Sequences, Series, and Financial Mathematics

Explicit and Recursive Sequences

Sequences are ordered lists of numbers that follow a pattern. In 11th Grade Maths, sequences are important because they help us describe repeated change, predict future values, and model real-life situations such as saving money, loan balances, and population growth.

There are two common ways to define a sequence:

• Explicit form: gives the value of a term directly from its position number.
• Recursive form: gives each term using one or more earlier terms.

Understanding both forms is useful because sometimes one form is easier for calculating a specific term, while the other is better for showing how the pattern grows step by step.

1. What is a sequence?

A sequence is a list of numbers written in a specific order. Each number is called a term. The position of a term is usually represented by the variable \(n\).

For example, in the sequence

$$2, 5, 8, 11, 14, \dots$$

the first term is \(2\), the second term is \(5\), and the third term is \(8\).

We often use notation like \(a_n\) to mean “the \(n\)th term of the sequence.” So \(a_1\) is the first term, \(a_2\) is the second term, and so on.

2. Explicit sequences

An explicit sequence gives a formula for the \(n\)th term directly. This means you can find any term without needing the earlier ones.

A general explicit rule looks like:

$$a_n = \text{formula involving } n$$

For example, consider the sequence

$$4, 7, 10, 13, 16, \dots$$

This sequence increases by \(3\) each time. Its explicit rule is

$$a_n = 4 + (n-1)3$$

This can also be simplified:

$$a_n = 3n + 1$$

Now you can find any term directly. For example, the 10th term is

$$a_{10} = 3(10) + 1 = 31$$

Why explicit rules are useful:

• They let you find a term quickly.
• You do not need to list all earlier terms.
• They are especially helpful for large term numbers, such as the 50th or 100th term.

3. Recursive sequences

A recursive sequence defines a term using one or more previous terms. Because of this, a recursive rule must include at least one starting value.

A recursive sequence often looks like this:

$$a_1 = \text{starting term}$$

$$a_n = \text{rule using } a_{n-1}$$

For the sequence

$$4, 7, 10, 13, 16, \dots$$

the recursive rule is

$$a_1 = 4$$

$$a_n = a_{n-1} + 3 \quad \text{for } n \ge 2$$

This means:

• Start at \(4\).
• To get each next term, add \(3\).

Why recursive rules are useful:

• They clearly show how the pattern changes from one term to the next.
• They are useful in real-life situations where each value depends on the one before it.
• They are common in financial mathematics, such as repeated deposits or interest growth.

4. Arithmetic sequences

An arithmetic sequence is a sequence where the difference between consecutive terms is constant.

Example:

$$6, 10, 14, 18, 22, \dots$$

The common difference is \(4\).

For an arithmetic sequence:

• Recursive form: add the common difference each time.
• Explicit form: use the first term and the common difference.

The explicit formula for an arithmetic sequence is

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

where:

• \(a_1\) is the first term
• \(d\) is the common difference

The recursive formula is

$$a_1 = \text{first term}$$

$$a_n = a_{n-1} + d \quad \text{for } n \ge 2$$

5. Geometric sequences

A geometric sequence is a sequence where each term is found by multiplying the previous term by the same constant ratio.

Example:

$$3, 6, 12, 24, 48, \dots$$

Each term is multiplied by \(2\), so the common ratio is \(2\).

For a geometric sequence:

• Recursive form: multiply by the common ratio each time.
• Explicit form: multiply the first term by the ratio raised to a power.

The explicit formula for a geometric sequence is

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

where:

• \(a_1\) is the first term
• \(r\) is the common ratio

The recursive formula is

$$a_1 = \text{first term}$$

$$a_n = r a_{n-1} \quad \text{for } n \ge 2$$

6. How to tell whether a sequence is explicit or recursive

You can identify the form by looking at what the rule uses.

• If the rule uses only \(n\), it is explicit.
• If the rule uses an earlier term like \(a_{n-1}\), it is recursive.

Examples:

• \(a_n = 5n - 2\) is explicit.
• \(a_1 = 3, \; a_n = a_{n-1} + 5\) is recursive.
• \(a_n = 7(2)^{n-1}\) is explicit.
• \(a_1 = 7, \; a_n = 2a_{n-1}\) is recursive.

7. Worked Example 1: Finding terms from an explicit rule

Suppose a sequence is defined by

$$a_n = 2n + 5$$

Find the first four terms.

Step 1: Substitute \(n = 1\).

$$a_1 = 2(1) + 5 = 7$$

Step 2: Substitute \(n = 2\).

$$a_2 = 2(2) + 5 = 9$$

Step 3: Substitute \(n = 3\).

$$a_3 = 2(3) + 5 = 11$$

Step 4: Substitute \(n = 4\).

$$a_4 = 2(4) + 5 = 13$$

So the sequence is

$$7, 9, 11, 13, \dots$$

This is an arithmetic sequence with common difference \(2\).

8. Worked Example 2: Finding terms from a recursive rule

Suppose a sequence is defined by

$$a_1 = 6$$

$$a_n = a_{n-1} + 4 \quad \text{for } n \ge 2$$

Find the first five terms.

Step 1: Start with the first term.

$$a_1 = 6$$

Step 2: Add \(4\) each time.

$$a_2 = a_1 + 4 = 6 + 4 = 10$$

$$a_3 = a_2 + 4 = 10 + 4 = 14$$

$$a_4 = a_3 + 4 = 14 + 4 = 18$$

$$a_5 = a_4 + 4 = 18 + 4 = 22$$

So the first five terms are

$$6, 10, 14, 18, 22$$

9. Worked Example 3: Writing both forms of a sequence

Consider the sequence

$$12, 9, 6, 3, 0, \dots$$

Write both the recursive rule and the explicit rule.

Step 1: Identify the pattern.

Each term decreases by \(3\), so this is an arithmetic sequence with common difference \(-3\).

