Sequences and Series

Explore patterns in lists of numbers and their sums

Have you ever noticed how your savings account grows month after month, or counted the rows of seats in a theater where each row has a few more seats than the one in front? Maybe you have watched a bouncing ball lose height with each bounce, or wondered how much you would earn if your salary increased by a fixed percentage every year. If so, you have already been thinking about sequences and series without even knowing it.

A sequence is simply an ordered list of numbers that follows some pattern. A series is what you get when you add up the terms in a sequence. These concepts are everywhere: in loan payments, population growth, computer algorithms, and even in the spiraling patterns of sunflower seeds. By the end of this lesson, you will not only understand how to recognize and work with sequences, but you will also have powerful formulas for finding any term you want and for adding up as many terms as you need.

Core Concepts

What Is a Sequence?

A sequence is an ordered list of numbers that follows a specific pattern or rule. Each number in the sequence is called a term, and terms are usually labeled by their position: the first term, second term, third term, and so on.

For example: $$2, 4, 6, 8, 10, \ldots$$

This is a sequence where each term is 2 more than the previous one. The dots (called an ellipsis) indicate that the pattern continues.

Here is another sequence: $$3, 6, 12, 24, 48, \ldots$$

In this case, each term is twice the previous one.

The key idea is that sequences are ordered. The number 6 appearing second in a sequence is different from 6 appearing fifth. Position matters.

Explicit vs. Recursive Formulas

There are two main ways to describe how a sequence is built:

Explicit formulas give you a direct way to calculate any term based on its position number. If you want the 100th term, you just plug 100 into the formula. No need to know any other terms.

For the sequence 2, 4, 6, 8, 10, …, an explicit formula is: $$a_n = 2n$$

This tells you that the nth term equals 2 times n. Want the 50th term? Just calculate $a_{50} = 2(50) = 100$.

Recursive formulas define each term based on one or more previous terms. You need a starting point (usually the first term) and a rule for getting from one term to the next.

For the same sequence 2, 4, 6, 8, 10, …, a recursive formula would be: $$a_1 = 2, \quad a_n = a_{n-1} + 2$$

This says “start with 2, and each term is 2 more than the previous term.”

Both formulas describe the same sequence, just from different perspectives. Explicit formulas are great for jumping straight to any term. Recursive formulas are great for understanding how a sequence grows step by step.

Arithmetic Sequences

An arithmetic sequence is one where you add the same number to get from one term to the next. This constant amount is called the common difference, usually denoted by $d$.

Examples of arithmetic sequences:

  • $5, 8, 11, 14, 17, \ldots$ (common difference $d = 3$)
  • $20, 15, 10, 5, 0, \ldots$ (common difference $d = -5$)
  • $1, 1, 1, 1, 1, \ldots$ (common difference $d = 0$)

The explicit formula for arithmetic sequences:

$$a_n = a_1 + (n-1)d$$

Where:

  • $a_n$ is the nth term
  • $a_1$ is the first term
  • $n$ is the position number
  • $d$ is the common difference

Why does this formula work? Think about it: to get from the first term to the second, you add $d$ once. To get to the third term, you add $d$ twice. To get to the nth term, you add $d$ exactly $(n-1)$ times.

Geometric Sequences

A geometric sequence is one where you multiply by the same number to get from one term to the next. This constant multiplier is called the common ratio, usually denoted by $r$.

Examples of geometric sequences:

  • $2, 6, 18, 54, 162, \ldots$ (common ratio $r = 3$)
  • $100, 50, 25, 12.5, \ldots$ (common ratio $r = 0.5$)
  • $5, -10, 20, -40, 80, \ldots$ (common ratio $r = -2$)

The explicit formula for geometric sequences:

$$a_n = a_1 \cdot r^{n-1}$$

Where:

  • $a_n$ is the nth term
  • $a_1$ is the first term
  • $r$ is the common ratio
  • $n$ is the position number

Why does this formula work? To get from the first term to the second, you multiply by $r$ once. To get to the third, you multiply by $r$ twice. To get to the nth term, you multiply by $r$ exactly $(n-1)$ times.

