Notice that each term differs from its preceding and succeeding term by a constant difference of

**– this type of series can be represented as such:**

*3*Where

**is the**

*a***, and**

*first term***is the**

*d***. Notice that also that the**

*common difference***is given by**

*n-th term***. This is known as an**

*a + (n – 1)d***, or an**

*Arithmetic Progression***.**

*Arithmetic Sum*Now, can we find a

**? That means, can we find a general formula for the following summation of the first term to the last term:**

*sum to n terms*Well, of course we can! Let us see how; first, let us consider the sum to

**terms on its own:**

*n*We started off by writing the sum from the first term

**– but we could have done so by starting from the last term**

*a***as well, and thus the sum to**

*A***terms written backwards from the last term**

*n***is simply:**

*A*Now let’s put them side by side and add them up and let’s see if you notice something:

What do you notice? Well, notice that the ‘

**’s and the ‘**

*d***’s all cancel one another out, leaving us with:**

*-d*The

**is there because we have**

*n***terms in each sum, and adding up two sums should give us**

*n***of**

*n***and**

*A***of**

*n***. Now, what is the last term**

*a***in terms of the first term**

*A***? We’ve already said that we want the sum to**

*a***terms, and thus the last term**

*n***must be the**

*A***, given by:**

*n-th term*In which case we make a substitution:

And therefore the sum to

**terms follows nicely:**

*n*Sweet eh? :)

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