I only know it in the form of Sn = n/2[2a+(n1)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+(311)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!
Can someone explain the equation: S_n = n*(a_1 + a_n)/2 to me please?
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 1
 12102016 20:40
Post rating:2 
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 2
 12102016 20:45
(Original post by blobbybill)
I only know it in the form of Sn = n/2[2a+(n1)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+(311)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!
So if you split up 2a + (n1)d into a + (a + (n1)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.Post rating:2 
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 3
 12102016 20:50
(Original post by blobbybill)
I only know it in the form of Sn = n/2[2a+(n1)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+(311)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!
What they've done is they've taken the formula (which is the same as yours, I've just written and put some brackets). Notice that what is inside the curly brackets is the nth term of an arithmetic sequence and is the first term. This thus means that the sum can also be expression as .
That is what they have started with  and they have then realised, they don't have , 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 knowPost rating:3 
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 4
 12102016 20:52
(Original post by SeanFM)
Remember that the nth term of the sequence is a + (n1)d.
So if you split up 2a + (n1)d into a + (a + (n1)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.
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+(n1)d], should I be worried? 
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 5
 12102016 20:55
(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 + n1 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+(n1)d], should I be worried?
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. 
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 6
 12102016 20:56
(Original post by kingaaran)
refers to the nth term of a sequence, and so refers to the first term in a sequence  this matches with your , which you define as the first term in the sequence.
What they've done is they've taken the formula (which is the same as yours, I've just written and put some brackets). Notice that what is inside the curly brackets is the nth term of an arithmetic sequence and is the first term. This thus means that the sum can also be expression as .
That is what they have started with  and they have then realised, they don't have , 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
Also, just to clarify, a1 always equals a doesn't it, because it is the first term? 
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 7
 12102016 21:02
(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+(n1)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 + (n1)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 
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 8
 12102016 21:04
(Original post by blobbybill)
AHH, SOMETHING JUST CLICKED!!! Sn=n/2 (a + a+(n1)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 + (n1)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
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. 
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 9
 12102016 21:11
(Original post by blobbybill)
AHH, SOMETHING JUST CLICKED!!! Sn=n/2 (a + a+(n1)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 + (n1)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
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Updated: October 12, 2016
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