Arithmetic Series

A series is what you get when you add up terms of a sequence. An arithmetic series is the sum of terms in an arithmetic sequence.

If you want to add the first $n$ terms of an arithmetic sequence, you could add them one by one. But there is a much faster formula:

$$S_n = \frac{n}{2}(a_1 + a_n)$$

Or equivalently:

$$S_n = \frac{n}{2}(2a_1 + (n-1)d)$$

Where:

  • $S_n$ is the sum of the first $n$ terms
  • $n$ is the number of terms
  • $a_1$ is the first term
  • $a_n$ is the last term
  • $d$ is the common difference

Where does this formula come from? Legend has it that a young Carl Friedrich Gauss was asked to add the numbers from 1 to 100. While his classmates began tediously adding, Gauss noticed that if you pair the first term with the last (1 + 100 = 101), the second with the second-to-last (2 + 99 = 101), and so on, every pair sums to the same value. Since there are 50 such pairs, the total is $50 \times 101 = 5050$.

The formula $S_n = \frac{n}{2}(a_1 + a_n)$ captures this pairing idea: the average of the first and last terms, multiplied by the number of terms.

Geometric Series

A geometric series is the sum of terms in a geometric sequence. The formula for the sum of the first $n$ terms is:

$$S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad \text{(when } r \neq 1\text{)}$$

Or equivalently:

$$S_n = a_1 \cdot \frac{r^n - 1}{r - 1}$$

Where:

  • $S_n$ is the sum of the first $n$ terms
  • $a_1$ is the first term
  • $r$ is the common ratio
  • $n$ is the number of terms

If $r = 1$, then every term is the same, so $S_n = n \cdot a_1$.

Infinite Geometric Series

Here is something remarkable: sometimes you can add up infinitely many numbers and get a finite answer.

For a geometric series with $|r| < 1$ (that is, when $r$ is between $-1$ and $1$), the terms get smaller and smaller, approaching zero. As you add more and more terms, the sum approaches a specific value:

$$S = \frac{a_1}{1 - r} \quad \text{(when } |r| < 1\text{)}$$

This is called the sum of an infinite geometric series.

For example, the series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$ has $a_1 = 1$ and $r = \frac{1}{2}$. Its sum is:

$$S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$$

If you keep adding halves, quarters, eighths, and so on forever, you get exactly 2.

Why does this work? When $|r| < 1$, the term $r^n$ gets closer and closer to 0 as $n$ increases. In the finite sum formula $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$, as $n$ approaches infinity, $r^n$ approaches 0, leaving us with $S = \frac{a_1}{1 - r}$.

When $|r| \geq 1$, the terms do not shrink to zero, so the sum grows without bound. We say the series diverges.

Sigma Notation

When mathematicians want to write “add up a bunch of terms,” they use sigma notation, named after the Greek letter sigma ($\Sigma$).

$$\sum_{k=1}^{n} a_k$$

This reads as “the sum of $a_k$ from $k = 1$ to $n$.”

  • The letter below the sigma ($k = 1$) tells you where to start counting
  • The number above the sigma ($n$) tells you where to stop
  • The expression after the sigma ($a_k$) tells you what to add

For example:

$$\sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$$

Sigma notation is compact and precise. It is especially useful when dealing with patterns that would take many lines to write out otherwise.

Notation and Terminology

Symbol/Term Meaning
$a_n$ The nth term of a sequence
$a_1$ The first term of a sequence
$d$ Common difference (arithmetic sequences)
$r$ Common ratio (geometric sequences)
$S_n$ Sum of the first n terms (partial sum)
$S$ or $S_\infty$ Sum of an infinite series
$\sum$ Sigma (summation symbol)
$\sum_{k=1}^{n}$ Sum from k=1 to n
Explicit formula A formula that gives $a_n$ directly in terms of $n$
Recursive formula A formula that defines $a_n$ in terms of previous terms
Converge When an infinite series has a finite sum
Diverge When an infinite series does not have a finite sum

Examples

Example 1: Identifying Sequence Types

Determine whether each sequence is arithmetic, geometric, or neither.

