arithmetic sequence is going to be equal to four. So given this recursive definition of our arithmetic
sequence right over here, what I challenge you to do is to find the sum of the first 650 terms of the sequence. Let me write that down. Find the sum of first 650 terms of the sequence, of this arithmetic sequence that we have just defined. And like always, pause the video and see if you can work that out. All right, so how can we think about this? Well, in many videos we give our intuition for the sum of an arithmetic sequence, and we came up with a formula for evaluating a sum of
an arithmetic sequence, which we call an arithmetic series, and that sum of the first n terms is going to be the first term plus the last term over two, so really the average of
the first and last terms, times the number of terms we have. And this is only the case
when it's an arithmetic series where each term that we're adding is a fixed amount larger or less than the term before it, or that
we have a fixed difference. So what about this one right over here? What is the first and the last term going to be, and what is our n? Well we know that n is 650, we know that n is 650, and we know what the first term is going to be. The first terms is going to be four. We need to figure out
what the nth term is, or we need to figure out
what the 650th term is going to be. Let's think about this a little bit. So this is going to be,
if we're taking the sum, it's going to be four plus the next term, the second term, so if a sub two is going
to be a sub one plus 11. So it's going to be four
plus 11, which is 15. We're going to add 11,
which is going to get us to 26, and we're going to keep adding 11. Now how many times are we going to add 11? Well to get to the second term, we add 11 once. To get to the third term, we add 11 twice. So to get to the 650th term, so this is a sub 650, a sub 650, we're going to have to add 11, look, to get to the second term, we added 11 once, third term, add 11 twice. So to get to the 650th term, we are going to add 11, we are going to add 11
650 minus one times, or 649 times. Notice, to get to each term, to get to the first term, you added one minus one, you had an 11 one minus one times, you added 11 zero times. You started with the four, didn't add 11 at all. Then the second term, you added 11 once. Third term, you added 11 twice. Fourth term, you added 11 three times. 650th term, you added 11 649 times. And so if you add 11 649 times, what do you get? So four plus 649 times 11 is going to be equal to, I'll get my calculator out for this, so this is going to be equal to 649 times 11 is equal to, now plus four is equal to 7,143, 7,143. So that's the 650th term. 7, 143. And so now we can just evaluate this. So I'll get the calculator out for that. So we have 7,143 plus four plus the first term plus four is equal to that. We're going to divide by two. So divide by two gets us 3,573.5. We're going to multiply that times 650. That's how many terms we have. Times 650 is equal to, that's a pretty large number, is going to be equal to 2,322,775. 2,322,775. I already forgot it. I have trouble remembering things when I take it off my screen. All right, 2,322,775. I'm glad I had a calculator in hand for that one, but you could do it by hand. I always encourage you to do it. It never hurts to practice the arithmetic.