# Sums of Arithmetic Sequences

## Key Questions

• To aid in teaching this, I'll use the following arithmetic sequence (technically, it's called a series if you're finding the sum):

Example A: $3 + 7 + 11 + 15 + 19 \ldots + {t}_{20}$

Example B: $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15$

To start, you should know the following equations:

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

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

Note: The first equation can only be used if you are given the last term (like in Example B). The second equation can be used with no restrictions.

Now, we'll find the sum of Example A, and because we don't know the last term , we have to use equation 2. Sub in all the known values: n = 20 (20 terms), a = 3 (first term is 3), and d = 4 (difference between terms is 4).

${S}_{20} = \left(\frac{20}{2}\right) \left(2 \left(3\right) + \left(4\right) \left(20 - 1\right)\right)$

Simplify: ${S}_{20} = \left(10\right) \left(6 + 76\right)$

${S}_{20} = \left(10\right) \left(82\right)$

${S}_{20} = 820$ $\to$ Therefore the sum of the series is 820!

Say you wanted to find the sum of Example B, where you know the last term, but don't know the number of terms. You would do the exact same process, but you would have to SOLVE for "n" (number of terms) first.

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

Then, sub in all known values. ${t}_{n} = 15$ (last term of the sequence), a = 1 (first term), d = 2 (difference between terms) and solve for n like so:

$15 = 1 + 2 \left(n - 1\right)$

Expand and simplify:

$15 = 1 + 2 n - 2$

$16 = 2 n$

$n = 8$ $\to$ Therefore, the series has 8 terms. Now, you can use this information to substitute known values into the first equation since you have the first/last terms, and the number of terms:

${S}_{8} = \frac{\left(8\right) \left(1 + 15\right)}{2}$

${S}_{8} = 64$ $\to$ Therefore, the sum of the series is 64!

Sorry this was so lengthy! I just wanted to cover everything you may see on a test/quiz or in-class! Hopefully you've understood all this and hopefully I was of some help! :)

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

#### 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

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