(a) $4, 7, 10, 13, 16, \ldots$

(b) $3, 9, 27, 81, \ldots$

(c) $1, 4, 9, 16, 25, \ldots$

Solution:

(a) Check for a common difference: $7 - 4 = 3$, $10 - 7 = 3$, $13 - 10 = 3$, $16 - 13 = 3$. Each term is 3 more than the previous. This is arithmetic with $d = 3$.

(b) Check for a common ratio: $\frac{9}{3} = 3$, $\frac{27}{9} = 3$, $\frac{81}{27} = 3$. Each term is 3 times the previous. This is geometric with $r = 3$.

(c) Check for a common difference: $4 - 1 = 3$, $9 - 4 = 5$, $16 - 9 = 7$. The differences are not constant. Check for a common ratio: $\frac{4}{1} = 4$, $\frac{9}{4} = 2.25$. The ratios are not constant. This is neither. (It is the sequence of perfect squares: $1^2, 2^2, 3^2, 4^2, 5^2, \ldots$)

Example 2: Finding Terms of an Arithmetic Sequence

An arithmetic sequence has first term $a_1 = 5$ and common difference $d = 4$. Find the 20th term.

Solution:

Use the explicit formula: $a_n = a_1 + (n-1)d$

$$a_{20} = 5 + (20-1)(4)$$ $$a_{20} = 5 + (19)(4)$$ $$a_{20} = 5 + 76$$ $$a_{20} = 81$$

Answer: The 20th term is 81.

Example 3: Finding Terms of a Geometric Sequence

A geometric sequence has first term $a_1 = 3$ and common ratio $r = 2$. Find the 8th term.

Solution:

Use the explicit formula: $a_n = a_1 \cdot r^{n-1}$

$$a_8 = 3 \cdot 2^{8-1}$$ $$a_8 = 3 \cdot 2^7$$ $$a_8 = 3 \cdot 128$$ $$a_8 = 384$$

Answer: The 8th term is 384.

Example 4: Sum of an Arithmetic Series

Find the sum of the first 50 positive even numbers: $2 + 4 + 6 + \ldots + 100$.

Solution:

This is an arithmetic series with:

  • First term: $a_1 = 2$
  • Last term: $a_{50} = 100$
  • Number of terms: $n = 50$

Use the formula: $S_n = \frac{n}{2}(a_1 + a_n)$

$$S_{50} = \frac{50}{2}(2 + 100)$$ $$S_{50} = 25 \cdot 102$$ $$S_{50} = 2550$$

Answer: The sum is 2,550.

Example 5: Sum of a Geometric Series

Find the sum: $5 + 10 + 20 + 40 + \ldots$ for the first 10 terms.

Solution:

This is a geometric series with:

  • First term: $a_1 = 5$
  • Common ratio: $r = 2$
  • Number of terms: $n = 10$

Use the formula: $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$

$$S_{10} = 5 \cdot \frac{1 - 2^{10}}{1 - 2}$$ $$S_{10} = 5 \cdot \frac{1 - 1024}{-1}$$ $$S_{10} = 5 \cdot \frac{-1023}{-1}$$ $$S_{10} = 5 \cdot 1023$$ $$S_{10} = 5115$$

Answer: The sum is 5,115.

Example 6: Infinite Geometric Series

Find the sum of the infinite series: $12 + 4 + \frac{4}{3} + \frac{4}{9} + \ldots$

Solution:

First, identify the common ratio: $$r = \frac{4}{12} = \frac{1}{3}$$

Since $|r| = \frac{1}{3} < 1$, the series converges.

Use the formula: $S = \frac{a_1}{1 - r}$

$$S = \frac{12}{1 - \frac{1}{3}}$$ $$S = \frac{12}{\frac{2}{3}}$$ $$S = 12 \cdot \frac{3}{2}$$ $$S = 18$$

Answer: The sum is 18.

