Series

1.1: Arithmetic Series and Sequences

Definition (Arithmetic Series and Sequences)

An arithmetic series is $a_{1}+a_{2}+a_{3}\ldots$ where $a_{i}-a_{i-1}=d$ for some constant $d$ and all $i\ge 2$ . Note that an arithmetic series refers to the sum while an arithmetic sequence refers to the ordered list of terms.¹

Using this definition, we can now derive some formulas for each term and the series.

Theorem (Formula For Term)

If $a_{i}$ is $i$ th term of an arithmetic sequence with common difference $d$ , then $a_{i}=a_{1}+(i-1)d$ .

Now, we'll present some common formulas for arithmetic series and sequences. These all easily follow from the previous formula but it's useful to know them to quickly manipulate arithmetic series.

Theorem (Arithmetic Average)

In an arithmetic series $a_{i}$ , $a_{n}+a_{m}=2a_{\frac{n+m}{2}}$ if $n+m$ is even.

In other words, the arithmetic average of two terms is the term with an index that is the arithmetic average of the two other terms' indexes. Visually, the term in the very middle of the two terms is the arithmetic average.

Proof

We have $a_{n}=a_{1}+(n-1)d$ and $a_{m}=a_{1}+(m-1)d$ . Then, $a_{n}+a_{m}=2a_{1}+(n+m-2)d=2(a_{1}+(\frac{n+m}{2}-1)d)=2a_{\frac{n+m}{2}}$

Theorem (Arithmetic Sequence and Series)

For an arithmetic sequence $a_{i}$ , $a_{1}+a_{2}\ldots + a_{n}=\frac{a_{1}+a_{n}}{2}\cdot n$ .

Proof

Note that we can pair up $a_{i}$ and $a_{n+1-i}$ to get $a_{1}+(i-1)d+a_{1}+(n-i)d=2a_{1}+(n-1)d$ . There is a special case that if $n$ is odd and $i=\frac{n+1}{2}$ , we can't pair up that term with itself. However, $a_{\frac{n+1}{2}}=a_{1}+\frac{n-1}{2}d$ so it still averages out as $\frac{1}{2}$ of a pair.

Then, we have $\frac{n}{2}$ pairs (if $n$ is odd, we have a "half" pair which is just the term $a_{\frac{1+n}{2}}=\frac{a_{1}+a_{n}}{2}$ ). So, our total sum is $\frac{n}{2}\cdot (a_{1}+a_{n})=\frac{a_{1}+a_{n}}{2}\cdot n$ .

Theorem (Arithmetic Series)

If the common difference is $d$ in arithmetic sequence $a_{i}$ , $a_{1}+a_{2}\ldots + a_{n}=n\cdot a_{1}+d(\frac{(n-1)(n)}{2})$ .

Proof

We use the previous theorem $a_{1}+a_{2}\ldots + a_{n}$ $=\frac{a_{1}+a_{n}}{2}\cdot n$ $=\frac{a_{1}+a_{1}+(n-1)d}{2}\cdot n$ $=a_{1}\cdot n + d(\frac{(n-1)(n)}{2})$

Most of time, problems with arithmetic series require only simple algebraic manipulation to solve.

Example

(NanoMath Fall Meet 2020)

Let $a_0, a_1, a_2, \ldots$ be an arithmetic sequence of positive integers. If $a_0 + a_1 + \cdots + a_{10} = 209$ and $a_{a_{0}} + a_{a_{1}} + \cdots + a_{a_{10}} = 671$ , then find $a_0$ .

Solution

Using our series formula on $a_0 + a_1 + \cdots + a_{10} = 209$ gives that $11\cdot \frac{a_{0}+a_{10}}{2}=209\implies a_{0}+a_{10}=38\implies 2a_{0}+10d=38$ .

Now, $a_{a_{0}}+a_{a_{1}} + \cdots + a_{a_{10}} = 671$ $11a_{0}+(a_{0}+a_{1}\cdots + a_{10})d=671$ $11a_{0}+209d=671$

Now, using the first equation, $11a_{0}+55d=209$ . Then, subtracting from the second equation, $154d=462\implies d =3$ . So, $a_{0}={4}$ .

Sometimes, an arithmetic sequence or series starts with index $0$ , making it zero-indexed. The same formulas still apply but you will have to adjust the indexes accordingly. ↩

Series

Footnotes