Recursive form:

Start with the first term \(12\), then subtract \(3\) each time.

$$a_1 = 12$$

$$a_n = a_{n-1} - 3 \quad \text{for } n \ge 2$$

Explicit form:

Use the arithmetic formula:

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

Substitute \(a_1 = 12\) and \(d = -3\):

$$a_n = 12 + (n-1)(-3)$$

$$a_n = 12 - 3(n-1)$$

$$a_n = 15 - 3n$$

So the explicit rule is

$$a_n = 15 - 3n$$

10. Worked Example 4: Geometric sequence in a financial context

A savings amount starts at \(\$200\) and doubles each year. Let \(a_n\) be the amount after \(n\) years, with \(a_1 = 200\).

Find a recursive rule, an explicit rule, and the amount in year 5.

Step 1: Identify the pattern.

Doubling means multiplying by \(2\) each time. This is a geometric sequence.

Recursive rule:

$$a_1 = 200$$

$$a_n = 2a_{n-1} \quad \text{for } n \ge 2$$

Explicit rule:

Use the geometric formula:

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

Here, \(a_1 = 200\) and \(r = 2\), so

$$a_n = 200(2)^{n-1}$$

Step 2: Find the amount in year 5.

$$a_5 = 200(2)^{5-1}$$

$$a_5 = 200(2)^4$$

$$a_5 = 200(16) = 3200$$

So the amount in year 5 is \(\$3200\).

11. Converting between recursive and explicit forms

You should be able to move between the two forms, especially for arithmetic and geometric sequences.

For arithmetic sequences:

• Recursive: identify the amount added or subtracted each time.
• Explicit: use \(a_n = a_1 + (n-1)d\).

For geometric sequences:

• Recursive: identify the number multiplied each time.
• Explicit: use \(a_n = a_1 r^{n-1}\).

12. Common mistakes to avoid

• Forgetting the starting term in a recursive sequence. Without it, the sequence is not fully defined.
• Mixing up \(n\) and \(n-1\) in explicit formulas. The exponent or multiplier often involves \(n-1\), not just \(n\).
• Confusing arithmetic and geometric sequences. Arithmetic uses addition or subtraction. Geometric uses multiplication or division.
• Using the wrong sign for a common difference. If terms decrease, the difference is negative.

13. Quick comparison

• Explicit sequence: find a term directly from \(n\).
• Recursive sequence: find a term from earlier terms.
• Arithmetic sequence: constant difference.
• Geometric sequence: constant ratio.

14. Summary

Sequences can be written in explicit form or recursive form. An explicit rule tells you the value of the \(n\)th term directly, while a recursive rule tells you how to get each term from a previous term.

Arithmetic sequences change by a constant difference, and geometric sequences change by a constant ratio. For arithmetic sequences, use

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

and for geometric sequences, use

$$a_n = a_1r^{n-1}$$

When solving problems, first look for the pattern: are you adding the same number each time or multiplying by the same number each time? That will help you decide which kind of sequence you have and whether an explicit or recursive rule is most useful.

Arithmetic Sequences are one of the most important types of sequences in 11th Grade Maths. They help us describe patterns that change by the same amount each time. This makes them a discrete version of a linear pattern.

For example, if a sequence goes 5, 8, 11, 14, 17, ... then each term increases by 3. Because the change is constant, this is an arithmetic sequence.

Understanding arithmetic sequences is useful in algebra, series, and financial mathematics because many real-life situations involve regular increases or decreases over equal steps.

What is an arithmetic sequence?

An arithmetic sequence is a sequence in which the difference between consecutive terms is always the same. This fixed amount is called the common difference.

If the first term is written as \(a_1\) and the common difference is \(d\), then the sequence is formed like this:

$$a_1,\ a_1+d,\ a_1+2d,\ a_1+3d,\ \dots$$

This shows that each new term is found by adding \(d\) to the previous term.

Common difference

To find the common difference, subtract one term from the next:

$$d = a_2-a_1$$

or more generally,

$$d = a_n-a_{n-1}$$

If this difference stays the same throughout the sequence, then the sequence is arithmetic.

For example, in the sequence \(12, 9, 6, 3, 0, \dots\):

• \(9-12=-3\)
• \(6-9=-3\)
• \(3-6=-3\)

The common difference is \(-3\), so this is an arithmetic sequence.

The nth term formula

The most important formula for arithmetic sequences is the formula for the \(n\)th term:

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

This formula lets you find any term in the sequence without listing all the earlier terms.

Here:

• \(a_n\) = the \(n\)th term
• \(a_1\) = the first term
• \(d\) = the common difference
• \(n\) = the term number

Why does the formula work?

Start with the first term, \(a_1\). To get to the second term, you add \(d\) once. To get to the third term, you add \(d\) twice. To get to the \(n\)th term, you add \(d\) a total of \(n-1\) times.

That is why:

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

Worked Example 1: Finding the common difference

Determine whether the sequence \(4, 10, 16, 22, 28, \dots\) is arithmetic, and find the common difference.

Step 1: Subtract consecutive terms.

$$10-4=6$$

$$16-10=6$$

$$22-16=6$$

Step 2: Check whether the difference is constant.

Since the difference is always 6, the sequence is arithmetic.

Answer: The common difference is \(d=6\).

Worked Example 2: Finding the nth term

Find a formula for the \(n\)th term of the arithmetic sequence \(7, 12, 17, 22, \dots\).

Step 1: Identify the first term and common difference.

• \(a_1=7\)
• \(d=12-7=5\)

Step 2: Use the formula.

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

Substitute the values:

$$a_n = 7 + (n-1)(5)$$

Step 3: Simplify.

$$a_n = 7 + 5n - 5$$

$$a_n = 5n + 2$$

Answer: The \(n\)th term is

$$a_n = 5n+2$$

You can check it:

• When \(n=1\), \(a_1=5(1)+2=7\)
• When \(n=2\), \(a_2=5(2)+2=12\)

Worked Example 3: Finding a specific term

Find the 20th term of the arithmetic sequence \(3, 8, 13, 18, \dots\).

Step 1: Identify \(a_1\) and \(d\).

• \(a_1=3\)
• \(d=5\)

Step 2: Use the formula.

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

For the 20th term, \(n=20\):

$$a_{20} = 3 + (20-1)(5)$$

$$a_{20} = 3 + 19\cdot 5$$

$$a_{20} = 3 + 95$$

$$a_{20} = 98$$

Answer: The 20th term is \(98\).