Example 7: Using Sigma Notation

Evaluate: $\displaystyle\sum_{k=1}^{6} (3k - 2)$

Solution:

Write out each term by substituting $k = 1, 2, 3, 4, 5, 6$:

When $k = 1$: $3(1) - 2 = 1$ When $k = 2$: $3(2) - 2 = 4$ When $k = 3$: $3(3) - 2 = 7$ When $k = 4$: $3(4) - 2 = 10$ When $k = 5$: $3(5) - 2 = 13$ When $k = 6$: $3(6) - 2 = 16$

This is an arithmetic sequence with $a_1 = 1$, $a_6 = 16$, and $n = 6$.

$$S_6 = \frac{6}{2}(1 + 16) = 3 \cdot 17 = 51$$

Answer: The sum is 51.

Example 8: Finding the Number of Terms

The arithmetic sequence $7, 11, 15, 19, \ldots$ has a sum of 945. How many terms are in this sum?

Solution:

We have:

  • First term: $a_1 = 7$
  • Common difference: $d = 4$
  • Sum: $S_n = 945$
  • Find: $n$

Use the formula: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$

$$945 = \frac{n}{2}(2(7) + (n-1)(4))$$ $$945 = \frac{n}{2}(14 + 4n - 4)$$ $$945 = \frac{n}{2}(10 + 4n)$$ $$1890 = n(10 + 4n)$$ $$1890 = 10n + 4n^2$$ $$4n^2 + 10n - 1890 = 0$$ $$2n^2 + 5n - 945 = 0$$

Using the quadratic formula: $$n = \frac{-5 \pm \sqrt{25 + 7560}}{4} = \frac{-5 \pm \sqrt{7585}}{4} = \frac{-5 \pm 87.1}{4}$$

$$n = \frac{-5 + 87.1}{4} \approx 20.5 \quad \text{or} \quad n = \frac{-5 - 87.1}{4} \approx -23$$

Since $n$ must be a positive integer, let us verify $n = 21$:

$$S_{21} = \frac{21}{2}(2(7) + (20)(4)) = \frac{21}{2}(14 + 80) = \frac{21}{2}(94) = 21 \times 47 = 987$$

That is too large. Let us try $n = 20$:

$$S_{20} = \frac{20}{2}(2(7) + (19)(4)) = 10(14 + 76) = 10(90) = 900$$

And $n = 21$ gives 987. Since 945 falls between these, let us recalculate more carefully.

Actually, the quadratic $2n^2 + 5n - 945 = 0$ factors: checking if $n = 21$ works: $2(441) + 5(21) = 882 + 105 = 987 \neq 945$.

Let us try $n = 20$: $2(400) + 5(20) = 800 + 100 = 900 \neq 945$.

The sum of 945 is not achievable with an integer number of terms. Let us re-examine the problem. If the question intended an integer solution, let us try $S_n = 945$ with a different approach.

If $n = 21$: $a_{21} = 7 + 20(4) = 87$, so $S_{21} = \frac{21}{2}(7 + 87) = \frac{21 \cdot 94}{2} = 987$.

Answer: The sum of 945 cannot be achieved with a whole number of terms. The sum of the first 20 terms is 900, and the sum of the first 21 terms is 987.

Key Properties

Properties of Arithmetic Sequences

  1. Any term can be found from any other term: If you know $a_m$, then $a_n = a_m + (n - m)d$

  2. The average of any two terms equals the middle term: In an arithmetic sequence, if $a$, $b$, $c$ are consecutive terms, then $b = \frac{a + c}{2}$

  3. The sum of terms equidistant from the ends is constant: $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \ldots$

Properties of Geometric Sequences

  1. Any term can be found from any other term: If you know $a_m$, then $a_n = a_m \cdot r^{n-m}$

  2. The product property: In a geometric sequence, if $a$, $b$, $c$ are consecutive terms, then $b^2 = ac$, which means $b = \sqrt{ac}$ (the geometric mean)

