Day 28 – Arithmetic Sequence
Name ____________________________________
Find the next two terms of each sequence. Then describe the pattern. The equations will be completed
later.
1, 3, 5, 7, 9, _____, _____
Description: ____________
Equation: ___________
2, 7, 12, 17, 22, ______, ______
Description: ____________
Equation: ___________
-416, -323, -230, -137, _____, _____
Description: ____________
Equation: ___________
-2, -5, -8, -11, _____, ______
Description: ____________
Equation: ___________
All of the patterns above are called arithmetic sequences. Hopefully you noticed something about their
pattern that makes them similar. Complete the sentence below by writing a description of the pattern
you noticed above. (If you need help, look in the textbook in section 11.1).
Arithmetic sequences are sequences of numbers where_______________________________________
____________________________________________________________________________________.
Let’s look more closely at the first pattern 1, 3, 5, 7, 9… Suppose the domain is the position of a term (1,
2, 3, 4, etc.) and the range is the term (1, 3, 5, 7, 9, etc.).
Make a graph of the points that are made (position, term)
with the pattern.
What quadrant(s) are these points in? Why?
What kind of graph do you have?
Write an equation for the graph.
How does this equation relate to the graph? How does this equation relate to the pattern?
Do you think the graphs of other arithmetic sequences would look similar? ______ Why or why not?
__________________________________________________________________________________
Checkpoint 1: Stop at this point for class comparison. If you are done before others, make equations for
the other three patterns listed at the top.
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Now, everyone should have the same equation ___________________ for the pattern 1, 3, 5, 7, 9…
However, we have a problem. This equation makes us use a number that is not on our pattern (-1).
Let’s say we want to use 1 as a starting point instead of -1 (since 1 is our first term in our sequence).
So, suppose our equation is now y = 2x + 1.
See if this works for the pattern.
Try x = 1 (for the first term).
Try x = 2 (for the second term).
Try x = 3 (for the third term).
What do you notice?
Our new equation y = 2x + 1 makes our pattern shift over one term (one x value). This means we are
adding one too many times!
Let’s alter the equation slightly to y = 2(x – 1) + 1. This will shift all the x values (just like we’ve done
before) and we won’t be adding the extra value of d.
See if this works for the pattern.
Try x = 1 (for the first term).
Try x = 2 (for the second term).
Try x = 3 (for the third term).
What do you notice?
Now, we have an equation y = 2 (x – 1) + 1 that uses the first term and the common difference (slope).
This can be used to make any equation for any arithmetic sequence. Let’s use d = common difference,
a1 = first term, and an = nth term.
So, the nth term of any arithmetic sequence can be found by
an = a1 + (n – 1) d
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Checkpoint 2: Find the rule/equation for the 2nd pattern using the formula above.
Now that you know arithmetic sequences need a common difference (number added or subtracted to
the pattern) and you know how to find the nth term (or equation) for any arithmetic sequence, let’s try
some problems.
Example 1: Is the sequence arithmetic? If so, what’s the common difference? If not, why not?
A) 2, -3, -8, -13,…
B) 1, 5/4, 3/2, 7/4,…
C) an = n2
D) an = 4n + 3
Example 2: Write the first 5 terms if a1  2 and 𝑑 = −7.
Checkpoint 3: Let’s make sure we are on the right track with examples 1 and 2.
Example 3: Write the rule/equation for the given information.
A) a1 = 2, d = 3
B) a1  2, a2  9
Example 4: Find the indicated term of each arithmetic sequence. First find the equation, then plug in
your n.
A) a1 = -4, d = 6, n = 9
B) a20 for a1 = 15, d = -8
Checkpoint 4: Let’s make sure we got the answers to examples 3 and 4.
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Example 5: Write the equation for the nth term of each arithmetic sequence.
A) 31, 17, 3, ….
Now, the next two are slightly different. I will give you a term and the d – but the term isn’t the first
one. You need to work backwards to find the first term.
B) a7 = 21, d = 5
We know that an = a1 + (n – 1) d
So, using the given information, we have 21 = a1 + (7 – 1) 5
Simplify and solve for a1.
Now, find the equation.
C) Follow the steps with this information: a6 =12, d = 8.
Checkpoint 5: Did we follow that?
Example 6: Find the arithmetic means (missing terms) in each sequence.
A) 6, ____, ____, ____, 42
B) 24, _____, _____, _____, _____, -1
Challenge: Let’s do this one together.
Use the given information to write an equation that represents the nth term in each arithmetic
sequence.
The 19th term of the sequence is 131. The 61st term is 509.
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Answer Key
Find the next two terms of each sequence. Then describe the pattern. The equations will be completed
later.
1, 3, 5, 7, 9, 11, 13 Description: Add 2 to the previous term
2, 7, 12, 17, 22, 27, 32 Description: Add 5 to the previous term
-416, 323, 230, 137, 44, 49, Description: Add 93 to the previous term
-2, 5, 8, 11, 14, 17,
All of the patterns above are called arithmetic sequences. Hopefully you noticed something about their
pattern that makes them similar. Complete the sentence below by writing a description of the pattern
you noticed above.
Arithmetic sequences are sequences of numbers where the difference between one term and the next is
a constant.
Let’s look more closely at the first pattern 1, 3, 5, 7, 9… Suppose the domain is the position of a term (1,
2, 3, 4, etc.) and the range is the term (1, 3, 5, 7, 9, etc.).
