Can someone explain the equation: S_n = n*(a_1 + a_n)/2 to me please?

Announcements Posted on
    • Thread Starter
    Offline

    3
    ReputationRep:
    I only know it in the form of Sn = n/2[2a+(n-1)d]

    I have an equation where I am calculating the sum, where a=4, d=3.5, and n=31

    I know it as S31 = 31/2[2x4+(31-1)x3.5]

    Online, I see a solution with the equation: S_n = n*(a_1 + a_n)/2 which has the following steps:

    S_31 = 31*(a_1 + a_31)/2
    S_31 = 31*(k + k + 30d)/2
    S_31 = 31*(4 + 4 + 30*3.5)/2
    S_31 = 31*(113)/2
    S_31 = 1,751.5

    Can you please explain how this formula and solution is the same as my formula and solution? I don't get what it means by a_1 and a_n, but also I don't get how that equation from the solution matches mine and converts.

    I would just like to be walked through that equation and solution as it is completely different to mine and I havent seen a_1 and a_n before so also don't know what they mean.
    Thanks!
    Offline

    3
    ReputationRep:
    (Original post by blobbybill)
    I only know it in the form of Sn = n/2[2a+(n-1)d]

    I have an equation where I am calculating the sum, where a=4, d=3.5, and n=31

    I know it as S31 = 31/2[2x4+(31-1)x3.5]

    Online, I see a solution with the equation: S_n = n*(a_1 + a_n)/2 which has the following steps:

    S_31 = 31*(a_1 + a_31)/2
    S_31 = 31*(k + k + 30d)/2
    S_31 = 31*(4 + 4 + 30*3.5)/2
    S_31 = 31*(113)/2
    S_31 = 1,751.5

    Can you please explain how this formula and solution is the same as my formula and solution? I don't get what it means by a_1 and a_n, but also I don't get how that equation from the solution matches mine and converts.

    I would just like to be walked through that equation and solution as it is completely different to mine and I havent seen a_1 and a_n before so also don't know what they mean.
    Thanks!
    Remember that the nth term of the sequence is a + (n-1)d.

    So if you split up 2a + (n-1)d into a + (a + (n-1)d) you get the first term plus the last term (i.e a1 + an).

    They're both right, you use them interchangebly depending on what info you have.
    Offline

    3
    ReputationRep:
    (Original post by blobbybill)
    I only know it in the form of Sn = n/2[2a+(n-1)d]

    I have an equation where I am calculating the sum, where a=4, d=3.5, and n=31
    I know it as S31 = 31/2[2x4+(31-1)x3.5]
    Online, I see a solution with the equation: S_n = n*(a_1 + a_n)/2 which has the following steps:
    S_31 = 31*(a_1 + a_31)/2
    S_31 = 31*(k + k + 30d)/2
    S_31 = 31*(4 + 4 + 30*3.5)/2
    S_31 = 31*(113)/2
    S_31 = 1,751.5

    Can you please explain how this formula and solution is the same as my formula and solution? I don't get what it means by a_1 and a_n, but also I don't get how that equation from the solution matches mine and converts.

    I would just like to be walked through that equation and solution as it is completely different to mine and I havent seen a_1 and a_n before so also don't know what they mean.
    Thanks!
     a_n refers to the nth term of a sequence, and so  a_1 refers to the first term in a sequence - this matches with your  a , which you define as the first term in the sequence.

    What they've done is they've taken the formula  S_n=\frac{n}{2}[a+(a+(n-1)d)] (which is the same as yours, I've just written  2a = a+a and put some brackets). Notice that what is inside the curly brackets is the nth term of an arithmetic sequence and  a is the first term. This thus means that the sum can also be expression as  S_n = \frac{n}{2} [a_1 + a_n] .

    That is what they have started with - and they have then realised, they don't have  a_{31} , but they do have a common difference, so they reverted to your formula instead. It's a bit of a roundabout method, but it's important to know that both formulae exist and they're just algebraic manipulations of each other.

    Hope that somewhat helps. If you still don't understand, just let me know
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by SeanFM)
    Remember that the nth term of the sequence is a + (n-1)d.

    So if you split up 2a + (n-1)d into a + (a + (n-1)d) you get the first term plus the last term (i.e a1 + an).