Worked Example 4: Finding the term number

In the arithmetic sequence \(15, 19, 23, 27, \dots\), which term is equal to 63?

Step 1: Identify the first term and common difference.

• \(a_1=15\)
• \(d=4\)

Step 2: Set the nth term equal to 63.

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

$$63 = 15 + (n-1)(4)$$

Step 3: Solve for \(n\).

$$63 = 15 + 4n - 4$$

$$63 = 11 + 4n$$

$$52 = 4n$$

$$n = 13$$

Answer: 63 is the 13th term.

Connecting arithmetic sequences to linear patterns

An arithmetic sequence is a discrete linear pattern. That means it behaves like a linear relationship, but only at whole-number term positions such as \(n=1,2,3,4,\dots\).

If you write the nth term in the form

$$a_n = dn + (a_1-d)$$

you can see it looks like the equation of a line. The common difference \(d\) acts like the slope, because it tells you how much the terms change from one step to the next.

Increasing and decreasing arithmetic sequences

• If \(d>0\), the sequence increases.
• If \(d<0\), the sequence decreases.
• If \(d=0\), the sequence stays constant.

Examples:

• \(2, 5, 8, 11, \dots\) has \(d=3\), so it increases.
• \(20, 16, 12, 8, \dots\) has \(d=-4\), so it decreases.
• \(9, 9, 9, 9, \dots\) has \(d=0\), so it stays constant.

How to recognize an arithmetic sequence

When you are given a list of terms, ask yourself:

1. Do I get the same result each time I subtract consecutive terms?
2. If yes, what is that common difference?
3. Can I use \(a_n = a_1 + (n-1)d\)?

If the difference is not constant, then the sequence is not arithmetic.

Common mistakes to avoid

• Using the wrong difference: Always subtract consecutive terms in the same order.
• Forgetting \((n-1)\): The formula is \(a_n = a_1 + (n-1)d\), not \(a_1+nd\).
• Mixing up term number and term value: In \(a_{10}\), the 10 tells you the position, not the value.
• Sign errors with negative differences: If the sequence is decreasing, keep the negative sign in \(d\).

Quick practice ideas

Try these on your own:

• Find the common difference of \(11, 15, 19, 23, \dots\).
• Write a formula for \(2, 7, 12, 17, \dots\).
• Find the 30th term of \(50, 47, 44, 41, \dots\).
• Find which term of \(6, 10, 14, 18, \dots\) is equal to 86.

Brief summary

An arithmetic sequence is a sequence where the difference between consecutive terms is constant. This constant change is called the common difference, \(d\).

The key formula is

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

Use this formula to find any term, create a rule for the sequence, or determine which term has a given value. Once you can identify the first term and common difference, you can solve most arithmetic sequence problems confidently.

Arithmetic Series is the sum of the terms of an arithmetic sequence.

An arithmetic sequence is a list of numbers where the difference between consecutive terms is always the same. This fixed difference is called the common difference.

For example, in the sequence \(3, 7, 11, 15, 19\), each term increases by \(4\). So this is an arithmetic sequence with common difference \(d = 4\).

When we add the terms of an arithmetic sequence, we get an arithmetic series. For example,

$$3 + 7 + 11 + 15 + 19$$

is an arithmetic series.

In this lesson, you will learn how to recognize an arithmetic series, how to find the sum of a certain number of terms, and how the famous Gaussian summation idea helps us do this quickly.

1. Review: Arithmetic Sequence Formula

Before finding the sum, it helps to know how to find individual terms.

The formula for the \(n\)th term of an arithmetic sequence is

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

where:

• \(a_1\) is the first term
• \(d\) is the common difference
• \(a_n\) is the \(n\)th term

Example: If \(a_1 = 5\) and \(d = 3\), then

$$a_n = 5 + (n-1)3$$

To find the 6th term:

$$a_6 = 5 + (6-1)3 = 5 + 15 = 20$$

2. What Is an Arithmetic Series?

An arithmetic series is written as a sum:

$$a_1 + a_2 + a_3 + \cdots + a_n$$

If the terms come from an arithmetic sequence, then the series is an arithmetic series.

The sum of the first \(n\) terms is called the partial sum, often written as \(S_n\).

So, \(S_n\) means:

$$S_n = a_1 + a_2 + a_3 + \cdots + a_n$$

3. Gaussian Summation Idea

A famous way to find the sum of an arithmetic series comes from pairing terms.

Suppose we want to add:

$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$$

Write the sum forward and backward:

$$S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$$

$$S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$$

Now add vertically:

$$2S = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11$$

There are 10 pairs, and each pair sums to 11. So:

$$2S = 10(11) = 110$$

$$S = 55$$

This idea works for any arithmetic series because the first and last terms always pair to the same total, the second and second-last pair to the same total, and so on.

4. Formula for the Sum of an Arithmetic Series

Using the pairing idea, the sum of the first \(n\) terms is:

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

This formula says:

• add the first and last terms
• multiply by the number of terms
• divide by 2

Another useful version comes from replacing \(a_n\) with \(a_1 + (n-1)d\):

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

You can use:

• $$S_n = \frac{n}{2}(a_1 + a_n)$$ when you know the first term and last term
• $$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$ when you know the first term and common difference

5. Steps for Finding an Arithmetic Series Sum

1. Identify the first term \(a_1\).
2. Find the common difference \(d\), if needed.
3. Find the number of terms \(n\).
4. If necessary, find the last term using \(a_n = a_1 + (n-1)d\).
5. Choose the sum formula.
6. Simplify carefully.

Worked Example 1: Sum of a Simple Arithmetic Series

Find the sum:

$$4 + 7 + 10 + 13 + 16$$

Step 1: Identify the information.

• First term: \(a_1 = 4\)
• Last term: \(a_n = 16\)
• Number of terms: \(n = 5\)

Step 2: Use the sum formula.

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

$$S_5 = \frac{5}{2}(4 + 16)$$

$$S_5 = \frac{5}{2}(20)$$

$$S_5 = 50$$

Answer: The sum is \(50\).

Worked Example 2: Find the Sum of the First 20 Terms

Find the sum of the first 20 terms of the arithmetic sequence

$$6, 11, 16, 21, \dots$$

Step 1: Identify the information.

