# Sum of Arithmetic Progression

## Theorem

Let $\sequence {a_k}$ be an arithmetic progression defined as:

$a_k = a + k d$ for $n = 0, 1, 2, \ldots, n - 1$

Then its closed-form expression is:

 $\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}$ $=$ $\displaystyle n \paren {a + \frac {n - 1} 2 d}$ $\displaystyle$ $=$ $\displaystyle \frac {n \paren {a + l} } 2$ where $l$ is the last term of $\sequence {a_k}$

## Proof

We have that:

$\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$

Then:

 $\displaystyle 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}$ $=$ $\displaystyle 2 \paren {a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d} }$ $\displaystyle$ $=$ $\displaystyle \paren {a + \paren {a + d} + \dotsb + \paren {a + \paren {n - 1} d} }$ $\displaystyle$ $=$ $\, \displaystyle + \,$ $\displaystyle \paren {\paren {a + \paren {n - 1} d} + \paren {a + \paren {n - 2} d} + \dotsb + \paren {a + d} + a}$ $\displaystyle$ $=$ $\displaystyle \paren {2 a + \paren {n - 1} d}_1 + \paren {2 a + \paren {n - 1} d}_2 + \dotsb + \paren {2 a + \paren {n - 1} d}_n$ $\displaystyle$ $=$ $\displaystyle n \paren {2 a + \paren {n - 1} d}$

So:

 $\displaystyle 2 \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}$ $=$ $\displaystyle n \paren {2 a + \paren {n - 1} d}$ $\displaystyle \leadsto \ \$ $\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d}$ $=$ $\displaystyle \frac {n \paren {2 a + \paren {n - 1} d} } 2$ $\displaystyle$ $=$ $\displaystyle \frac {n \paren {a + l} } 2$ Definition of Last Term $l$

Hence the result.

$\blacksquare$

## Examples

### Sum of $j$ from $m$ to $n$

 $\displaystyle \sum_{j \mathop = m}^n j$ $=$ $\displaystyle m \left({n - m + 1}\right) + \frac 1 2 \left({n - m}\right) \left({n - m + 1}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {n \left({n + 1}\right)} 2 - \frac {\left({m - 1}\right) m} 2$

### Sum of $i + k \left({2 + 2 i}\right)$

Let $A_n$ be the arithmetic progression of $n$ terms defined as:

 $\displaystyle A_n$ $=$ $\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a_0 + \paren {2 + 2 i} k}$ $\displaystyle$ $=$ $\displaystyle i + \paren {2 + 3 i} + \paren {4 + 5 i} + \paren {6 + 7 i} + \dotsb + \paren {2 n - 2 + \paren {2 n - 1} i}$

Then:

$A_n = n \paren {n - 1} + n^2 i$

## Historical Note

Doubt has recently been cast on the accuracy of the tale about how Carl Friedrich Gauss supposedly discovered this technique at the age of 8.

## Linguistic Note

In the context of an arithmetic progression or arithmetic-geometric progression, the word arithmetic is pronounced with the stress on the first and third syllables: a-rith-me-tic, rather than on the second syllable: a-rith-me-tic.

This is because the word is being used in its adjectival form.