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# Arithmetic series

Walk through a guided practice where you'll start by finding a simple sum and end by evaluating finite arithmetic series.

### Find the sum of $1 + 3 + 5 + 7 + 9$1, plus, 3, plus, 5, plus, 7, plus, 9.

Awesome! You just found the sum of a small arithmetic series. It only had 5 terms. But, what if it had one million terms? We'd want a formula for sure. Thankfully, we've already learned of such a formula.
Identify the formula for the sum of an arithmetic series.

Sweet! So you remember the formula. Now let's make sure we remember how to apply it.
Choose the answer that shows the formula correctly used to find the sum you found.

Alright, so we're feeling good so far. Let's try to use the formula to find the sum of an arithmetic series that would be tedious to calculate by hand.

## Consider the series $3 + 5 + 7 + ... + 401$3, plus, 5, plus, 7, plus, point, point, point, plus, 401.

Find the values of a, start subscript, 1, end subscript and a, start subscript, n, end subscript for this series.
a, start subscript, 1, end subscript, equals
a, start subscript, n, end subscript, equals

Find the value of n for this series.
n, equals

Find the sum of 3, plus, 5, plus, 7, plus, point, point, point, plus, 401

Wow! Okay, looks like you've got this.

## Try it yourself

Problem 1
Find the sum.
11, plus, 20, plus, 29, plus, point, point, point, plus, 4052, equals

Nice! Try another one!
Problem 2
Find the sum.
10, plus, left parenthesis, minus, 1, right parenthesis, plus, left parenthesis, minus, 12, right parenthesis, plus, point, point, point, plus, left parenthesis, minus, 10, comma, 979, right parenthesis, equals

## Want to join the conversation?

• I do not understand how to find the n value. This did not make any sense to me. I need a formula or an explanation on how to find the n.
• Another way you can do it is to come up with a function, and plug in your final value and solve for x. For example, when finding the sum of 3 + 5 + 7 + .... + 401 It might help to start out with a little chart:

x f(x)
1 3
2 5
3 7
4 9

And from that you might intuit
f(x) = 2x + 1

Then plug in your final term, which is 401, and solve for x
401 = 2x + 1
400 = 2x

Sometimes you’ll see a slightly different, but equivalent function. For example, the same 3 + 5 + 7 + .... + 401 you might first identify as
f(x) = 3 + 2(x-1)
and that’s fine, because mathematically it’s the same function. Even if you don’t simplify it to 2x + 1 you still find x = 200 when you plug in your final term in the sequence.
401 = 3 + 2(x-1)
401-3 = 2(x-1)
398 = 2 (x-1)
398/2 = x-1
199 = x - 1
x = 200
• How do we calculate the value of n
• Take the LAST number in the sequence MINUS the FIRST number in the sequence.
DIVIDE that value by the pattern in the sequence.
How do you find the pattern? Ask, "How do I get from the first term, to the second term?"
Hope that helps
• How do you determine the value of n? I don't think this was every explained.
• n is the number of terms. Take the difference of the first and last term over the common difference and add 1 to get n.
• What's exactly the difference between "progression", "sequence" and "series"?
• For problem 2: how did you find out the n was 450? I tried doing An-A1/2 and adding one for the first term and I got a different number.
• it's like this...
let's say we have an arithmetic sequence that goes like 2, 4, 6, 8, ..., 262
lets take first term as a=2
common difference as d=2
last term or nth term as a(n)=262
we know the formula for nth term is a(n)=a+(n-1)d
so here, 262=2+(n-1)2
so n-1 = (262-2)/2 = 260/2 = 130
so n= 130+1 =131
Sal explained it in a non-formula, less mathematical more logic-based kind of way, but this is the mathematical basis for it
hope it helps :)
(1 vote)
• How is it 1000 for the last one?
• can somebody pleeeease explain how to find n like in question 4b I read the comments and the explanation but couldn't conclude a formula
• it's like this...
let's say we have an arithmetic sequence that goes like 2, 4, 6, 8, ..., 262
lets take first term as a=2
common difference as d=2
last term or nth term as a(n)=262
we know the formula for nth term is a(n)=a+(n-1)d
so here, 262=2+(n-1)2
so n-1 = (262-2)/2 = 260/2 = 130
so n= 130+1 =131
Sal explained it in a non-formula, less mathematical more logic-based kind of way, but this is the mathematical basis for it
hope it helps :)
(1 vote)
• on problem 1, could someone tell me how they found out the number of terms was 450?
• 11+20+29+...+4052

They found n (the last term) by set 4052 into the explicit formula. So to find n you must know how to formulate the formula from the sequence.

This has the initial of 11 and common difference of 9, so a(n)=11+9(n-1). So to find what n is when a(n) = 4052, you set 4052=11+9(n-1) and solve for n. They didn't explain but that's how you would find n in that problem.
• What the heck does this mean: Find the sum of first 335 terms

A(sub)1 = 2
A(sub)i = A(sub< i -1 >) -3

What does the "i" mean?
• The subscripted numbers denote the number of the term in series A. The subscripted i can be any number other than 1. (The first term is separately defined.) The subscripted i-1 refers to the term immediately before term i.

Based on the example you gave, here are the first few terms. (The underscore is used to show a subscript.)
A_1 = 2
A_2 = A_1 - 3 = 2 - 3 = -1
A_3 = A_2 - 3 = -1 - 3 = -4
A_4 = A_3 - 3 = -4 - 3 = -7
• I did not understand this sum:
Find the sum; 11+20+29+...+4052.

Please let me know,how to solve this?