Make a graph of the points that are made (position, term) with the pattern.
What quadrant(s) are these points in? Why?
We see all points are in the first quadrant. That’s because all domain and range values are positive
numbers.
What kind of graph do you have?
We have a linear graph here.
Write an equation for the graph
𝑦 = 2𝑥 – 1
How does this equation relate to the graph? How does this equation relate to the pattern?
This equation represents the pattern of the sequence.
Do you think the graphs of other arithmetic sequences would look similar? Why or why not?
Yes, the graphs of other arithmetic sequences would look similar, because all graphs of arithmetic
sequences are linear graphs.
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Now, everyone should have the same equation 𝑦 = 2𝑥 − 1 for the pattern 1, 3, 5, 7, 9…
However, we have a problem. This equation makes us use a number that is not on our pattern (1).
Let’s say we want to use 1 as a starting point instead of 1 (since 1 is our first term in our sequence).
So, suppose our equation is now y = 2x + 1.
See if this works for the pattern.
Try x = 1 (for the first term).
Try x = 2 (for the second term).
Try x = 3 (for the third term).
What do you notice?
Our new equation y = 2x + 1 makes our pattern shift over one term (one x value). This means we are
adding one too many times!
Let’s alter the equation slightly to y = 2(x – 1) + 1. This will shift all the x values (just like we’ve done
before) and we won’t be adding the extra value of d.
See if this works for the pattern.
Try x = 1 (for the first term).
Try x = 2 (for the second term).
Try x = 3 (for the third term).
What do you notice?
We notice it works now
Now, we have an equation 𝑦 = 2(𝑥 − 1) + 1 that uses the first term and the common difference
(slope).
This can be used to make any equation for any arithmetic sequence.
Let’s use d = common difference, a1= first term, and an= nth term.
So, the nth term of any arithmetic sequence can be found by 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑
Checkpoint 2: Find the rule/equation for the 2nd pattern using the formula above.
𝑎𝑛 = 2 + 5(𝑛 − 1)
Now that you know arithmetic sequences need a common difference (number added or subtracted to
the pattern) and you know how to find the nth term (or equation) for any arithmetic sequence, let’s try
some problems.
Example 1: Is the sequence arithmetic? If so, what’s the common difference? If not, why not?
A) 2, 3, 8, 13
This sequence is arithmetic. Its common difference is 5.
B) 1, 5/4, 3/2, 7/4
The sequence is arithmetic. The common difference is ¼.
C) 𝑎𝑛 = 𝑛2
The sequence is not arithmetic, because the common the difference between one term and the next is
not a constant.
D) 𝑎𝑛 = 4𝑛 + 3
The sequence is arithmetic. The common difference is 4.
Example 2: Write the first 5 terms if a and 1 = 2 d =− 7
𝑎1 = 2
𝑎2 = −5
𝑎3 = −12
𝑎4 = −19
𝑎5 = −26
Algebra1Teachers @ 2015
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Day 28 – Arithmetic Sequence
Name ____________________________________
Checkpoint 3: Let’s make sure we are on the right track with examples 1 and 2.
Example 3: Write the rule/equation for the given information.
A) 𝑎1 = 2, 𝑑 = 3
𝑎𝑛 = 2 + 3(𝑛 − 1)
B) 𝑎1 = 2, 𝑎2 = 9
𝑎𝑛 = 2 + 7(𝑛 − 1)
Example 4: Find the indicated term of each arithmetic sequence. First find the equation, then plug in
your n.
A) 𝑎1 = −4, 𝑑 = 6, 𝑛 = 9
𝑎𝑛 = − 4 + 6(𝑛 − 1)
𝑎9 = − 4 + 6(9 − 1) = 44
B) 𝑎20 𝑓𝑜𝑟 𝑎1 = 15, 𝑑 = −8
𝑎𝑛 = 15 − 8(𝑛 − 1)
𝑎20 = 15 − 8(20 − 1) = −137
Checkpoint 4: Let’s make sure we got the answers to examples 3 and 4.
Example 5: Write the equation for the nth term of each arithmetic sequence.
A) 31, 17, 3…
𝑎𝑛 = 31 − 14(𝑛 − 1)
Now, the next two are slightly different. I will give you a term and the d – but the term isn’t the first one.
You need to work backwards to find the first term.
B) 𝑎7 = 21, 𝑑 = 5
We know that 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑
So, using the given information, we have 21 = 𝑎1 + (7 − 1)5
Simplify and solve for 𝑎1 .
𝑎1 = −9
Now, find the equation.
𝑎𝑛 = −9 + 5(𝑛 − 1)
C) Follow the steps with this information: 𝑎6 = 12, 𝑑 = 8
12 = 𝑎1 + (6 − 1)8
𝑎1 = −28
𝑎𝑛 = −28 + 8(𝑛 − 1)
Checkpoint 5: Did we follow that?
Yes, we did!
Example 6: Find the arithmetic means (missing terms) in each sequence.
A) 6, 15, 24, 33, 42
B) 24, 19, 14, 9, 4, 1
Challenge: Let’s do this one together.
Use the given information to write an equation that represents the nth term in each arithmetic
sequence. The 19th term of the sequence is 131. The term is th 61st 509.
𝑎𝑛 = −31 + (𝑛 − 1)9
Algebra1Teachers @ 2015
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