Like what you saw?
Create FREE Account and:
 Watch all FREE content in 21 subjects(388 videos for 23 hours)
 FREE advice on how to get better grades at school from an expert
 Attend and watch FREE live webinar on useful topics
Arithmetic Series  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding the partial sum of an arithmetic sequence; so what we're looking at right here is we have an arithmetic sequence we know it's arithmetic because each time we are adding 4 to get to the next term and we are trying to find s of 40 which is the sum of the first 40 terms.
So we have a formula that is s of n is equal to n over 2 a sub 1 plus a sub n. We know that the general term a sub n is equal to a1 plus n minus 1 times d, so we can do is actually substitute that in here. And why I'm doing that is this particular equation we need to know the first term and the last term. For this particular example, we don't know the last term, so we have to figure that out. So by plugging in what our general term for a sub n, we can rewrite this equation a sub n is equal to 8 over 2, 2a1 plus n minus 1 times d.
So what I've done just there is I've changed it from having to know my last term to just having to know my difference which by looking at this I can know. So all we have to do now is plug in some information.
We are looking for s of 40, the sum of the first 40 terms, that means n is 40, we can just plug that in, 2 times a1, a1 is just our first term which is just going to be 3, plus n minus 1, n is the number of terms, so this is just going to be 39 and times our difference, our difference is what we do from one term to get to the next, so our difference here is just 4.
So what we have is this fairly elaborate equation, but it's all numbers, we can just plug into our calculator starting at our parenthesis 39 times 4 is 156 plus 6 is 162 times 40 over 2, so times 20 which leaves us with 3240.
So by adding up our first 40 terms, we end up with our sum, using either one of our two summation equations either one would work, but for this one I find the second one to work a little bit better because we have the difference and not the last term.
Please enter your name.
Are you sure you want to delete this comment?
Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”
BRIGHTSTORM IS A REVOLUTION !!!”
because of you i ve got a 100/100 in my test thanks”
Concept (1)
Sample Problems (12)
Need help with a problem?
Watch expert teachers solve similar problems.

Arithmetic Series
Problem 1 7,465 viewsFind S_{40}
3 + 7 + 11 + 15 + ... 
Arithmetic Series
Problem 2 6,054 views3 + 3 + 0 + ... 30 2 
Arithmetic Series
Problem 3 5,758 views_{20} ∑(1 − 2i) ^{i = 0} 
Arithmetic Series
Problem 4 5,321 viewsHow many terms in 5 + 7 + 9 + ... must be added to make 572.

Arithmetic Series
Problem 5 723 views 
Arithmetic Series
Problem 6 746 views 
Arithmetic Series
Problem 8 835 views 
Arithmetic Series
Problem 9 662 views 
Arithmetic Series
Problem 10 655 views 
Arithmetic Series
Problem 11 706 views 
Arithmetic Series
Problem 12 619 views 
Arithmetic Series
Problem 13 735 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete