# What is the formula for the sum of an arithmetic sequence?

Jun 27, 2018

${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

#### Explanation:

Suppose we have an AP with first term $a$ and common difference $d$, then we can write the sum of the first $n$ terms as:

${S}_{n} = a + \left(a + d\right) + \left(a + 2 d\right) + \ldots + \left(a + \left(n - 1\right) d\right)$

Writing the same sum, but in reverse, we get:

${S}_{n} = \left(a + \left(n - 1\right) d\right) + \ldots \left(a + 2 d\right) + \left(a + d\right) + a$

If we add both of these we get:

$2 {S}_{n} = \left(2 a + \left(n - 1\right) d\right) + \left(2 a + \left(n - 1\right) d\right) + \ldots + \left(2 a + \left(n - 1\right) d\right)$
$\setminus \setminus \setminus \setminus \setminus = n \left(2 a + \left(n - 1\right) d\right)$

Leading to the standard AP summation formula

${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

Jun 27, 2018

The sum of an arithmetic sequence is given by ${S}_{n} = {\sum}_{i = 1}^{n} {a}_{i} = \frac{n}{2} \left({a}_{1} + {a}_{n}\right)$.

#### Explanation:

Let ${S}_{n}$ be the sum of the arithmetic sequence and the ${n}^{\text{th}}$ term of the sequence be ${a}_{n} = {a}_{1} + d \left(n - 1\right)$ where $d$ is the common difference.
${S}_{n} = {a}_{1} + \left({a}_{1} + d\right) + \left({a}_{1} + 2 d\right) + \ldots + \left({a}_{n} - d\right) + {a}_{n}$
${S}_{n} = {a}_{n} + \left({a}_{n} - d\right) + \left({a}_{n} - 2 d\right) + \ldots + \left({a}_{1} + d\right) + {a}_{1}$
Adding up the two equations term by term we get
$2 {S}_{n} = \left({a}_{1} + {a}_{n}\right) + \left({a}_{1} + {a}_{n}\right) + \ldots + \left({a}_{1} + {a}_{n}\right) + \left({a}_{1} + {a}_{n}\right)$
*Notice how the $d - d$ and $2 d - 2 d$ all cancel out
Since we have $n$ terms of $\left({a}_{1} + {a}_{n}\right)$ on the right side, we can re-write the equation as
$2 {S}_{n} = n \left({a}_{1} + {a}_{n}\right)$
$\therefore {S}_{n} = \frac{n}{2} \left({a}_{1} + {a}_{n}\right)$