The constant is called a common difference and is usually denoted by d. The sequences will increase when d is positive or decrease when d is negative. Because the terms are increasing or decreasing by a constant amount, they will form a straight line when plotted on a graph.

Similarly, an arithmetic sequence is represented in explicit notation by the formula:

\displaystyle a_n=a_{0}+dn

\bm{n}

term number

\bm{a_n}

nth term

\bm{a_{0}}

initial (first) term

\bm{d}

common difference

The domain of any arithmetic sequence is a subset of the integers. The domain can begin from any non-negative integer but will most often begin at 0 or 1.

For an arithmetic sequence where a term and the common difference are known we can use a similar formula:

\displaystyle a_n=a_k+d\left(n-k\right)

\bm{a_n}

nth term

\bm{a_k}

a known (kth) term

\bm{d}

common difference

\bm{n}

term number (of the nth term)

\bm{k}

term number of the known (kth) term

Examples

Example 1

The points on this graph represent an arithmetic sequence:

-8

-6

-4

-2

a_n

Complete the following table of values:

n	1	2	3	4
a_n

Worked Solution

Create a strategy

The second row of the table represents the values of a_n, so we are looking for the output values of the points.

Apply the idea

n	1	2	3	4
a_n	3	-1	-5	-9

Reflect and check

These values of a_n represent the terms in the arithmetic sequence.

Write the explicit rule that represents the sequence.

Worked Solution

Create a strategy

The explicit rule of an arithmetic sequence is a_n=a_1+d\left(n-1\right). We need to find the common difference and the first term and then substitute them into the formula.

Apply the idea

To find the common difference, we can subtract any term from the following term:

d=-5-\left(-1\right)=-4

From the table or the graph, we see the first term of the sequence is a_1=3.

The explicit rule for finding any term in this sequence is a_n=3-4\left(n-1\right).

Reflect and check

We can simplify this formula if we want:

\displaystyle a_n	\displaystyle =	\displaystyle 3-4\left(n-1\right)	Given equation
	\displaystyle =	\displaystyle 3-4n+4	Distributive property
	\displaystyle =	\displaystyle -4n+7	Evaluate the addition

Find the 15th term of the sequence.

Worked Solution

Create a strategy

Since we are looking for the 15th term, n=15.

Apply the idea

\displaystyle a_n	\displaystyle =	\displaystyle 3-4\left(n-1\right)	Explicit formula
\displaystyle a_{15}	\displaystyle =	\displaystyle 3-4\left(15-1\right)	Substitute n=15
	\displaystyle =	\displaystyle 3-4\left(14\right)	Evaluate the subtraction
	\displaystyle =	\displaystyle 3-56	Evaluate the multiplication
	\displaystyle =	\displaystyle -53	Evaluate the subtraction

The 15th term is -53.

Reflect and check

Using the simplified form of the equation instead:

\displaystyle a_n	\displaystyle =	\displaystyle -4n+7	Simplified explicit formula
\displaystyle a_{15}	\displaystyle =	\displaystyle -4\left(15\right)+7	Substitute n=15
	\displaystyle =	\displaystyle -60+7	Evaluate the multiplication
	\displaystyle =	\displaystyle -53	Evaluate the addition

Example 2

An arithmetic sequence is defined by the rule T_n=T_{n-1}+1.95 where T_1=1.5.

Write the explicit rule of this sequence.

Worked Solution

Create a strategy

When given a recursive rule, we are given the common difference and the first term. We can take these from the recursive notation and substitute them into the explicit formula.

Apply the idea

In recursive notation, T_n=T_{n-1}+d, the common difference is the number being added. For this rule, d=1.95.

The explicit rule is T_n=1.5+1.95(n-1).

Find T_{10}.

Worked Solution

Create a strategy

It is much quicker to find T_{10} using the explicit rule. If we used the recursive rule, we would need to find all the terms between T_1 and T_{10}.

Apply the idea

\displaystyle T_n	\displaystyle =	\displaystyle 1.5+1.95\left(n-1\right)	Explicit formula from part (a)
\displaystyle T_{10}	\displaystyle =	\displaystyle 1.5+1.95\left(10-1\right)	Substitute n=10
\displaystyle T_{10}	\displaystyle =	\displaystyle 1.5+1.95\left(9\right)	Evaluate the subtraction
\displaystyle T_{10}	\displaystyle =	\displaystyle 1.5+17.55	Evaluate the multiplication
\displaystyle T_{10}	\displaystyle =	\displaystyle 19.05	Evaluate the addition

Example 3

In an arithmetic sequence, a_7=43 and a_{14}=85.

Find the common difference.

Worked Solution

Create a strategy

One way to solve this is by substituting the known information about each term into the explicit formula. Since there are 2 unknowns, a_1 and d, this will create a system of equations that we can solve.

Apply the idea

From the 7th term, we know a_7=43, n=7. Substituting this into the explicit formula, we have 43=a_1+d\left(7-1\right).

From the 14th term, we know a_{14}=85, n=14. Substituting this into the explicit formula, we have 85=a_1+d\left(14-1\right).

After simplifying, the system of equations is:

\begin{aligned}43&=a_1+6d\\85&=a_1+13d\end{aligned}

Subtracting the second equation from the first one, we have:

\begin{aligned}43&=a_1+6d\\-(85&=a_1+13d) \\ \hline -42&=-7d \\ 6&=d \\ d&=6 \end{aligned}

Reflect and check

Another way to solve this is by determining how many times we need to add the common difference to the 7th term to get the 14th term.

We would need to add the common difference 7 times. Using this, we can set up an equation.

\displaystyle 43+d+d+d+d+d+d+d	\displaystyle =	\displaystyle 85
\displaystyle 43+7d	\displaystyle =	\displaystyle 85
\displaystyle 7d	\displaystyle =	\displaystyle 42
\displaystyle d	\displaystyle =	\displaystyle 6

Find the first term.

Worked Solution

Create a strategy

We can use either of the equations from the system we set up before because the answer for the first term will be the same.

Apply the idea

For this problem, we will use the equation for the 7th term and substitute d=6.

\displaystyle 43	\displaystyle =	\displaystyle a_1+6\left(7-1\right)	Equation for 7th term
\displaystyle 43	\displaystyle =	\displaystyle a_1+36	Evaluate the subtraction and multiplication
\displaystyle 7	\displaystyle =	\displaystyle a_1	Subtract 36 from both sides

The first term of this sequence is a_1=7.

Reflect and check

Another method we could use is to determine how many times we would need to subtract the common difference to get from the 7th term to the 1st term.

We would need to subtract the common difference 6 times.

\displaystyle a_1	\displaystyle =	\displaystyle 43-d-d-d-d-d-d
	\displaystyle =	\displaystyle 43-6d
	\displaystyle =	\displaystyle 43-6\left(6\right)
	\displaystyle =	\displaystyle 43-36
	\displaystyle =	\displaystyle 7

Write the explicit rule for this sequence.

Worked Solution

Create a strategy

Now that we know the first term and the common difference, we can substitute them into the explicit formula.

Apply the idea

a_n=7+6\left(n-1\right)

Reflect and check

We can check the rule by finding the 7th and 14th terms.

a_7=7+6\left(7-1\right)=43

a_{14}=7+6\left(14-1\right)=85

These are the same as the given information, so our rule is correct.

Example 4

Tiles were stacked in a pattern as shown:

An image showing tiles stacked in a pattern with increasing stack height and corresponding number of tiles. Stack 1 has 1 tile. Stack 2 has 3 tiles. Stack 3 has 5 tiles. And stack 4 has 7 tiles

Describe the recursive pattern and write the explicit equation for the sequence using function notation.

Worked Solution

Create a strategy

The recursive pattern will describe the change in the number of tiles going from each stack to the next.

The explicit equation is a function that can be used to find the number of tiles for a stack of any height.

Apply the idea

The number of tiles from one stack to the next increases by 2. This tells us that the common difference is +2. We can also see that the first stack has one tile. So, the explicit equation for this pattern is a\left(n\right)=1+2\left(n-1\right)where a\left(n\right) is the number of tiles and n is the height of the stack.

Reflect and check

We can verify our rule by using it to find the number of tiles in the 4th stack.

a(4)=1+2(4-1)=7

In the image, there are 7 tiles in the 4th stack, so our rule is accurate.

A table of values representing the relationship between the height of the stack and the number of tiles was partially completed.

Height of stack	1	2	3	4	5	10	100
Number of tiles	1	3

Complete the table of values representing the relationship between the height of the stack and the number of tiles.

Worked Solution

Create a strategy

The number of tiles in the 3rd and 4th stacks are shown, and we can use the explicit function we created in part (b) to find the number of tiles in the 5th, 10th, and 100th stacks.

Apply the idea

a(5)=1+2(5-1)=9

a(10)=1+2(10-1)=19

a(100)=1+2(100-1)=199

Height of stack	1	2	3	4	5	10	100
Number of tiles	1	3	5	7	9	19	199

Reflect and check

We can double-check that the pattern found in part (a) is correct by comparing some of the results in the table to the images of the sequence of stacks.

State the domain.

Worked Solution

Create a strategy

Since we are using function notation for our sequence, it is important to consider the domain. For this function, the domain represents the height of the stacks as defined in part (c). The height of the stacks begins at 1 and increases by 1 each time.

Apply the idea

The domain is the set of natural numbers.

Reflect and check

The domain of a sequence will always be a subset of the integers, and natural numbers are a subset of the integers.

Idea summary

An arithmetic sequence has a common difference.

When the first term and common difference are known:

\displaystyle a_n=a_{0}+dn

\bm{n}

term number

\bm{a_n}

nth term

\bm{a_{0}}

initial (first) term

\bm{d}

common difference

When any term and the common difference are known:

\displaystyle a_n=a_k+d\left(n-k\right)

\bm{a_n}

nth term

\bm{a_k}

a known (kth) term

\bm{d}

common difference

\bm{n}

term number (of the nth term)

\bm{k}

term number of the known (kth) term

Geometric sequences

A geometric sequence is a pattern of multiplication, where the same number is multiplied to each subsequent term.

Geometric sequence

A sequence of numbers in which each consecutive pair of numbers has a common ratio

Example:

1, \, 3, \, 9, \,27, \ldots

A geometric sequence is represented by the formula:

\displaystyle g_n=g_k r^{\left(n-k\right)}

\bm{g_n}

nth term

\bm{g_k}

a known (kth) term

\bm{r}

common ratio

\bm{n}

term number

\bm{k}

term number of the known (kth) term

Just like arithmetic sequences, the domain of any geometric sequence is a subset of the integers, usually starting from 0 or 1.

Examples

Example 5

Consider the following geometric sequence:

n	1	2	3	4
t_n	-9	3.6	-1.44	0.576

Write the explicit rule for this sequence.

Worked Solution

Create a strategy

From the table, we see the first term is t_1=-9. To find the common ratio, we need to divide any term by the previous term.

Apply the idea

r=\dfrac{3.6}{-9}=-0.4

Substituting t_1 and r into the explicit formula for geometric sequences, we get:

t_n=-9\left(-0.4\right)^{n-1}

Reflect and check

Let's verify the rule by using it to determine t_4.

\displaystyle t_4	\displaystyle =	\displaystyle -9(-0.4)^{4-1}	Substitute n=4
	\displaystyle =	\displaystyle -9(-0.4)^3	Evaluate the subtraction
	\displaystyle =	\displaystyle -9(-0.064)	Evaluate the exponent
	\displaystyle =	\displaystyle 0.576	Evaluate the multiplication

Example 6

A geometric sequence is defined by T_n=5\cdot T_{n-1} where T_1=-1.

Write the explicit rule for this geometric sequence.

Worked Solution

Create a strategy

In a recursive formula, the number being multiplied to the previous term is r, and the first term is given. Once we identify these values from the recursive rule, we can substitute them into the explicit formula.

Apply the idea

r=5, \, T_1=-1

The explicit rule for this sequence is T_n=-1(5)^{n-1}.

Find the 6th term.

Worked Solution

Create a strategy

We will use the explicit formula to find T_6 because it will take less work and less time than it would if we used the recursive formula.

Apply the idea

\displaystyle T_6	\displaystyle =	\displaystyle -1(5)^{6-1}	Substitute n=6
	\displaystyle =	\displaystyle -1(5)^5	Evaluate the subtraction
	\displaystyle =	\displaystyle -3125	Evaluate the exponent and multiplication

Example 7

In a geometric sequence, a_2=-\dfrac{3}{5} and a_5=\dfrac{3}{625}.

Find the common ratio.

Worked Solution

Create a strategy

One way to solve this is by determining how many times we need to multiply by the common ratio to get from the 2nd term to the 5th term. From there, we can set up an equation and solve for r.

Apply the idea

To get from the 2nd term to the 5th term, we would need to multiply by the common ratio 3 times.

