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This pdf covers the definition of Arithmetic Sequences and how to identify them from examples with step-by-step explanation and how to find the term in Arithmetic Sequences.

1.
Arithmetic Sequences

A simple way to generate a sequence is to start with a number a, and add to it a fixed

constant d, over and over again. This type of sequence is called an arithmetic sequence.

Definition: An arithmetic sequence is a sequence of the form

a, a + d, a + 2d, a + 3d, a + 4d, …

The number a is the first term, and d is the common difference of the

sequence. The nth term of an arithmetic sequence is given by

an = a + (n – 1)d

The number d is called the common difference because any two consecutive terms of an

arithmetic sequence differ by d, and it is found by subtracting any pair of terms an and

an+1. That is

d = an+1 – an

Is the Sequence Arithmetic?

Example 1: Determine whether or not the sequence is arithmetic. If it is arithmetic, find

the common difference.

(a) 2, 5, 8, 11, …

(b) 1, 2, 3, 5, 8, …

Solution (a): In order for a sequence to be arithmetic, the differences between

each pair of adjacent terms should be the same. If the differences

are all the same, then d, the common difference, is that value.

Step 1: First, calculate the difference between each pair of adjacent

terms.

5–2=3

8–5=3

11 – 8 = 3

Step 2: Now, compare the differences. Since each pair of adjacent terms

has the same difference 3, the sequence is arithmetic and the

common difference d = 3 .

By: Crystal Hull

A simple way to generate a sequence is to start with a number a, and add to it a fixed

constant d, over and over again. This type of sequence is called an arithmetic sequence.

Definition: An arithmetic sequence is a sequence of the form

a, a + d, a + 2d, a + 3d, a + 4d, …

The number a is the first term, and d is the common difference of the

sequence. The nth term of an arithmetic sequence is given by

an = a + (n – 1)d

The number d is called the common difference because any two consecutive terms of an

arithmetic sequence differ by d, and it is found by subtracting any pair of terms an and

an+1. That is

d = an+1 – an

Is the Sequence Arithmetic?

Example 1: Determine whether or not the sequence is arithmetic. If it is arithmetic, find

the common difference.

(a) 2, 5, 8, 11, …

(b) 1, 2, 3, 5, 8, …

Solution (a): In order for a sequence to be arithmetic, the differences between

each pair of adjacent terms should be the same. If the differences

are all the same, then d, the common difference, is that value.

Step 1: First, calculate the difference between each pair of adjacent

terms.

5–2=3

8–5=3

11 – 8 = 3

Step 2: Now, compare the differences. Since each pair of adjacent terms

has the same difference 3, the sequence is arithmetic and the

common difference d = 3 .

By: Crystal Hull

2.
Example 1 (Continued):

Solution (b):

Step 1: Calculate the difference between each pair of adjacent terms.

2–1=1

3–2=1

5–3=2

8–5=3

Step 2: Compare the differences. Since the differences between each

pair of adjacent terms are not all the same, the sequence is not

arithmetic.

An arithmetic sequence is determined completely by the first term a, and the common

difference d. Thus, if we know the first two terms of an arithmetic sequence, then we can

find the equation for the nth term.

Finding the Terms of an Arithmetic Sequence:

Example 2: Find the nth term, the fifth term, and the 100th term, of the arithmetic

sequence determined by a = 2 and d = 3.

Solution: To find a specific term of an arithmetic sequence, we use the formula

for finding the nth term.

Step 1: The nth term of an arithmetic sequence is given by

an = a + (n – 1)d.

So, to find the nth term, substitute the given values a = 2 and

d = 3 into the formula.

an = 2 + (n – 1)3

Step 2: Now, to find the fifth term, substitute n = 5 into the equation for

the nth term.

a5 = 2 + (5 – 1)3

= 14

Step 3: Finally, find the 100th term in the same way as the fifth term.

a100 = 2 + (100 – 1)3

= 299

By: Crystal Hull

Solution (b):

Step 1: Calculate the difference between each pair of adjacent terms.

2–1=1

3–2=1

5–3=2

8–5=3

Step 2: Compare the differences. Since the differences between each

pair of adjacent terms are not all the same, the sequence is not

arithmetic.

An arithmetic sequence is determined completely by the first term a, and the common

difference d. Thus, if we know the first two terms of an arithmetic sequence, then we can

find the equation for the nth term.

Finding the Terms of an Arithmetic Sequence:

Example 2: Find the nth term, the fifth term, and the 100th term, of the arithmetic

sequence determined by a = 2 and d = 3.

Solution: To find a specific term of an arithmetic sequence, we use the formula

for finding the nth term.

Step 1: The nth term of an arithmetic sequence is given by

an = a + (n – 1)d.

So, to find the nth term, substitute the given values a = 2 and

d = 3 into the formula.

an = 2 + (n – 1)3

Step 2: Now, to find the fifth term, substitute n = 5 into the equation for

the nth term.

a5 = 2 + (5 – 1)3

= 14

Step 3: Finally, find the 100th term in the same way as the fifth term.

a100 = 2 + (100 – 1)3

= 299

By: Crystal Hull

3.
Example 3: Find the common difference, the fifth term, the nth term, and the 100th

term of the arithmetic sequence.

(a) 4, 14, 24, 34, …

15 9 21

(b) t + 3, t + , t + , t + , ...

4 2 4

Solution (a): In order to find the nth and 100th terms, we will first have to

determine what a and d are. We will then use the formula for

finding the nth term.

Step 1: First, we will determine what a and d are. The number a is

always the first term of the sequence, so

a=4

The difference between any pair of adjacent terms should be the

same because the sequence is arithmetic, so we can choose any

one pair to find the common difference d. If we choose the first

two terms then

d = 14 – 4

= 10

Step 2: Since we are given the fourth term, we can add the common

difference d = 10 to it to get the fifth term.

a5 = 34 + 10

= 44

Step 3: Now to find the nth term, substitute a = 4 and d = 10 into the

formula for the nth term.

an = 4 + (n – 1)10

Step 4: Finally, substitute n = 100 into the equation for the nth term to

get the 100th term.

a100 = 4 + (100 – 1)10

= 994

By: Crystal Hull

term of the arithmetic sequence.

(a) 4, 14, 24, 34, …

15 9 21

(b) t + 3, t + , t + , t + , ...

4 2 4

Solution (a): In order to find the nth and 100th terms, we will first have to

determine what a and d are. We will then use the formula for

finding the nth term.

Step 1: First, we will determine what a and d are. The number a is

always the first term of the sequence, so

a=4

The difference between any pair of adjacent terms should be the

same because the sequence is arithmetic, so we can choose any

one pair to find the common difference d. If we choose the first

two terms then

d = 14 – 4

= 10

Step 2: Since we are given the fourth term, we can add the common

difference d = 10 to it to get the fifth term.

a5 = 34 + 10

= 44

Step 3: Now to find the nth term, substitute a = 4 and d = 10 into the

formula for the nth term.

an = 4 + (n – 1)10

Step 4: Finally, substitute n = 100 into the equation for the nth term to

get the 100th term.

a100 = 4 + (100 – 1)10

= 994

By: Crystal Hull

4.
Example 3 (Continued):

Solution (b):

Step 1: Calculate a and d.

a=t+3

⎛ 15 ⎞

d = ⎜ t + ⎟ − ( t + 3)

⎝ 4⎠

15

= t + −t −3

4

15

= −3

4

3

=

2

Step 2: The fifth term is the fourth term plus the common difference.

Therefore,

⎛ 21 ⎞ 3

a5 = ⎜ t + ⎟ +

⎝ 4⎠ 2

24

=t+

4

=t+6

3

Step 3: Now, substitute a = t + 3, d = into the formula for the nth term.

2

3

an = ( t + 3) + ( n − 1)

2

Step 4: Finally, substitute n = 100 into the equation for the nth term that

we just found.

3

an = ( t + 3) + (100 − 1)

2

3

= t + 3 + ( 99 )

2

303

=t+

2

By: Crystal Hull

Solution (b):

Step 1: Calculate a and d.

a=t+3

⎛ 15 ⎞

d = ⎜ t + ⎟ − ( t + 3)

⎝ 4⎠

15

= t + −t −3

4

15

= −3

4

3

=

2

Step 2: The fifth term is the fourth term plus the common difference.

Therefore,

⎛ 21 ⎞ 3

a5 = ⎜ t + ⎟ +

⎝ 4⎠ 2

24

=t+

4

=t+6

3

Step 3: Now, substitute a = t + 3, d = into the formula for the nth term.

2

3

an = ( t + 3) + ( n − 1)

2

Step 4: Finally, substitute n = 100 into the equation for the nth term that

we just found.

3

an = ( t + 3) + (100 − 1)

2

3

= t + 3 + ( 99 )

2

303

=t+

2

By: Crystal Hull

5.
Partial Sums of an Arithmetic Sequence:

To find a formula for the sum, Sn, of the first n terms of an arithmetic sequence, we can

write out the terms as

S n = a + ( a + d ) + ( a + 2d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦ .

This same sum can be written in reverse as

S n = an + ( an − d ) + ( an − 2d ) + ... + ⎡⎣ an − ( n − 1) d ⎤⎦

Now, add the corresponding terms of these two expressions for Sn to get

Sn = a + ( a + d ) + ( a + 2d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦

Sn = an + ( an − d ) + ( an − 2d ) + ... + ⎡⎣ an − ( n − 1) d ⎤⎦

2 Sn = ( a + an ) + ( a + an ) + ( a + an ) + ... + ( a + an )

The right hand side of this expression contains n terms, each equal to a + an, so

2 S n = n ( a + an )

n

Sn = ( a + an ) .

2

Definition: For the arithmetic sequence an = a + ( n − 1) d , the nth partial sum

S n = a + ( a + d ) + ( a + 2d ) + ( a + 3d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦

is given by either of the following formulas.

n

1. S n = ⎡ 2a + ( n − 1) d ⎤⎦

2⎣

⎛ a + an ⎞

2. S n = n ⎜ ⎟

⎝ 2 ⎠

By: Crystal Hull

To find a formula for the sum, Sn, of the first n terms of an arithmetic sequence, we can

write out the terms as

S n = a + ( a + d ) + ( a + 2d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦ .

This same sum can be written in reverse as

S n = an + ( an − d ) + ( an − 2d ) + ... + ⎡⎣ an − ( n − 1) d ⎤⎦

Now, add the corresponding terms of these two expressions for Sn to get

Sn = a + ( a + d ) + ( a + 2d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦

Sn = an + ( an − d ) + ( an − 2d ) + ... + ⎡⎣ an − ( n − 1) d ⎤⎦

2 Sn = ( a + an ) + ( a + an ) + ( a + an ) + ... + ( a + an )

The right hand side of this expression contains n terms, each equal to a + an, so

2 S n = n ( a + an )

n

Sn = ( a + an ) .

2

Definition: For the arithmetic sequence an = a + ( n − 1) d , the nth partial sum

S n = a + ( a + d ) + ( a + 2d ) + ( a + 3d ) + ... + ⎡⎣ a + ( n − 1) d ⎤⎦

is given by either of the following formulas.

n

1. S n = ⎡ 2a + ( n − 1) d ⎤⎦

2⎣

⎛ a + an ⎞

2. S n = n ⎜ ⎟

⎝ 2 ⎠

By: Crystal Hull

6.
The nth partial sum of an arithmetic sequence can also be written using summation

n

∑ ki − c

i =1

represents the sum of the first n terms of an arithmetic sequence having the first term

a = k(1) + c = k + c and the nth term an = k(n) + c = kn + c. We can find this sum with

the second formula for Sn given above.

Example 4: Find the partial sum Sn of the arithmetic sequence that satisfies the given

conditions.

(a) a = 6, d = 3, and n = 7

14

(b) ∑ 2i − 7

i =1

Solution (a): To find the nth partial sum of an arithmetic sequence, we can use

either of the formulas

n ⎛ a + an ⎞

Sn = ⎡⎣ 2a + ( n − 1) d ⎤⎦ or S n = n ⎜ ⎟

2 ⎝ 2 ⎠

Step 1: To use the first formula for the nth partial sum, we only need to

substitute the given values a = 6, d = 3, and n = 7 into the

equation.

n

Sn = ⎡ 2a + ( n − 1) d ⎤⎦

2⎣

7

S7 = ⎡ 2 ( 6 ) + ( 7 − 1) 3⎤⎦

2⎣

7

= [12 + 18]

2

= 105

By: Crystal Hull

n

∑ ki − c

i =1

represents the sum of the first n terms of an arithmetic sequence having the first term

a = k(1) + c = k + c and the nth term an = k(n) + c = kn + c. We can find this sum with

the second formula for Sn given above.

Example 4: Find the partial sum Sn of the arithmetic sequence that satisfies the given

conditions.

(a) a = 6, d = 3, and n = 7

14

(b) ∑ 2i − 7

i =1

Solution (a): To find the nth partial sum of an arithmetic sequence, we can use

either of the formulas

n ⎛ a + an ⎞

Sn = ⎡⎣ 2a + ( n − 1) d ⎤⎦ or S n = n ⎜ ⎟

2 ⎝ 2 ⎠

Step 1: To use the first formula for the nth partial sum, we only need to

substitute the given values a = 6, d = 3, and n = 7 into the

equation.

n

Sn = ⎡ 2a + ( n − 1) d ⎤⎦

2⎣

7

S7 = ⎡ 2 ( 6 ) + ( 7 − 1) 3⎤⎦

2⎣

7

= [12 + 18]

2

= 105

By: Crystal Hull

7.
Example 4 (Continued):

Solution (b): This is the sum of the first fourteen terms of the arithmetic

sequence having an = 2n – 7.

