# 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

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

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

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

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

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 $\color{blue}a+a$.

Notice that $a+(n-1)d$ is actually the Arithmetic Sequence Formula.

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.

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