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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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# Arithmetic series worksheet

Try a few problems where you'll be asked to evaluate finite arithmetic series.

# Practice evaluating arithmetic series

## Problem 1

## Problem 2

## Problem 3

## Want to join the conversation?

- Why do you always add one when solving for n?(17 votes)
- You have to add one because you're working out how many items there are in the series by counting how many hops it takes to get from the first to the last.

Imagine there is a line of 5 squares on the ground. You are standing in the first. You hop from the first to the second: 1 hop. Then from the second to the third: 2 hops. Third to fourth: 3 hops. Fourth to fifth: 4 hops. You have to hop 4 times to get from your initial position in the first square to your final position in the fifth square. More generally, for each hop you take, the number of hops is always one less than the number of squares you've been standing in.

This is what we do when we divide by the difference. We start out on the first item in the sequence, then work out how many times we have to add our constant to get to the final item. That show us how many times we've added.

But, as (I hope) is shown in the example with the squares, there's always going to be one more in the series than the number of hops, because it doesn't count your starting position, only the movement between positions(51 votes)

- In problem #2 what does ai-1 refer too?(6 votes)
- The "i-1" is a subscript. "i" is the "index", which tells you which term you are in. Another way to write it is "a(i-1)". "a(i-1) is the term before "a(i)". "a(i+2)" would be the 2nd term after "a(i)".

When i is 1, what's i-1? when i is 335, what's i-1? What's i for the first term? The 335th term?

0, 334, 1, 335.(8 votes)

- In the first question, how do you get n=275?(1 vote)
- The problem is written in sigma notation. 275 being above the Greek letter means there are 275 terms in the series. I'd recommend the following links to review and practice.

https://www.khanacademy.org/math/algebra2/sequences-and-series/copy-of-sigma-notation/v/sigma-notation-sum

https://www.khanacademy.org/math/algebra2/sequences-and-series/copy-of-sigma-notation/e/evaluating-basic-sigma-notation(12 votes)

- Hi guys I need help! since long break I have forgotten math a little bit!

The sum of five numbers in arithmetic progression is 40 and the sum of their squares is

410; find the five numbers(2 votes)- The five number are:
*a, (a + k), (a + 2k), (a + 3k), (a + 4k)*

a + (a + k) + (a + 2k) + (a + 3k) + (a + 4k) = 40

5a + 10k = 40

a^2 + (a + k)^2 + (a + 2k)^2 + (a + 3k)^2 + (a + 4k)^2 = 410

5a^2 + 30k^2 + 20ak = 410

We know that 5a + 10k = 40, then a = 8 - 2k

5(8-2k)^2+30k^2+20(8-2k)k=410

k = -3 or 3

a = 14 or 2

The numbers are either*14, 11, 8, 5, 2*or*2, 5, 8, 11, 14.*

Hope this helps.(4 votes)

- In problem after multipying -499 by 335 Im'm getting -15215. I guess I'm doing something wrong here.(2 votes)
- It appears you have a math error. -499 (335 ) = -167165

Without seeing the actual work that created your answer of -15215, it's hard to say where you made the error.(4 votes)

- In Problem #2. Can anyone explain the reasoning behind subtracting 1000 (EDIT: 1002, NOT 1000) from 2? Why would you want to "set it to zero"?(1 vote)
- To find the sum of terms in an arithmetic series we need to :

1. Find the first term (in this case, 2)

2. Find the last term (in this case, - 1000)

3. Take the average of their sum : (in this case, { 2 + - 1000) / 2 }

That's why it looks like 1000 is being subtracted from 2.

As for your "set it to zero" question, I can't see where that is in the video.(6 votes)

- How to prove a^2(b+c) + b^2(a+c) + c^2(a+b) = (a+b+c)^3 * 2/9(2 votes)
- Simplify both sides of the expression and see if they become the same expression in simplified form.(2 votes)

- In problem #1:

How come when I use this Formula: An= A1 + (n-1)d to find n it wont give me the right answer?

It comes out to: (((-1,363)-7)/5) +1 = 273

-1,363 = 7+(n-1)5 -> -1,370=(n-1)5 -> -274= n-1 -> here i should add one, but according the the formula I have to subtract, ending with 273.

Am I missing a rule here? when its negative you add?(2 votes)- In this problem, what's the first term of the sequence?

Right, it's 12-5 = 7. So the first term already has 5 subtracted from it. The subtraction starts with the first term, not the second. So, the formula is:

"a(k) = 12 - 5k",

not "a(k) = 12 - 5(k-1)".,

because the sum is "sigma (k=1 to 275) of (-5k + 12)".(2 votes)

- how come in question 2, you're decreasing by -3 for each new term, but yet when you multiplied how many terms are in the sequence with the -3, you took out the negative in the -3 to make it 334 multiplied by 3 and have the answer come out to be 1002(1 vote)
- What they did there is actually still correct. This is because they are not "decreasing by -3" but are actually just decreasing by 3, or alternatively increasing by -3. So you are right, 334 times -3 is -1002, which is why they do 2 - 1002. You will note they say "decreases by a total of 334 * 3".

This way of writing it might be a little confusing but if you look at it carefully you can usually figure it out.(3 votes)

- How do you get to the 350 part of the equation?(1 vote)
- 2044 is the 350th term of the series.

The nth term is given by:

T(n) = a+(n-1)d and T(n) is 2044 so we should find n

2044= -50+(n-1)6

solving this we get n=350

Therefore 2044 is the 350th term(1 vote)