Derivation of 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 \large{d} is the common difference.

first term = \large{a}

second term = \large{a+d}

third term = \large{a+2d}

nth term or last term = \large{a + \left( {n - 1} \right)d}

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 {2a+(n-1)d}.

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

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

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

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

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 2a as \color{blue}a+a.

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

Notice that a+(n-1)d 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 a+(n-1)d by \large{a_n}. This is the standard form of the Arithmetic Series Formula.

Sn=n(a+an)/2

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