• \(a_1 = 6\)
• \(d = 5\)
• \(n = 20\)

Step 2: Use the formula with \(a_1\), \(d\), and \(n\).

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

$$S_{20} = \frac{20}{2}[2(6) + (20-1)(5)]$$

$$S_{20} = 10[12 + 19 \cdot 5]$$

$$S_{20} = 10[12 + 95]$$

$$S_{20} = 10(107)$$

$$S_{20} = 1070$$

Answer: The sum of the first 20 terms is \(1070\).

Worked Example 3: Find the Number of Terms from a Given Sum

The arithmetic sequence

$$2, 5, 8, 11, \dots$$

has a partial sum of \(152\). How many terms were added?

Step 1: Identify the information.

• \(a_1 = 2\)
• \(d = 3\)
• \(S_n = 152\)

Step 2: Use the sum formula.

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

$$152 = \frac{n}{2}[2(2) + (n-1)3]$$

$$152 = \frac{n}{2}[4 + 3n - 3]$$

$$152 = \frac{n}{2}(3n + 1)$$

Multiply both sides by 2:

$$304 = n(3n + 1)$$

$$304 = 3n^2 + n$$

$$3n^2 + n - 304 = 0$$

Now solve the quadratic. Factor:

$$3n^2 + n - 304 = (3n + 32)(n - 9)$$

So:

$$n = -\frac{32}{3} \quad \text{or} \quad n = 9$$

The number of terms must be positive, so:

$$n = 9$$

Answer: \(9\) terms were added.

Worked Example 4: Arithmetic Series in a Financial Context

A student saves money each week. In the first week, they save \(\$10\). Each week after that, they save \(\$5\) more than the previous week. How much will they save in the first 12 weeks?

Step 1: Identify the arithmetic sequence.

The weekly savings form the sequence:

$$10, 15, 20, 25, \dots$$

• \(a_1 = 10\)
• \(d = 5\)
• \(n = 12\)

Step 2: Use the sum formula.

$$S_{12} = \frac{12}{2}[2(10) + (12-1)(5)]$$

$$S_{12} = 6[20 + 11 \cdot 5]$$

$$S_{12} = 6[20 + 55]$$

$$S_{12} = 6(75)$$

$$S_{12} = 450$$

Answer: The student saves \(\$450\) in the first 12 weeks.

This is one way arithmetic series connect to financial mathematics: repeated amounts that increase or decrease by a constant amount can often be modeled with an arithmetic series.

6. Common Mistakes to Avoid

• Mixing up sequence and series: A sequence is a list of terms. A series is the sum of those terms.
• Using the wrong value for \(n\): \(n\) is the number of terms, not the last term.
• Forgetting to find the last term first: If you use $$S_n = \frac{n}{2}(a_1 + a_n),$$ make sure you know \(a_n\).
• Using the wrong common difference: Check the pattern carefully. In \(12, 9, 6, 3\), the common difference is \(-3\), not \(3\).
• Arithmetic errors: Because the formula has several steps, simplify carefully.

7. Quick Check

Try these on your own:

1. Find the sum: $$8 + 12 + 16 + 20 + 24$$
2. Find the sum of the first 15 terms of: $$3, 7, 11, 15, \dots$$
3. The arithmetic sequence $$5, 9, 13, 17, \dots$$ has a sum of \(230\). How many terms were added?

Answers:

1. \(80\)
2. \(465\)
3. \(10\)

8. Key Ideas to Remember

• An arithmetic series is the sum of terms from an arithmetic sequence.
• The sum of the first \(n\) terms is called a partial sum.
• The main formulas are:

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

and

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

9. Brief Summary

Arithmetic series involve adding terms from a sequence with a constant difference.

The Gaussian pairing idea shows why the sum can be found efficiently. Instead of adding every term one by one, you can use a formula based on the number of terms and the first and last terms, or the first term and common difference.

Once you identify \(a_1\), \(d\), and \(n\), you can solve many problems involving patterns, totals, and financial situations.

Geometric Sequences are sequences where each term is found by multiplying the previous term by the same constant number.

That constant number is called the common ratio. Geometric sequences are important because they model repeated growth or repeated decay, which appears in finance, science, and many real-world situations.

For example, in the sequence 3, 6, 12, 24, 48, ... each term is multiplied by 2 to get the next term. So this is a geometric sequence with common ratio 2.

In 81, 27, 9, 3, 1, ... each term is multiplied by \(\frac{1}{3}\). This is also geometric, even though the terms are getting smaller.

How to recognize a geometric sequence

• Check whether the ratio between consecutive terms is constant.
• Divide a term by the term before it.
• If the same ratio appears each time, the sequence is geometric.

If a sequence has first term \(a_1\) and common ratio \(r\), then its terms look like:

\(a_1, a_1r, a_1r^2, a_1r^3, \dots\)

This leads to the nth term formula:

In this formula:

• \(a_n\) is the term you want to find
• \(a_1\) is the first term
• \(r\) is the common ratio
• \(n\) is the term number

Finding the common ratio

To find the common ratio, divide any term by the term before it:

For example, in the sequence 5, 15, 45, 135, ...

\(15 \div 5 = 3\), \(45 \div 15 = 3\), and \(135 \div 45 = 3\). Since the ratio is always 3, the common ratio is \(r=3\).

What the common ratio tells you

• If \(r > 1\), the sequence grows.
• If \(0 < r < 1\), the sequence decreases but stays positive if the first term is positive.
• If \(r < 0\), the signs of the terms alternate.
• If \(r = 1\), every term is the same.

For example:

• \(2, 6, 18, 54, ...\) has \(r=3\), so it grows.
• \(64, 32, 16, 8, ...\) has \(r=\frac{1}{2}\), so it decreases.
• \(5, -10, 20, -40, ...\) has \(r=-2\), so the signs alternate.

Worked Example 1: Identify whether a sequence is geometric

Consider the sequence: 4, 12, 36, 108, ...

Check the ratios:

The ratio is always 3, so the sequence is geometric.

The first term is \(a_1=4\) and the common ratio is \(r=3\).

Worked Example 2: Find the nth term

Find a formula for the sequence: 7, 14, 28, 56, ...

Step 1: Identify the first term and common ratio.

• \(a_1=7\)
• \(r=\frac{14}{7}=2\)

Step 2: Use the geometric formula.

So the nth term is:

Let us check the 4th term:

This matches the sequence, so the formula is correct.

Worked Example 3: Find a specific term

Find the 6th term of the geometric sequence with \(a_1=3\) and \(r=4\).

Use the formula:

So the 6th term is:

Worked Example 4: Find the missing information from later terms

A geometric sequence has \(a_3=20\) and common ratio \(r=2\). Find \(a_1\) and then write the nth term.

Use the formula for the 3rd term:

Now use the nth term formula:

So the first term is 5, and the nth term is \(a_n = 5(2)^{n-1}\).

Geometric sequences and real-life patterns

Geometric sequences model situations where the same multiplier is applied repeatedly.

• A population that doubles each time period
• The value of an item that loses the same percentage each year
• Money in an account growing by the same factor each period

For example, if a quantity starts at 200 and increases by 5% each period, then each term is multiplied by \(1.05\). The sequence is geometric with:

Its nth term is:

If a value decreases by 20% each period, then 80% remains. That means the multiplier is \(0.8\), so the sequence is geometric with common ratio \(0.8\).

Common mistakes to avoid

• Do not add or subtract to test for a geometric sequence. Geometric sequences are based on multiplying by a constant ratio.
• Be careful with fractions and decimals. A ratio like \(\frac{1}{2}\) or 0.5 is still a valid common ratio.
• Remember the exponent is \(n-1\), not just \(n\). This is because the first term must come out correctly when \(n=1\).
• Watch negative ratios. If \(r\) is negative, the signs alternate.

Quick check

1. Is the sequence 2, 10, 50, 250, ... geometric?
   Yes, because the ratio is 5 each time.
2. What is the common ratio of 48, 24, 12, 6, ...?
   \(r=\frac{1}{2}\)
3. What is the nth term of 9, 27, 81, 243, ...?
   \(a_n = 9(3)^{n-1}\)

Summary

A geometric sequence is a sequence where each term is found by multiplying the previous term by the same number, called the common ratio.

You can recognize a geometric sequence by checking that consecutive terms have the same ratio. The main formula is:

This formula allows you to find any term in the sequence once you know the first term and common ratio.

Geometric sequences are especially useful for modeling repeated growth and decay, including many financial situations.

Geometric Series is an important topic in sequences and series. It helps us add terms from a geometric pattern quickly without writing out every term one by one.

A geometric sequence is a sequence where each term is found by multiplying the previous term by the same constant number. This constant is called the common ratio, written as \(r\).

For example, in the sequence \(3, 6, 12, 24, \dots\), each term is multiplied by \(2\). So the common ratio is \(r=2\).

A geometric series is the sum of the terms of a geometric sequence. So if the sequence is \(3, 6, 12, 24\), then the geometric series is

In this lesson, you will learn how to recognize a geometric series, find its finite sum, and use the formula correctly.

1. Geometric Sequence Review

If the first term is \(a\) and the common ratio is \(r\), then the terms of a geometric sequence are:

The \(n\)th term of a geometric sequence is

This formula is useful because in a geometric series, you often need to know how many terms there are and what pattern they follow.

2. What Is a Finite Geometric Series?

A finite geometric series has a limited number of terms. Its sum is written as

Here:

• \(a\) = first term
• \(r\) = common ratio
• \(n\) = number of terms
• \(S_n\) = sum of the first \(n\) terms

Adding all these terms directly can be slow, especially when \(n\) is large. That is why we use a formula.

3. Finite Geometric Series Formula

For \(r \ne 1\), the sum of the first \(n\) terms is

An equivalent form of the formula is

These two formulas mean the same thing. You can use either one, as long as you use it carefully.

If \(r=1\), then every term is the same. In that special case, the sum is simply

4. Why the Formula Works

It is helpful to see where the formula comes from.

Start with the series

Multiply both sides by \(r\):

Now subtract the second equation from the first:

Factor both sides:

Now divide by \(1-r\):

This subtraction works because most terms cancel out. That is what makes the formula so powerful.

5. Steps for Finding the Sum

1. Identify the first term \(a\).
2. Identify the common ratio \(r\).
3. Count the number of terms \(n\).
4. Substitute into the formula \(S_n = \frac{a(1-r^n)}{1-r}\).
5. Simplify carefully.

6. Worked Examples

Example 1: Find the sum of \(2+6+18+54\).

First, check that it is geometric. Each term is multiplied by \(3\), so \(r=3\).

We have:

• \(a=2\)
• \(r=3\)
• \(n=4\)

Use the formula:

Now simplify:

So the sum is

You can check by direct addition:

Example 2: Find the sum of the first 6 terms of the geometric sequence \(5, -10, 20, -40, \dots\).

Find the first term and common ratio:

• \(a=5\)
• \(r=-2\)
• \(n=6\)

Use the formula:

Simplify:

So the sum of the first 6 terms is

This example shows that geometric series can have a negative common ratio. The terms alternate signs, but the same formula still works.

Example 3: Find the sum

This is geometric because each term is multiplied by \(\frac{1}{2}\).

So:

• \(a=4\)
• \(r=\frac{1}{2}\)
• \(n=5\)

Use the formula:

Now simplify:

So the sum is

As a decimal, this is \(15.5\).

Example 4: A financial-style application

A student saves money in a pattern: \(\$10\) in the first week, \(\$20\) in the second week, \(\$40\) in the third week, and so on, doubling each week. How much is saved in the first 5 weeks?

This forms a geometric series:

Identify the values:

• \(a=10\)
• \(r=2\)
• \(n=5\)

Use the formula:

Simplify:

So the total amount saved is

7. Common Mistakes to Avoid

• Confusing sequence and series: A sequence is a list of terms. A series is the sum of those terms.
• Using the wrong common ratio: Always divide one term by the previous term to check \(r\).
• Forgetting the number of terms: Count terms carefully. Do not confuse the last term with \(n\).
• Sign errors: Be careful when \(r\) is negative or when subtracting expressions like \(1-r^n\).
• Using the formula when \(r=1\): If every term is the same, just use \(S_n=an\).

8. Quick Check: Is It Geometric?

To decide if a series is geometric, ask: Is each term multiplied by the same number to get the next term?

• \(3+9+27+81\) → yes, \(r=3\)
• \(8+4+2+1\) → yes, \(r=\frac{1}{2}\)
• \(2+5+8+11\) → no, this is not geometric because terms are increased by adding \(3\), not multiplying

9. When the Formula Is Most Useful

The finite geometric series formula is especially useful when:

• there are many terms,
• the terms get very large,
• the terms are fractions, or
• the series appears in financial situations such as repeated growth or repeated doubling.

10. Summary

A geometric series is the sum of terms from a geometric sequence. If the first term is \(a\), the common ratio is \(r\), and there are \(n\) terms, then the finite sum is

If \(r=1\), then

To solve problems, identify \(a\), \(r\), and \(n\), then substitute carefully into the formula. With practice, geometric series become a fast and reliable way to add patterned terms.

Infinite Convergent Geometric Series

In mathematics, a geometric sequence is a sequence where each term is found by multiplying the previous term by the same number. That number is called the common ratio, written as \(r\).

For example, in the sequence \(8, 4, 2, 1, \dots\), each term is multiplied by \(\frac{1}{2}\). So the common ratio is \(r = \frac{1}{2}\).

When we add the terms of a geometric sequence, we get a geometric series. If the series goes on forever, it is called an infinite geometric series.

An example is

This lesson explains when an infinite geometric series has a finite sum, and how to find that sum.

1. Geometric series review

A geometric series has the form

where:

• \(a\) is the first term
• \(r\) is the common ratio

For example, in

the first term is \(a = 3\), and the common ratio is \(r = 2\).

2. What does convergence mean?

An infinite series is called convergent if the sum of more and more terms gets closer and closer to one fixed number.

If the sum does not approach a fixed number, the series is called divergent.

Look at this series:

The partial sums are:

• First term: \(1\)
• First two terms: \(1 + \frac{1}{2} = 1.5\)
• First three terms: \(1.75\)
• First four terms: \(1.875\)

These sums keep getting closer to \(2\). So this infinite geometric series converges to \(2\).

Now look at

The partial sums are \(2, 6, 14, 30, \dots\). They keep growing and do not approach a fixed value. So this series is divergent.

3. The condition for convergence

An infinite geometric series

is convergent only when

This means the common ratio must be between \(-1\) and \(1\), not including \(-1\) or \(1\).

In other words:

• If \(-1 < r < 1\), the series converges.
• If \(|r| \ge 1\), the series diverges.

Why does this happen?

When \(|r| < 1\), the powers of \(r\) get smaller and smaller:

They approach \(0\), so the added terms become tiny. This allows the sum to settle to a fixed number.

For example, if \(r = \frac{1}{2}\), then

These terms get closer to \(0\).

If \(r = 2\), the powers get larger. If \(r = 1\), the terms stay the same. If \(r = -1\), the terms keep switching sign without settling. In all of these cases, the infinite series does not converge.

4. Formula for the sum to infinity

If an infinite geometric series converges, its sum to infinity is

This formula can only be used when \(|r| < 1\).

So the steps are:

1. Identify the first term \(a\).
2. Identify the common ratio \(r\).
3. Check that \(|r| < 1\).
4. Use $$S_\infty = \frac{a}{1-r}$$

5. Worked Examples

Example 1: A basic convergent series

Find the sum to infinity of

Step 1: Identify \(a\) and \(r\)

The first term is

The common ratio is

Step 2: Check convergence

Since

the series converges.

Step 3: Use the formula

Answer: The sum to infinity is \(12\).

Example 2: A series with a negative ratio

Find the sum to infinity of

Step 1: Identify \(a\) and \(r\)

The first term is

The common ratio is

Step 2: Check convergence

Since

the series converges.

Step 3: Use the formula

Answer: The sum to infinity is \(\frac{10}{3}\).

Notice that the terms alternate between positive and negative, but the series still converges because \(|r| < 1\).

Example 3: Decide if the series converges

Determine whether this series converges, and if it does, find the sum:

Step 1: Identify \(a\) and \(r\)

Step 2: Check convergence

Since

the series does not converge.

Answer: The series is divergent, so it has no sum to infinity.

Example 4: Finding the missing values from a sum to infinity

An infinite geometric series has first term \(a = 8\) and common ratio \(r = \frac{1}{4}\). Find the sum to infinity.

Since

the series converges.

Use the formula:

Answer: The sum to infinity is \(\frac{32}{3}\).

6. Common mistakes to avoid

• Using the formula when the series does not converge. Always check that \(|r| < 1\) first.
• Mixing up \(a\) and \(r\). \(a\) is the first term, and \(r\) is found by dividing a term by the one before it.
• Forgetting brackets with negative ratios. For example, if \(r = -\frac{1}{2}\), then $$1-r = 1-\left(-\frac{1}{2}\right)$$ not just \(1-\frac{1}{2}\).
• Thinking all infinite series have a finite sum. Only some do. Infinite geometric series converge only when \(|r| < 1\).

7. Connection to financial mathematics

Infinite geometric series are useful in financial mathematics because they model repeated payments or values that shrink by the same factor each time.

For example, if a payment or value is reduced by a constant percentage each period, the total amount over a very long time can sometimes be modeled by an infinite geometric series.

This is one reason why understanding convergence matters: it tells us whether the total stays finite or keeps growing without bound.

8. Quick checklist

• Write the series in the form \(a + ar + ar^2 + ar^3 + \dots\)
• Find the first term \(a\)
• Find the common ratio \(r\)
• Check whether \(|r| < 1\)
• If yes, use $$S_\infty = \frac{a}{1-r}$$
• If no, the series is divergent

Summary

An infinite geometric series has the form \(a + ar + ar^2 + ar^3 + \dots\). It only has a finite sum when the common ratio satisfies \(|r| < 1\).

When that condition is true, the sum to infinity is found using

Always check the value of \(r\) before using the formula. If \(|r| \ge 1\), the series is divergent and has no sum to infinity.