  3. Signs alternate when $r < 0$: The terms switch between positive and negative

  4. Growth behavior:

    • If $r > 1$, terms grow without bound
    • If $0 < r < 1$, terms shrink toward zero
    • If $-1 < r < 0$, terms alternate in sign and shrink toward zero
    • If $r < -1$, terms alternate in sign and grow in magnitude

Properties of Series

  1. Linearity of summation:

    • $\sum (a_k + b_k) = \sum a_k + \sum b_k$
    • $\sum c \cdot a_k = c \cdot \sum a_k$
  2. Useful formulas for common sums:

    • $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ (sum of first $n$ positive integers)
    • $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ (sum of squares)
    • $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$ (sum of cubes)

Real-World Applications

Saving Money

Suppose you save $100 in the first month, $110 in the second month, $120 in the third month, and so on, increasing by $10 each month. How much will you have saved after 2 years (24 months)?

This is an arithmetic series with $a_1 = 100$, $d = 10$, and $n = 24$.

$$S_{24} = \frac{24}{2}(2 \cdot 100 + 23 \cdot 10) = 12(200 + 230) = 12 \cdot 430 = \$5,160$$

Compound Interest and Investments

If you invest $1,000 and it grows by 5% each year, your balance forms a geometric sequence:

Year 1: $\$1,000$ Year 2: $\$1,000 \times 1.05 = \$1,050$ Year 3: $\$1,000 \times 1.05^2 = \$1,102.50$

After $n$ years: $\$1,000 \times 1.05^{n-1}$

Bouncing Ball

A ball is dropped from a height of 10 meters. Each bounce reaches 80% of the previous height. What is the total vertical distance the ball travels before coming to rest?

The ball falls 10 m, then bounces up $10 \times 0.8 = 8$ m and falls 8 m, then bounces up $8 \times 0.8 = 6.4$ m and falls 6.4 m, and so on.

Total distance = Initial drop + 2 $\times$ (sum of all bounce heights)

The bounce heights form an infinite geometric series with $a_1 = 8$ and $r = 0.8$: $$S = \frac{8}{1 - 0.8} = \frac{8}{0.2} = 40 \text{ m}$$

Total distance = $10 + 2 \times 40 = 90$ meters.

Depreciation

A car worth $30,000 depreciates by 15% each year. Its value forms a geometric sequence with $r = 0.85$.

After $n$ years: Value $= \$30,000 \times 0.85^{n-1}$

After 5 years: Value $= \$30,000 \times 0.85^4 \approx \$15,660$

Repeating Decimals

The repeating decimal $0.333\ldots$ can be written as an infinite geometric series:

$$0.333\ldots = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \ldots$$

This is a geometric series with $a_1 = 0.3$ and $r = 0.1$:

$$S = \frac{0.3}{1 - 0.1} = \frac{0.3}{0.9} = \frac{1}{3}$$

This proves that $0.333\ldots = \frac{1}{3}$!

Stadium Seating

A stadium has 20 seats in the first row. Each subsequent row has 2 more seats than the row before it. If there are 50 rows, how many total seats are in the stadium?

This is an arithmetic series with $a_1 = 20$, $d = 2$, and $n = 50$.

The 50th row has: $a_{50} = 20 + 49 \times 2 = 118$ seats.

Total seats: $S_{50} = \frac{50}{2}(20 + 118) = 25 \times 138 = 3,450$ seats.

Self-Test Problems

Problem 1: Find the 15th term of the arithmetic sequence: $8, 13, 18, 23, \ldots$

Show Answer

First, identify $a_1 = 8$ and $d = 13 - 8 = 5$.

Using $a_n = a_1 + (n-1)d$:

$$a_{15} = 8 + (15-1)(5) = 8 + 14 \times 5 = 8 + 70 = 78$$

The 15th term is 78.

Problem 2: Find the 6th term of the geometric sequence: $4, 12, 36, 108, \ldots$

Show Answer

First, identify $a_1 = 4$ and $r = \frac{12}{4} = 3$.

