How to Derive the Arithmetic Series Formula

In this lesson, we are going to derive the Arithmetic Series Formula. This is a good way to appreciate why the formula works.

Suppose we have the following terms where [latex]\large{d}[/latex] is the common difference.

first term = [latex]\large{a}[/latex]

second term = [latex]\large{a+d}[/latex]

third term = [latex]\large{a+2d}[/latex]

nth term or last term = [latex]\large{a + \left( {n – 1} \right)d}[/latex]

If we add the terms together, we get

Sn=a+(a+d)=(a+2d)+...+a+(n-1)d

Let’s add the terms in reverse or descending order.

Sn=a+(n-1)d+a+(n-2)d+a+(n-3)d+...+a

Here’s the “trick”. Now, we sum up the two arithmetic series above – the ones with ascending and descending terms. Notice that the sum of each column is always [latex]{2a+(n-1)d}[/latex].

2S=2a+(n-1)d+2a+(n-1)d+2a+(n-1)d+...+2a+(n-1)d

Notice that there are [latex]\large{n}[/latex] numbers of [latex]{2a+(n-1)d}[/latex] on the right side of the equation. Therefore, we can rewrite the equation above as

2Sn=n[2a+(n-1)d}

Finally, solve [latex]\large{S_n}[/latex] by dividing both sides of the equation by [latex]2[/latex].

Sn=n/2[2a+(n-1)d]

The formula above is the expanded version of the arithmetic series formula. We can manipulate it to make it a bit more compact.

Let’s rewrite [latex]2a[/latex] as [latex]\color{blue}a+a[/latex].

Sn=n/2[a+a+(n-1)d]

Notice that [latex]a+(n-1)d[/latex] is actually the Arithmetic Sequence Formula.

Sn=(n/2)[a+a+(n-1)d]
an=a+(n-1)d

Simplify the formula by substituting the expression [latex]a+(n-1)d[/latex] by [latex]\large{a_n}[/latex]. This is the standard form of the Arithmetic Series Formula.

Sn=n(a+an)/2

You may also be interested in these related math lessons or tutorials:

Arithmetic Series Formula

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Arithmetic Sequence Formula

Arithmetic Sequence Formula Practice Problems