1. Arithmetic Sequences

2. Aim: What are arithmetic sequences & how can we write functions to represent them?
Do Now: Write the next 3 numbers that continue the pattern.
9, 15, 21, 27, …
12, 7, 2, -3, …

3. A is an ordered list of numbers that form a pattern.
Sequence
Each number in a sequence is called a
term.
Definition
To find the common difference (d), subtract any term from one that follows it.
A(1)
A(2)
A(3)
A(4)
A(5) 3 3 3 3

4. Definition In an the distance between consecutive terms is constant.
Arithmetic Sequence,
The constant (d) is called the
Definition
Common Difference.
To find the common difference (d), subtract any term from one that follows it.
A(1)
A(2)
A(3)
A(4)
A(5) 3 3 3 3

5. Find the first term and the common difference of each arithmetic sequence.
Examples:
First term A(1): 4
Common difference (d):
= 9 – 4 = 5
First term A(1): 34
Common difference (d): -7
BE CAREFUL: ALWAYS CHECK TO MAKE SURE THE DIFFERENCE IS THE SAME BETWEEN EACH TERM !

6. Now you try!
Find the first term and the common difference of each of these arithmetic sequences.
a) 1, -4, -9, -14, ….
b) 11, 23, 35, 47, ….
c) 3x-8, x-8, -x-8, -3x-8
d) s-4, 3s-3, 5s-2, 7s-1, …..

7. Answers with solutions
A(1) = and
d = A(2) - A(1) = = - 5
b) 11, 23, 35, 47, ….
A(1) = and
d = A(2) - A(1) = = 12
c) 3x-8, x-8, -x-8, -3x-8
A(1) = 3x and
d = A(2) - A(1)= x – 8 – (3x – 8) = - 2x
A(1) = s and
d = A(2) - A(1)= 3s – 3 – (s – 4) = 2s + 1
d) s-4, 3s-3, 5s-2, 7s-1, …..

8. We can write arithmetic sequences as functions in two ways:
A recursive formula is a function that relates each term to the term before it.
3, 7, 11, 15, … We know A(1) = 3 and d = 4
A(1) = 3
A(2) = A(1) + d = = 7
A(3) = A(2) + d = = 11
A(4) = A(3) + d = = 15
A(n) = A(n - 1) + 4

9. The first term of an arithmetic sequence is A(1)
The first term of an arithmetic sequence is A(1) . We add (d) to get the next term. There is a pattern, therefore there is a formula we can use to give use any term that we need without listing the whole sequence .
An explicit formula is a function that relates each term in the sequence to the term number.
3, 7, 11, 15, … We know A(1) = 3 and d = 4
A(1) = 3
A(2) = A(1) + d = = 7
A(3) = A(1) + d + d = A(1) + 2d = 3+2(4) = 11
A(4) = A(1) + d + d + d = A(1) + 3d = 3+3(4) = 15

10. The first term of an arithmetic sequence is A(1)
The first term of an arithmetic sequence is A(1) . We add (d) to get the next term. There is a pattern, therefore there is a formula we can use to give use any term that we need without listing the whole sequence .
The nth term of an arithmetic sequence is given by:
A(n) = A(1) + (n – 1) d
The last # in the sequence/or the # you are looking for
The position the term is in
First term
The common difference

11. The 14th term in this sequence is the number 43!
Examples:
Find the 14th term of the arithmetic sequence
4, 7, 10, 13,……
A(n) = A(1) + (n – 1) d
You are looking for the term!
A(14) =
The 14th term in this sequence is the number 43!

12. Now you try! a) 1, 7, 13, 19 …. b) x+10, x+7, x+4, x+1, ….
Find the 10th and 25th term given the following information. Make sure to derive the general formula first and then list ehat you have been provided.
a) 1, 7, 13, 19 ….
b) x+10, x+7, x+4, x+1, ….
c) The first term is 3 and the common difference is -21
d) The second term is 8 and the common difference is 3

13. Answers with solutions
A(1) = and d = A(2) – A(1) = 7 – 1 = 6
A(n) = A(1)+(n-1)d = 1 + (n-1) 6 = 1+6n-6 So A(n) = 6n-5
A(10) = 6(10) – 5 = 55
A(25) = 6(25)-5 = 145
a) 1, 7, 13, 19 …. ….
A(1) = x and d = A(2) – A(1) = x+7-(x+10) = -3
A(n) = A(1)+(n-1)d = x+10 + (n-1)(-3) = x+10-3n+3 So A(n)= x-3n+13
A(10) = x -3(10)+13 = x - 17
A(25) = x -3(25)+13 = x - 62
b) x+10, x+7, x+4, x+1,.
A(1) = and d = -21
A(n) = A(1)+(n-1)d = 3 + (n-1) -21 = 3-21n+21 So A(n)= 24-21n
A(10) = 24-21(10) = = 24-21(25) = -501
c) The first term is 3 and the common difference is -21
d) The second term is 8 and the common difference is 3
A(1) = = and d = 3
A(n) = A(1)+(n-1)d = 5 + (n-1) 3 = 5+3n So A(n) = 3n+2
A(10) = 3(10) +2 = A(25) = 3(25)+2 = 77

14. The 14th term in this sequence is the number -73!
Examples:
Find the 14th term of the arithmetic sequence with first term of 5 and the common difference is –6.
A(1) = 5 and d = -6
A(14) = a + (n – 1) d
You are looking for the term! List which variables from the general term are provided!
A(14) = -6 5
= (13) * -6 =
= -73
The 14th term in this sequence is the number -73!

15. The 100th term in this sequence is 301!
Examples:
In the arithmetic sequence 4,7,10,13,…, which term has a value of 301?
A(n) = a + (n – 1) d
You are looking for n!
The 100th term in this sequence is 301!

16. HMMM! Two equations you can solve!
Examples:
In an arithmetic sequence, term 10 is 33 and term 22 is –3. What are the first four terms of the sequence?
A(10) = 33
A(22) = -3
Use what you know!
A(n) = A(1) + (n – 1) d
For term 22: -3= A(1) + 21d
A(n) = A(1) + (n – 1) d
For term 10: 33= A(1) + 9d
HMMM! Two equations you can solve!
SOLVE:
33 = A(1) + 9d
-3 = A(1) + 21d
By elimination = 12d
-3 = d
SOLVE:
33 = A(1) + 9d
33 = A(1) + 9(-3)
33 = A(1) – 27
60 = A(1)
The sequence is 60, 57, 54, 51, …….

17. Paired Practice – Follow the directions carefully
Paired Practice – Follow the directions carefully. Raise your hand if you need help.
Write the first four terms of any arithmetic sequence on a separate sheet of paper.
In your notebook, find the first 10 terms of that sequence.
Swap with your partner. Write a recursive formula that could be used to find A(6).
Write an explicit formula that could be used to find A(100).
Use your explicit formula to find A(10).
Check with your partner to see if your answer matches theirs.
HW – Chapter 4 Review #12 – 16