Using $a_n = a_1 \cdot r^{n-1}$:

$$a_6 = 4 \cdot 3^{6-1} = 4 \cdot 3^5 = 4 \cdot 243 = 972$$

The 6th term is 972.

Problem 3: Find the sum of the arithmetic series: $3 + 7 + 11 + 15 + \ldots + 99$

Show Answer

First, find how many terms there are. With $a_1 = 3$, $d = 4$, and $a_n = 99$:

$$99 = 3 + (n-1)(4)$$ $$96 = (n-1)(4)$$ $$24 = n - 1$$ $$n = 25$$

Now use $S_n = \frac{n}{2}(a_1 + a_n)$:

$$S_{25} = \frac{25}{2}(3 + 99) = \frac{25}{2}(102) = 25 \times 51 = 1275$$

The sum is 1,275.

Problem 4: Find the sum of the first 8 terms of the geometric series: $2 + 6 + 18 + 54 + \ldots$

Show Answer

With $a_1 = 2$, $r = 3$, and $n = 8$:

$$S_8 = 2 \cdot \frac{1 - 3^8}{1 - 3} = 2 \cdot \frac{1 - 6561}{-2} = 2 \cdot \frac{-6560}{-2} = 2 \cdot 3280 = 6560$$

The sum is 6,560.

Problem 5: Find the sum of the infinite geometric series: $24 + 12 + 6 + 3 + \ldots$

Show Answer

With $a_1 = 24$ and $r = \frac{12}{24} = \frac{1}{2}$:

Since $|r| = \frac{1}{2} < 1$, the series converges.

$$S = \frac{24}{1 - \frac{1}{2}} = \frac{24}{\frac{1}{2}} = 24 \times 2 = 48$$

The sum is 48.

Problem 6: Evaluate: $\displaystyle\sum_{k=1}^{5} 2^k$

Show Answer

Write out the terms:

$$\sum_{k=1}^{5} 2^k = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 = 2 + 4 + 8 + 16 + 32$$

This is a geometric series with $a_1 = 2$, $r = 2$, and $n = 5$:

$$S_5 = 2 \cdot \frac{1 - 2^5}{1 - 2} = 2 \cdot \frac{1 - 32}{-1} = 2 \cdot 31 = 62$$

The sum is 62.

Problem 7: A geometric sequence has $a_3 = 20$ and $a_6 = 160$. Find the first term and common ratio.

Show Answer

Using $a_n = a_1 \cdot r^{n-1}$:

$$a_3 = a_1 \cdot r^2 = 20$$ $$a_6 = a_1 \cdot r^5 = 160$$

Divide the second equation by the first:

$$\frac{a_1 \cdot r^5}{a_1 \cdot r^2} = \frac{160}{20}$$ $$r^3 = 8$$ $$r = 2$$

Substitute back to find $a_1$:

$$a_1 \cdot 2^2 = 20$$ $$a_1 \cdot 4 = 20$$ $$a_1 = 5$$

The first term is 5 and the common ratio is 2.

Summary

  • A sequence is an ordered list of numbers following a pattern. Each number is a term.

  • An explicit formula gives any term directly: plug in the position number and calculate.

  • A recursive formula defines each term based on previous terms.

  • Arithmetic sequences add a constant difference $d$ between terms:

    • Explicit formula: $a_n = a_1 + (n-1)d$
  • Geometric sequences multiply by a constant ratio $r$ between terms:

    • Explicit formula: $a_n = a_1 \cdot r^{n-1}$
  • A series is the sum of terms in a sequence.

  • Arithmetic series sum: $S_n = \frac{n}{2}(a_1 + a_n)$ or $S_n = \frac{n}{2}(2a_1 + (n-1)d)$

  • Geometric series sum: $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$ (when $r \neq 1$)

  • Infinite geometric series converge when $|r| < 1$: $S = \frac{a_1}{1 - r}$

  • Sigma notation ($\Sigma$) provides a compact way to write sums.

  • Sequences and series appear everywhere: finance, physics, computer science, and everyday counting problems.