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11.2 Arithmetic Sequences & Series
Chapter 11 Further Topics in Algebra
Concepts and Objectives
⚫ Arithmetic Sequences & Series
⚫ Find the nth term of an arithmetic sequence
⚫ Find the index of a given term of a sequence
⚫ Given an arithmetic series, be able to calculate Sn, the nth
partial sum and vice versa
Arithmetic Sequences
⚫ An arithmetic sequence is a sequence in which one term
equals a constant added to the preceding term.
⚫ The constant for an arithmetic sequence is called the
common difference, d, because the difference between any
two adjacent terms equals this constant.
Arithmetic Sequences (cont.)
⚫ Formulas for calculating tn for arithmetic sequences can
be found by linking the term number to the term value.
Example: The arithmetic sequence 3, 10, 17, 24, 31, …,
has a first term t1 = 3, and common difference d = 7:
Arithmetic Sequences (cont.)
⚫ Formulas for calculating tn for arithmetic sequences can
be found by linking the term number to the term value.
Example: The arithmetic sequence 3, 10, 17, 24, 31, …,
has a first term t1 = 3, and common difference d = 7:
1 3t =
2 3 7t = +
( )( )3 3 7 7 3 2 7t = + + = +
( )( )4 3 7 7 7 3 3 7t = + + + = +
( )( )3 1 7nt n= + −
Arithmetic Sequences (cont.)
⚫ The nth term of an arithmetic sequence equals the first
term plus (n – 1) common differences. That is,
⚫ We can see that this is a linear function (where n is the
independent variable), and if we compare this to the
slope-intercept form, we can see that the slope is d, and
the y-intercept would be t0, or t1 – d if zero were in the
domain of the function (remember that our domain is
).
( )1 1nt t n d= + −
Sequences - Examples
1. Calculate t100 for the arithmetic sequence
17, 22, 27, 32, …
Sequences - Examples
1. Calculate t100 for the arithmetic sequence
17, 22, 27, 32, …
d = 5
Therefore, ( )( )100 17 100 1 5t = + −
17 495 512= + =
Sequences - Examples
2. The number 68 is a term in the arithmetic sequence
with t1 = 5 and d = 3. Which term is it?
Sequences - Examples
2. The number 68 is a term in the arithmetic sequence
with t1 = 5 and d = 3. Which term is it?
( )( )68 5 1 3n= + −
( )( )63 1 3n= −
21 1n= −
22n =
Arithmetic Series
⚫ An arithmetic series is a series which results from adding
the terms of an arithmetic sequence.
⚫ While it is possible to find a partial sum by writing down
all of the terms and adding them up, it would easier to
find a formula for Sn.
Arithmetic Series (cont.)
⚫ Consider the arithmetic sequence 7, 13, 19, 25, 31, …
What if we wanted to find the sum of the first 100
terms?
We can find the 100th term using the formula from
before:
Since d = 6, we can work backwards from 601 to find the
last few terms as well, so our sum looks like:
( )( )= + − =100 7 100 1 6 601t
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S
Arithmetic Series (cont.)
Now, what do you notice if we add the terms together
starting from the outside and working inside?
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S
Arithmetic Series (cont.)
Now, what do you notice if we add the terms together
starting from the outside and working inside?
We have 50 sums (100  2), so our sum is
50(608)=30400. This leads us to our formula.
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S
608
608
608
608
Arithmetic Series (cont.)
⚫ The formula for the nth partial sum of an arithmetic
series is
⚫ Since if we don’t know tn, we can find it by the sequence
formula, we can substitute that into the above formula
and get
( )
+ 
= + =  
 
1
1 or
2 2
n
n n n
t tn
S t t S n
( )( )= + + −1 1 1
2
n
n
S t t n d
( )( )= + −12 1
2
n
t n d
Arithmetic Series (cont.)
⚫ Example: Find S33 for the series with t1 = 9 and d = –2.
Arithmetic Series (cont.)
⚫ Example: Find S33 for the series with t1 = 9 and d = –2.
Notice that even though n is odd, we still get a whole
number. Why does the formula work this way?
( )( )= + −12 1
2
n
n
S t n d
( ) ( )( )( )= + −
33
2 9 32 2
2
( )= − = −
33
46 759
2
Partial Sums
⚫ Example: 30702 is a partial sum in the arithmetic series
with first term 17 and common difference 3. Which
partial sum is it? (What is n?)
Partial Sums
⚫ Example: 30277 is a partial sum in the arithmetic series
with first term 17 and common difference 3. Which
partial sum is it? (What is n?)
( )( )= + −12 1
2
n
n
S t n d
( )( )( )= + −30277 2 17 1 3
2
n
n
( )= +30277 31 3
2
n
n
= + 2
60554 31 3n n
Partial Sums
(cont.)
So n = 137.
+ − =2
3 31 60554 0n n
− + − =2
3 411 442 60554 0n n n
( ) ( )− + − =3 137 442 137 0n n n
( )( )+ − =3 442 137 0n n
= −
442
, 137
3
n
Classwork
⚫ College Algebra
⚫ Page 1014: 14-26; page 1004: 52-66; page 978: 38-46

