First Formula

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

S₁ = a₁

The sum of the first 2 terms is:

S₂ = a₁ + ( a₁ + d) = 2a₁ + d

Of course, S₂ also equals:

S₂ = a₁ + a₂

We can write this answer as 2/2[a₁ + a₂]. (You'll see why we do this soon). Putting the above 2 results together, we get:

S₂ = 2a₁ + d = 2/2[a₁ + a₂]

Now, the sum of the first 3 terms is:

S₃ = a₁ + ( a₁ + d) +  ( a₁ + 2d) = 3a₁ + 3d

And we examine some more relationships:

a₃ = a₁ + 2d

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

and

a₁ + a₃ = a₁ + ( a₁ + 2d) = 2a₁ + 2d

And now, look at these two results:

S₃ = 3a₁ + 3d

a₁ + a₃ = 2a₁ + 2d

Multiplying the last line throughout by 3/2 gives

3/2(a₁ + a₃) = 3/2(2a₁ + 2d) = 3a₁ + 3d = S₃

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

S₃ = 3a₁ + 3d = 3/2[a₁ + a₃]

The sum of the first 4 terms is:

S₄ = a₁ + ( a₁ + d) +  ( a₁ + 2d) + ( a₁ + 3d) = 4a₁ + 6d

Now, a₄ = a₁ + 3d and

a₁ + a₄ = a₁ + ( a₁ + 3d) = 2a₁ + 3d

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

S₄ = 4a₁ + 6d = 4/2[a₁ + a₄]

In general, the sum to n terms is:

S_n = n/2[a₁ + a_n]

Second Formula

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

S₃ = a₁ + ( a₁ + d) +  ( a₁ + 2d) = 3a₁ + 3d

We can write the answer as 3/2[2a₁ + (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:

S₄ = 4a₁ + 6d

We can write the answer as 4/2[2a₁ + (4 − 1)d]

The sum of the first 5 terms is:

S₅ = 5a₁ + 10d which can be written 5/2[2a₁ + (5 − 1)d]

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

S_n = n/2[2a₁ + (n − 1)d]