\displaystyle -\dfrac{3}{5}\cdot r \cdot r \cdot r	\displaystyle =	\displaystyle \dfrac{3}{625}	Given equation
\displaystyle -\dfrac{3}{5} r^3	\displaystyle =	\displaystyle \dfrac{3}{625}	Rewrite using exponents
\displaystyle r^3	\displaystyle =	\displaystyle -\dfrac{1}{125}	Divide both sides by -\dfrac{3}{5}
\displaystyle r	\displaystyle =	\displaystyle -\dfrac{1}{5}	Cube root property

Reflect and check

Another way to solve this is by creating a system of equations using the given information about each term.

From the second term, we know a_2=-\dfrac{3}{5}, n=2. Substituting this information into the explicit formula, we get:

-\dfrac{3}{5}=a_1(r)^{1}

From the fifth term, we know a_5=\dfrac{3}{625}, n=5. Substituting this into the explicit formula, we get:

\dfrac{3}{625}=a_1(r)^{4}

Next, we will solve the first equation for a_1.

a_1=\dfrac{-3}{5r}

Then, substitute this into the second equation and solve.

\displaystyle \dfrac{3}{625}	\displaystyle =	\displaystyle -\dfrac{3}{5r}\cdot r^4	Substitute a_1=\dfrac{-3}{5r}
\displaystyle \dfrac{3}{625}	\displaystyle =	\displaystyle -\dfrac{3}{5}r^3	Quotient of powers property
\displaystyle -\dfrac{1}{125}	\displaystyle =	\displaystyle r^3	Divide both sides by -\dfrac{3}{5}
\displaystyle -\dfrac{1}{5}	\displaystyle =	\displaystyle r	Cube root property

Find the first term.

Worked Solution

Create a strategy

We can use the explicit formula with the information given about one of the terms to solve for the common ratio. For this problem, we will use a_2=-\dfrac{3}{5}, \, n=2, \, r=-\dfrac{1}{5}.

Apply the idea

\displaystyle a_n	\displaystyle =	\displaystyle a_1\left(-\dfrac{1}{5}\right)^{n-1}	Explicit formula
\displaystyle -\dfrac{3}{5}	\displaystyle =	\displaystyle a_1\left(-\dfrac{1}{5}\right)^{2-1}	Substitute a_2=-\dfrac{3}{5}, n=2, and r=-\dfrac{1}{5}
\displaystyle -\dfrac{3}{5}	\displaystyle =	\displaystyle -\dfrac{1}{5}a_1	Evaluate the subtraction
\displaystyle 3	\displaystyle =	\displaystyle a_1	Multiply both sides by -5

The first term is 3.

Reflect and check

Since we knew the second term and the common ratio, we could have divided the second term by the common ratio to find the first term.

\displaystyle a_1	\displaystyle =	\displaystyle -\dfrac{3}{5}\div r
	\displaystyle =	\displaystyle -\dfrac{3}{5}\div -\dfrac{1}{5}
	\displaystyle =	\displaystyle -\dfrac{3}{5}\times -\dfrac{5}{1}
	\displaystyle =	\displaystyle 3

Write the explicit rule for the sequence.

Worked Solution

Create a strategy

Now that we know the first term and the common ratio, we can substitute them into the explicit formula for geometric sequences.

Apply the idea

a_n=3\left(-\dfrac{1}{5}\right)^{n-1}

Reflect and check

We can verify our rule by using it to confirm the 2nd and 5th terms.

a_2=3\left(-\dfrac{1}{5}\right)^{2-1}=-\dfrac{3}{5}

\begin{aligned} a_5&=3\left(-\dfrac{1}{5}\right)^{5-1} \\ &=3\left(-\dfrac{1}{5}\right)^4\\ &=3\left(\dfrac{1}{625}\right) \\ &=\dfrac{3}{625} \end{aligned}

Idea summary

A geometric sequence has a common ratio.

\displaystyle g_n=g_k r^{\left(n-k\right)}

\bm{g_n}

nth term

\bm{g_k}

a known (kth) term

\bm{r}

common ratio

\bm{n}

term number

\bm{k}

term number of the known (kth) term

Outcomes

2.1.A

Express arithmetic sequences found in mathematical and contextual scenarios as functions of the whole numbers.

2.1.B

Express geometric sequences found in mathematical and contextual scenarios as functions of the whole numbers.

2.1 Change in arithmetic and geometric sequences

Introduction

Ideas

Arithmetic sequences

Examples

Example 1

Example 2

Example 3

Example 4

Geometric sequences

Examples

Example 5

Example 6

Example 7

Outcomes

2.1.A

2.1.B

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