Lesson: Sigma Notation

When mathematicians want to write a long sum in a short, clear way, they often use sigma notation. This is a compact way to represent adding many terms of a sequence.

Sigma notation is especially useful in sequences and series because it helps us describe patterns, calculate totals, and write formulas efficiently. In financial mathematics, it can also be used to represent repeated payments, deposits, or interest-related sums.

The symbol used is the Greek letter sigma: \(\Sigma\). It stands for sum.

A general sigma notation expression looks like this:

Here is what each part means:

• \(\Sigma\): add the terms
• \(i\): the index, or counting variable
• \(m\): the starting value of the index
• \(n\): the ending value of the index
• \(a_i\): the rule for the terms being added

For example,

means: substitute \(i=1,2,3,4\) into the rule \(i\), then add the results.

This shows the main idea of sigma notation: follow the rule, list the terms, then add them.

How to expand sigma notation

To expand a summation, use these steps:

1. Identify the starting and ending values of the index.
2. Substitute each index value into the term rule.
3. Write all terms in order.
4. Add them.

For example, consider

The index starts at \(k=2\) and ends at \(k=5\). Substitute each value:

• When \(k=2\): \(2(2)+1=5\)
• When \(k=3\): \(2(3)+1=7\)
• When \(k=4\): \(2(4)+1=9\)
• When \(k=5\): \(2(5)+1=11\)

So,

Notice that the index letter does not matter by itself. You could use \(i\), \(k\), \(n\), or another variable, as long as it is used consistently.

For example, these mean the same thing:

Both expand to

Reading sigma notation carefully

A common mistake is forgetting that the expression beside sigma is a rule, not just one term. You must substitute each index value into that rule.

For instance,

does not mean \(1+3+4\). Instead, it means:

which simplifies to

Worked Example 1: Basic expansion

Evaluate

Step 1: List the values of \(r\): \(1,2,3,4,5\)

Step 2: Substitute into the rule \(r\):

Step 3: Add:

Answer: \(15\)

Worked Example 2: Using a formula inside the summation

Evaluate

Step 1: Substitute each value of \(t\):

• \(t=1\): \(3(1)-2=1\)
• \(t=2\): \(3(2)-2=4\)
• \(t=3\): \(3(3)-2=7\)
• \(t=4\): \(3(4)-2=10\)

Step 2: Write the sum:

Step 3: Add:

Answer: \(22\)

Worked Example 3: Powers in sigma notation

Evaluate

Step 1: Substitute values of \(m\):

Step 2: Simplify:

Step 3: Add:

Answer: \(30\)

Worked Example 4: Starting from a number other than 1

Evaluate

Step 1: The index values are \(3,4,5,6\).

Step 2: Substitute:

Step 3: Simplify:

Step 4: Add:

Answer: \(36\)

Writing a series in sigma notation

You should also be able to go in the other direction: from a written sum to sigma notation.

Example:

This is an arithmetic sequence with first term \(4\) and common difference \(3\). A rule for the terms is

Check it:

• \(n=1\): \(3(1)+1=4\)
• \(n=2\): \(3(2)+1=7\)
• \(n=3\): \(3(3)+1=10\)

There are 5 terms, so the sigma notation is

Another example:

This can be written as

Useful properties of sigma notation

Sigma notation follows patterns that make calculations easier. These are especially useful when simplifying sums algebraically.

1. Constant multiple rule

This means a constant factor can be taken outside the summation.

Example:

2. Sum of two expressions

Example:

Now evaluate each part:

So,

3. Sum of a constant

If you add a constant value repeatedly, multiply the constant by the number of terms.

The number of terms from \(i=m\) to \(i=n\) is

So in general,

Changing the index

Sometimes the same sum can be written with a different index variable. This does not change the value, as long as the terms generated stay the same.

For example,

Both expand to

Be careful: changing the letter of the index is fine, but changing the starting or ending values changes the sum unless you also adjust the rule correctly.

Connection to sequences and series

A sequence is an ordered list of numbers. A series is what you get when you add the terms of a sequence.

Sigma notation is the language used to write a series compactly.

For example, the sequence

can be described by the rule \(2n\), and the corresponding series is

In sigma notation, that series is

In financial mathematics, repeated equal deposits or payments can be written as a sum. For example, if \$100 is deposited at the end of each month for 6 months, the total amount deposited can be written as

This equals

This is a simple example, but it shows how sigma notation can represent repeated financial actions.

Common mistakes to avoid

• Forgetting to include all index values. If the sum is from \(1\) to \(4\), there are 4 terms.
• Using the rule incorrectly. Substitute each index value carefully.
• Stopping too early. Make sure you include the last value of the index.
• Confusing a sequence with a series. A sequence lists terms; a series adds them.
• Ignoring brackets. In expressions like \((2n+1)\), substitute into the whole expression.

Quick practice ideas

Try expanding and evaluating these on your own:

• \(\sum_{n=1}^{3} (n+4)\)
• \(\sum_{k=2}^{5} k^2\)
• \(\sum_{j=1}^{4} 5\)

Try writing these in sigma notation:

• \(3+6+9+12\)
• \(2^2+3^2+4^2+5^2\)

Summary

Sigma notation is a compact way to write sums. The symbol \(\Sigma\) tells you to add terms generated by a rule, while the index shows where to start and stop.

To evaluate a sum, substitute each index value into the rule, write the terms, and add them. To write a series in sigma notation, identify the pattern, write a formula for the terms, and give the correct starting and ending values.

Once you understand how to expand, interpret, and rewrite sigma notation, you have an important tool for working with sequences, series, and financial mathematics.

Lesson: Annuities and Financial Series

In financial mathematics, money is often added or paid in regular equal amounts. These repeating payments create patterns that can be modeled using geometric sequences and geometric series. This is why annuities are closely connected to series.

An annuity is a situation where the same amount of money is paid or received at equal time intervals. Examples include saving a fixed amount every month, repaying a loan with equal monthly installments, or receiving a pension payment each year.

To understand annuities well, we need to connect them to how money grows with compound interest. When interest is added regularly, each payment grows by a factor of \((1+i)\), where \(i\) is the interest rate per payment period.