    They're both right, you use them interchangebly depending on what info you have.
    So when you say (a1 + an), you basically mean U1 (as in the first term - because first term is just a) and Un (as in the last term - because last term is given by a + n-1 d)??

    You say you use them interchangeably depending on which info you have. When would you use one compared to the other?

    My teacher hasn't told us about that equation, we have only been taught n/2[2a+(n-1)d], should I be worried?
    Offline

    3
    ReputationRep:
    (Original post by blobbybill)
    So when you say (a1 + an), you basically mean U1 (as in the first term - because first term is just a) and Un (as in the last term - because last term is given by a + n-1 d)??

    You say you use them interchangeably depending on which info you have. When would you use one compared to the other?

    My teacher hasn't told us about that equation, we have only been taught n/2[2a+(n-1)d], should I be worried?
    Yes, except you used a1 and an so I stuck to the same notation

    Depends on which info you have in your question given you have all the pieces of information so could do either. But as you don't have Un and would have to work it out it's less effort to use the one you've been taught.

    Depends on whether you've finished the topic/chapter or not.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by kingaaran)
     a_n refers to the nth term of a sequence, and so  a_1 refers to the first term in a sequence - this matches with your  a , which you define as the first term in the sequence.

    What they've done is they've taken the formula  S_n=\frac{n}{2}[a+(a+(n-1)d)] (which is the same as yours, I've just written  2a = a+a and put some brackets). Notice that what is inside the curly brackets is the nth term of an arithmetic sequence and  a is the first term. This thus means that the sum can also be expression as  S_n = \frac{n}{2} [a_1 + a_n] .

    That is what they have started with - and they have then realised, they don't have  a_{31} , but they do have a common difference, so they reverted to your formula instead. It's a bit of a roundabout method, but it's important to know that both formulae exist and they're just algebraic manipulations of each other.

    Hope that somewhat helps. If you still don't understand, just let me know
    So to use the method that they have used, they would need the first term (a, to put into the "a1" part of the equation), and they would also need the last term (given by (a + (n-1)d)) for that equation to work?? Is this correct? So you need the first term, the last term (and obviously n) to work out the sum using that equation that they used?

    Also, just to clarify, a1 always equals a doesn't it, because it is the first term?
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by SeanFM)
    Yes, except you used a1 and an so I stuck to the same notation

    Depends on which info you have in your question given you have all the pieces of information so could do either. But as you don't have Un and would have to work it out it's less effort to use the one you've been taught.

    Depends on whether you've finished the topic/chapter or not.

    AHH, SOMETHING JUST CLICKED!!! Sn=n/2 (a + a+(n-1)d), ie the equation that they used in that example, is the same as writing it as Sn=n/2(a+l) isn't it? Because l, for the last term, is given by a + (n-1)d. I'm right in realising this aren't I?

    Also, when you say "an" for the last term in that equation, is that the same as saying a31, or u31 in this instance (for the 31st term), because n=31 in this case? Am I right in thinking this too?
    Thanks
    Offline

    3
    ReputationRep:
    (Original post by blobbybill)
    AHH, SOMETHING JUST CLICKED!!! Sn=n/2 (a + a+(n-1)d), ie the equation that they used in that example, is the same as writing it as Sn=n/2(a+l) isn't it? Because l, for the last term, is given by a + (n-1)d. I'm right in realising this aren't I?

    Also, when you say "an" for the last term in that equation, is that the same as saying a31, or u31 in this instance (for the 31st term), because n=31 in this case? Am I right in thinking this too?
    Thanks
    Correct

    And also correct. If you had n = 1 then the first and last term would be a1. If n=2 then you have a1 and a2... n= 31 you have ... up to a31.
    Offline

    3
    ReputationRep:
    (Original post by blobbybill)
    AHH, SOMETHING JUST CLICKED!!! Sn=n/2 (a + a+(n-1)d), ie the equation that they used in that example, is the same as writing it as Sn=n/2(a+l) isn't it? Because l, for the last term, is given by a + (n-1)d. I'm right in realising this aren't I?

    Also, when you say "an" for the last term in that equation, is that the same as saying a31, or u31 in this instance (for the 31st term), because n=31 in this case? Am I right in thinking this too?
    Thanks
    Perfect - exactly this!
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: October 12, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.