Step 1: Since the partial sum is given in summation notation, we must

first find a and an. From the given information we know k = 2,

c = –7, and n = 14, so

a=k +c

= 2 + (−7)

= −5

an = kn + c

a14 = 2(14) + (−7)

= 21

Step 2: Now that we know a = -5, n = 14, and a14 = 21, we can substitute

these values into the second formula for the nth partial sum to

find the fourteenth partial sum.

⎛ a + a14 ⎞

S14 = n ⎜ ⎟

⎝ 2 ⎠

⎛ −5 + 21 ⎞

= 14 ⎜ ⎟

⎝ 2 ⎠

= 112

Example 5: Find the sum of the first 37 even numbers.

Solution:

Step 1: First, we must find the values a, d, and n. Since the first even

number is zero, a = 0. The next even number is 2, so

d = 2 – 0 = 2. Since we are told to find the sum of the first 37

even numbers, n = 37.

By: Crystal Hull

Solution (b): This is the sum of the first fourteen terms of the arithmetic

sequence having an = 2n – 7.

Step 1: Since the partial sum is given in summation notation, we must

first find a and an. From the given information we know k = 2,

c = –7, and n = 14, so

a=k +c

= 2 + (−7)

= −5

an = kn + c

a14 = 2(14) + (−7)

= 21

Step 2: Now that we know a = -5, n = 14, and a14 = 21, we can substitute

these values into the second formula for the nth partial sum to

find the fourteenth partial sum.

⎛ a + a14 ⎞

S14 = n ⎜ ⎟

⎝ 2 ⎠

⎛ −5 + 21 ⎞

= 14 ⎜ ⎟

⎝ 2 ⎠

= 112

Example 5: Find the sum of the first 37 even numbers.

Solution:

Step 1: First, we must find the values a, d, and n. Since the first even

number is zero, a = 0. The next even number is 2, so

d = 2 – 0 = 2. Since we are told to find the sum of the first 37

even numbers, n = 37.

By: Crystal Hull

8.
Example 5 (Continued):

Step 2: Now that we know a = 0, d = 2, and n = 37 we can solve this

problem the same way as in the previous example. First find

a37, and then substitute the values for a, d, and a37 into the

equation for the nth partial sum. Thus,

a37 = 0 + ( 37 − 1) 2

= 18

⎛ 0 + 18 ⎞

S37 = 37 ⎜ ⎟

⎝ 2 ⎠

= 363

Example 6: A partial sum of an arithmetic sequence is given. Find the sum.

1 + 8 + 15 + … + 78

Solution:

Step 1: As in the previous example, we must first find a, d, and n. The

values a and d are easy to find.

a=1

d=8–1

=7

Now, finding n is a bit more work because we are not explicitly

told how many numbers we will be summing. We know a and d,

and we know the nth term, so we will substitute these values into

the formula for the nth term of a sequence.

an = a + ( n − 1) d

78 = 1 + ( n − 1) 7

Now solve for n.

77 = ( n − 1) 7

11 = n − 1

12 = n

Therefore, we will be summing twelve terms and 78 = a12 .

By: Crystal Hull

Step 2: Now that we know a = 0, d = 2, and n = 37 we can solve this

problem the same way as in the previous example. First find

a37, and then substitute the values for a, d, and a37 into the

equation for the nth partial sum. Thus,

a37 = 0 + ( 37 − 1) 2

= 18

⎛ 0 + 18 ⎞

S37 = 37 ⎜ ⎟

⎝ 2 ⎠

= 363

Example 6: A partial sum of an arithmetic sequence is given. Find the sum.

1 + 8 + 15 + … + 78

Solution:

Step 1: As in the previous example, we must first find a, d, and n. The

values a and d are easy to find.

a=1

d=8–1

=7

Now, finding n is a bit more work because we are not explicitly

told how many numbers we will be summing. We know a and d,

and we know the nth term, so we will substitute these values into

the formula for the nth term of a sequence.

an = a + ( n − 1) d

78 = 1 + ( n − 1) 7

Now solve for n.

77 = ( n − 1) 7

11 = n − 1

12 = n

Therefore, we will be summing twelve terms and 78 = a12 .

By: Crystal Hull

9.
Example 6 (Continued):

Step 2: Now that we know a = 1, n = 12, and a12 = 78 we can solve this

problem the same way as in example 4. Substitute the values for

a, d, and a12 into the formula for the nth partial sum.

⎛ 1 + 78 ⎞

S12 = 12 ⎜ ⎟

⎝ 2 ⎠

= 474

By: Crystal Hull

Step 2: Now that we know a = 1, n = 12, and a12 = 78 we can solve this

problem the same way as in example 4. Substitute the values for

a, d, and a12 into the formula for the nth partial sum.

⎛ 1 + 78 ⎞

S12 = 12 ⎜ ⎟

⎝ 2 ⎠

= 474

By: Crystal Hull