1. The basic idea of compound growth

If an amount \(P\) is invested at an interest rate of \(i\) per period, then after \(n\) periods its value is

This is a geometric pattern because each period multiplies the amount by the same factor \((1+i)\).

For example, if \(i=0.05\), then the growth factor each period is \(1.05\). After one period, the amount is multiplied by \(1.05\); after two periods, by \(1.05^2\); and so on.

2. What is an annuity?

An annuity involves a fixed payment, usually called \(R\), made every period.

• In a savings annuity, you deposit money regularly.
• In a loan annuity, you repay a loan with regular installments.

There are two common types:

• Ordinary annuity: payments are made at the end of each period.
• Annuity due: payments are made at the beginning of each period.

In most school problems, we begin with the ordinary annuity.

3. Future value of an ordinary annuity

Suppose you deposit \(R\) dollars at the end of every period for \(n\) periods, with interest rate \(i\) per period. At the end of the full term, each deposit has grown for a different number of periods.

The first deposit grows for \(n-1\) periods, the second grows for \(n-2\) periods, and the last deposit grows for 0 periods. So the future value is

This is a geometric series with first term \(R\) and common ratio \((1+i)\) if written in reverse order, or first term \(R(1+i)^{n-1}\) and ratio \(\frac{1}{1+i}\) if written forward. Using the sum formula for a geometric series, we get

This is the future value formula for an ordinary annuity.

Meaning of the variables:

• \(FV\): future value
• \(R\): regular payment each period
• \(i\): interest rate per period
• \(n\): number of payments

4. Present value of an ordinary annuity

Sometimes we know the regular future payments and want to know what they are worth today. This is called the present value.

For a loan, the amount borrowed now is the present value of all the repayments. Each payment is discounted back to the present using powers of \((1+i)\).

The present value of an ordinary annuity is

This is also a geometric series. Its sum is

This formula is very important in loan calculations.

5. Annuity due formulas

If payments happen at the beginning of each period instead of the end, each payment earns interest for one extra period. This means an annuity due is worth more than an ordinary annuity.

The formulas become:

So you can think of an annuity due as the ordinary annuity formula multiplied by \((1+i)\).

6. Important financial meanings

• Savings plan: use future value, because you want to know how much money will be in the account later.
• Loan or mortgage: use present value, because the loan amount today equals the present value of future repayments.
• Installments: the fixed payment amount \(R\) is usually what you solve for.

7. Matching the interest rate and the payment period

This is one of the most important skills in financial mathematics. The interest rate and the number of payments must use the same time unit.

For example:

• If payments are monthly, then \(i\) must be the monthly interest rate.
• If the annual interest rate is 12%, compounded monthly, then the monthly rate is \(\frac{0.12}{12}=0.01\).
• If payments are made for 3 years monthly, then \(n=3\times 12=36\).

8. Worked Example 1: Future value of a savings annuity

A student deposits \(\$200\) at the end of each month into a savings account that earns 6% per year, compounded monthly. Find the amount in the account after 2 years.

Step 1: Identify the values.

• \(R=200\)
• Annual rate = 6%, so monthly rate \(i=\frac{0.06}{12}=0.005\)
• 2 years monthly gives \(n=2\times 12=24\)

Step 2: Use the future value formula.

Step 3: Calculate.

Answer: After 2 years, the account will contain approximately \(\$5086.40\).

9. Worked Example 2: Present value of a loan

A borrower agrees to repay \(\$350\) at the end of each month for 4 years. The interest rate is 9% per year, compounded monthly. How much money was borrowed?

Step 1: Identify the values.

• \(R=350\)
• Monthly interest rate: \(i=\frac{0.09}{12}=0.0075\)
• Number of payments: \(n=4\times 12=48\)

Step 2: Use the present value formula.

Step 3: Calculate.

Answer: The borrower received approximately \(\$14,061.95\).

10. Worked Example 3: Finding the installment amount

A laptop costs \(\$1800\). A buyer wants to pay it off in 12 equal monthly payments. The interest rate is 12% per year, compounded monthly. What is the monthly payment?

This is a loan problem, so the price now is the present value.

Step 1: Identify the values.

• \(PV=1800\)
• \(i=\frac{0.12}{12}=0.01\)
• \(n=12\)
• Need to find \(R\)

Step 2: Start with the present value formula.

Substitute values:

Step 3: Solve for \(R\).

Answer: The monthly payment is approximately \(\$159.93\).

11. Worked Example 4: Annuity due

A person deposits \(\$500\) at the beginning of each year into an account earning 8% per year. Find the amount after 5 years.

Because the payments are at the beginning of each period, this is an annuity due.

Step 1: Identify the values.

• \(R=500\)
• \(i=0.08\)
• \(n=5\)

Step 2: Use the annuity due future value formula.

Step 3: Calculate.

Answer: The account will contain approximately \(\$3167.96\).

12. Why geometric series appear in annuities

The reason annuity formulas work is that every payment is multiplied by a power of \((1+i)\) or divided by a power of \((1+i)\). These powers form a geometric pattern.

For future value, the terms look like

For present value, the terms look like

So annuities are a real-life application of geometric series.

13. Common mistakes to avoid

• Using the annual rate directly when payments are monthly. Always convert to rate per period.
• Using the wrong number of periods. If payments are monthly for 3 years, then \(n=36\), not 3.
• Mixing up present value and future value. Savings problems usually ask for future value; loan problems usually use present value.
• Forgetting whether payments happen at the beginning or end. Beginning means annuity due.
• Rounding too early. Keep more decimal places until the final answer.

14. Problem-solving steps for annuity questions

1. Decide whether it is a savings or loan problem.
2. Decide whether it is an ordinary annuity or an annuity due.
3. Write down \(R\), \(i\), and \(n\).
4. Make sure \(i\) and \(n\) match the payment interval.
5. Choose the correct formula.
6. Substitute carefully and solve.
7. Check whether the answer is reasonable.

15. Brief summary

An annuity is a series of equal payments made at regular intervals. Because each payment grows or is discounted by a factor involving \((1+i)\), annuities are modeled using geometric series.

The main formulas are:

For annuities due, multiply by \((1+i)\). These formulas help us model savings plans, installment payments, and loan repayments in real-life financial situations.