First Formula

The sum of the first term is, well, simply the first term:

`S_1 = a_1`

The sum of the first 2 terms is:

`S_2 = a_1 + (a_1 + d) = 2a_1 + d`

Of course, `S_2` also equals:

`S_2 = a_1 + a_2`

We can write this answer as `2/2[a_1 + a_2]. `

(You'll see why we do this soon).

Putting the above 2 results together, we get:

`S_2 = 2a_1 + d = 2/2[a_1 + a_2]`


Now, the sum of the first 3 terms is:

`S_3 = a_1 + ( a_1 + d) +  (a_1 + 2d)` ` = 3a_1 + 3d`

And we examine some more relationships:

`a_3 = a_1 + 2d `

(This follows from the definition of an arithmetic progression.)

and

`a_1 + a_3 = a_1 + (a_1 + 2d)` ` = 2a_1 + 2d`

And now, look at these two results:

`S_3 = 3a_1 + 3d`

`a_1 + a_3 = 2a_1 + 2d`

Multiplying the last line throughout by `3/2` gives

`3/2(a_1 + a_3) = 3/2(2a_1 + 2d)` ` = 3a_1 + 3d = S_3`

So we can write the answer for the sum of 3 terms as

`S_3 = 3a_1 + 3d` ` = 3/2[a_1 + a_3] `


The sum of the first 4 terms is:

`S_4 = a_1 + (a_1 + d) +` ` (a_1 + 2d) +` ` (a_1 + 3d)` ` = 4a_1 + 6d`

Now, `a_4 = a_1 + 3d` and

`a_1 + a_4 = a_1 + (a_1 + 3d)` ` = 2a_1 + 3d`

Using a similar idea to the step above, we can write the answer for the sum of 4 terms as

`S_4 = 4a_1 + 6d` ` = 4/2[a_1 + a_4]`


In general, the sum to n terms is:

`S_n = n/2[a_1 + a_n]`


Second Formula

From above, the sum of the first 3 terms is:

`S_3 = a_1 + ( a_1 + d) +  ( a_1 + 2d)` ` = 3a_1 + 3d`

We can write the answer as `3/2[2a_1 + (3-1)d]`

We'll see in a minute why we want to do this rather odd step.


The sum of the first 4 terms is:

`S_4 = 4a_1 + 6d`

We can write the answer as `4/2[2a_1 + (4 − 1)d]`


The sum of the first 5 terms is:

`S_5 = 5a_1 + 10d` which can be written `5/2[2a_1 + (5 − 1)d]`


Are you getting the idea? In general, the sum to n terms is

`S_n = n/2[2a_1 + (n − 1)d]`