Arithmetic & Geometric Series

Q: What is the difference between a sequence and a series?

A sequence lists terms in order, such as $3,\; 7,\; 11$. A series adds those terms: $3 + 7 + 11$. Same numbers, and the series is the total.

Q: How many terms are in $\displaystyle\sum_{k=3}^{9}$?

Seven. Count with upper limit minus lower limit plus $1$, so $9 - 3 + 1 = 7$. Subtracting alone gives $6$ and quietly drops a term.

A sequence lists numbers in order: $3,\; 7,\; 11,\; 15,\; \ldots$ . A series takes that same list and adds it up: $3 + 7 + 11 + 15 + \cdots$ . One operation is the entire distinction, and getting it straight early saves a lot of confusion later, because the two words look alike and the formulas do not.

Adding ten or fifty terms by hand is slow and easy to get wrong, so each type of sequence comes with a sum formula. An arithmetic series, where every step adds the same number, uses $S_n = \dfrac{n}{2}(a_1 + a_n)$ . A geometric series, where every step multiplies by the same number, uses $S_n = \dfrac{a_1(1 - r^n)}{1 - r}$ . Sigma notation is the shorthand that writes either one in a single line.

Arithmetic or geometric?

Look at what happens from one term to the next. If the same number is added every step, the sequence is arithmetic and that number is the common difference $d$ . Find it by subtracting consecutive terms: in $5,\; 9,\; 13,\; \ldots$ , $9 - 5 = 4$ and $13 - 9 = 4$ , so $d = 4$ . The $n$ th term is $a_n = a_1 + (n - 1)d$ .

If instead each term is the previous one multiplied by the same number, the sequence is geometric and that number is the common ratio $r$ , found by dividing consecutive terms. In $2,\; 6,\; 18,\; \ldots$ , $6 \div 2 = 3$ , so $r = 3$ and $a_n = a_1 r^{\,n-1}$ . Deciding which type you have is the first move on every problem, because it picks the sum formula for you.

Adding an arithmetic series

The arithmetic sum formula is $S_n = \dfrac{n}{2}(a_1 + a_n)$ : the number of terms times the average of the first and last terms. The reason is a pairing trick. Write $1 + 2 + \cdots + 99 + 100$ forwards and backwards, stack them, and every column adds to $101$ . There are $100$ columns, so twice the sum is $100 \cdot 101$ and the sum is $5{,}050$ .

Notice the formula needs the last term, so most problems have two stages: find $a_n$ with $a_n = a_1 + (n - 1)d$ , then add with $S_n$ . The classic slip is using $n$ where the formula wants $n - 1$ . Getting from the first term to the tenth takes nine steps, not ten.

Sigma notation

The symbol $\displaystyle\sum$ is an instruction to add. In $\displaystyle\sum_{k=1}^{4} (2k + 1)$ , the letter $k$ is the index, the number below is where the index starts, the number above is where it stops, and the expression on the right is the rule. Plug in $k = 1, 2, 3, 4$ and add the four results: $3 + 5 + 7 + 9 = 24$ .

Counting terms matters when the index does not start at $1$ : the count is upper limit minus lower limit, plus $1$ . To write a sum in sigma notation, work in reverse. $4 + 8 + 12 + 16 + 20$ has five terms that all come from multiplying by $4$ , so the rule is $4k$ and the sum is $\displaystyle\sum_{k=1}^{5} 4k$ .

Worked examples

Example 1: sum of an arithmetic series

Find the sum of the first $10$ terms of $3 + 7 + 11 + \cdots$

Find the common difference

d = 7 - 3 = 4

Find the tenth term with

a_n = a_1 + (n-1)d

a_{10} = 3 + 9 \cdot 4 = 39

Apply the sum formula

S_{10} = \dfrac{10}{2}(3 + 39)

Compute

5 \cdot 42 = 210

Answer: $S_{10} = 210$

Example 2: evaluating sigma notation

Evaluate $\displaystyle\sum_{k=1}^{5} (3k - 2)$ .

Plug in

k = 1

through

k = 5

1,\; 4,\; 7,\; 10,\; 13

It is arithmetic with

d = 3

, so use the sum formula

S_5 = \dfrac{5}{2}(1 + 13)

Compute

\dfrac{5}{2} \cdot 14 = 35

Answer: $35$

Example 3: writing sigma notation

Write $4 + 8 + 12 + 16 + 20$ in sigma notation.

Count the terms

n = 5

Each term is

4

times its position

4 \cdot 1,\; 4 \cdot 2,\; 4 \cdot 3,\; \ldots

Write the rule with index

k

from

1

5

\displaystyle\sum_{k=1}^{5} 4k

Answer: $\displaystyle\sum_{k=1}^{5} 4k$

Example 4: a geometric series

Find the sum $2 + 6 + 18 + 54$ .

Each term is

3

times the one before, so it is geometric

r = 3,\; a_1 = 2,\; n = 4

Use the geometric sum formula

S_4 = \dfrac{2(1 - 3^4)}{1 - 3}

Evaluate the power

\dfrac{2(1 - 81)}{-2}

Simplify

\dfrac{2(-80)}{-2} = 80

Answer: $80$

Try one yourself

Common questions

What is the difference between a sequence and a series?

A sequence lists terms in order, such as $3,\; 7,\; 11$ . A series adds those terms: $3 + 7 + 11$ . Same numbers, and the series is the total.

Do I need the last term to add an arithmetic series?

Yes, $S_n = \dfrac{n}{2}(a_1 + a_n)$ asks for it. If the problem only gives the first term and the common difference, find $a_n = a_1 + (n-1)d$ first, then add.

How many terms are in

\displaystyle\sum_{k=3}^{9}

Seven. Count with upper limit minus lower limit plus $1$ , so $9 - 3 + 1 = 7$ . Subtracting alone gives $6$ and quietly drops a term.

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