Arithmetic Series

When given an arithmetic sequence, you might be asked to find the sum of the sequence up to a certain term. This is called a series, and more specifically an arithmetic series. Let's say you are given the arithmetic sequence: 

{1, 5, 9, 13, ... 73}

the series version of this would be:

1+5+9+13 + ... + 73

As you've probably understood, summing a series by continuously adding the numbers together takes a lot of time and effort. So instead, we a couple of formulas we can use.

The first formula is:

Sn = (n/2) × (u1+un)

where Sn is the sum of the first n terms, n is the amount of terms being summed, u1 is the first term and un is the value of the final term of the series. The parenthesis here is just to show that you're not dividing n by 2*(u1+un), but rather the quotient of n/2 multiplied by (u1+un). As an example, we can try to sum the first four numbers of the sequence:

{1, 5, 9 ,13, …, 73}

For that we do:

S4 = (4/2) × (1 + 13)

S4 = 2 × 14

S4 = 28

Now, often you will not be given the sequence like that, where you get every number including un which is the last term in you series. un is often not included, but since we already have a formula for un, un = u1 + (n-1)*d ,we can substitute that into the first formula, to get a new one: 

Sn = (n/2) × [2 × u1 + (n-1)d]

Where Sn is the sum of n terms, n is the final term in the series (so the last term you add), u1 is the first term of the series and d is the difference. The brackets [] just means that it is a combination of multiple formulas. They operate the same way as parenthesis. As an example: given an arithmetic sequence where the 1st term is 6 and the the difference is 8. What is the sum of the first 14 terms? To solve this we do:

S14 = (14/2) × [2 × 6 + (14-1) × 8]

S14 = 7 × (12 + 13 × 8)

S14 = 7 × (12+104)

S14 =7 × 116

S14 =812

Here, we know that u1 is 9 and that S16 is 384. We don't have the final number in the series and we want to find the difference, so we the second formula. By plugging in what we know, we get:

384 = (16/2) × [2 × 9 + (16-1) × d]

We continue by solving the parentheses, and get

384 = 8 × (18 +15d)

384 = 144 + 120d

384-144 =120d

240 = 120d

d = 2

So the differnce is two. From here you can, if you  want, find the last term in series.