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11.2 Arithmetic Sequences and Series

  • 1. 11.2 Arithmetic Sequences & Series Chapter 11 Further Topics in Algebra
  • 2. Concepts and Objectives ⚫ Arithmetic Sequences & Series ⚫ Find the nth term of an arithmetic sequence ⚫ Find the index of a given term of a sequence ⚫ Given an arithmetic series, be able to calculate Sn, the nth partial sum and vice versa
  • 3. Arithmetic Sequences ⚫ An arithmetic sequence is a sequence in which one term equals a constant added to the preceding term. ⚫ The constant for an arithmetic sequence is called the common difference, d, because the difference between any two adjacent terms equals this constant.
  • 4. Arithmetic Sequences (cont.) ⚫ Formulas for calculating tn for arithmetic sequences can be found by linking the term number to the term value. Example: The arithmetic sequence 3, 10, 17, 24, 31, …, has a first term t1 = 3, and common difference d = 7:
  • 5. Arithmetic Sequences (cont.) ⚫ Formulas for calculating tn for arithmetic sequences can be found by linking the term number to the term value. Example: The arithmetic sequence 3, 10, 17, 24, 31, …, has a first term t1 = 3, and common difference d = 7: 1 3t = 2 3 7t = + ( )( )3 3 7 7 3 2 7t = + + = + ( )( )4 3 7 7 7 3 3 7t = + + + = + ( )( )3 1 7nt n= + −
  • 6. Arithmetic Sequences (cont.) ⚫ The nth term of an arithmetic sequence equals the first term plus (n – 1) common differences. That is, ⚫ We can see that this is a linear function (where n is the independent variable), and if we compare this to the slope-intercept form, we can see that the slope is d, and the y-intercept would be t0, or t1 – d if zero were in the domain of the function (remember that our domain is ). ( )1 1nt t n d= + −
  • 7. Sequences - Examples 1. Calculate t100 for the arithmetic sequence 17, 22, 27, 32, …
  • 8. Sequences - Examples 1. Calculate t100 for the arithmetic sequence 17, 22, 27, 32, … d = 5 Therefore, ( )( )100 17 100 1 5t = + − 17 495 512= + =
  • 9. Sequences - Examples 2. The number 68 is a term in the arithmetic sequence with t1 = 5 and d = 3. Which term is it?
  • 10. Sequences - Examples 2. The number 68 is a term in the arithmetic sequence with t1 = 5 and d = 3. Which term is it? ( )( )68 5 1 3n= + − ( )( )63 1 3n= − 21 1n= − 22n =
  • 11. Arithmetic Series ⚫ An arithmetic series is a series which results from adding the terms of an arithmetic sequence. ⚫ While it is possible to find a partial sum by writing down all of the terms and adding them up, it would easier to find a formula for Sn.
  • 12. Arithmetic Series (cont.) ⚫ Consider the arithmetic sequence 7, 13, 19, 25, 31, … What if we wanted to find the sum of the first 100 terms? We can find the 100th term using the formula from before: Since d = 6, we can work backwards from 601 to find the last few terms as well, so our sum looks like: ( )( )= + − =100 7 100 1 6 601t = + + + + + + + +100 7 13 19 25 ... 583 589 595 601S
  • 13. Arithmetic Series (cont.) Now, what do you notice if we add the terms together starting from the outside and working inside? = + + + + + + + +100 7 13 19 25 ... 583 589 595 601S
  • 14. Arithmetic Series (cont.) Now, what do you notice if we add the terms together starting from the outside and working inside? We have 50 sums (100  2), so our sum is 50(608)=30400. This leads us to our formula. = + + + + + + + +100 7 13 19 25 ... 583 589 595 601S 608 608 608 608
  • 15. Arithmetic Series (cont.) ⚫ The formula for the nth partial sum of an arithmetic series is ⚫ Since if we don’t know tn, we can find it by the sequence formula, we can substitute that into the above formula and get ( ) +  = + =     1 1 or 2 2 n n n n t tn S t t S n ( )( )= + + −1 1 1 2 n n S t t n d ( )( )= + −12 1 2 n t n d
  • 16. Arithmetic Series (cont.) ⚫ Example: Find S33 for the series with t1 = 9 and d = –2.
  • 17. Arithmetic Series (cont.) ⚫ Example: Find S33 for the series with t1 = 9 and d = –2. Notice that even though n is odd, we still get a whole number. Why does the formula work this way? ( )( )= + −12 1 2 n n S t n d ( ) ( )( )( )= + − 33 2 9 32 2 2 ( )= − = − 33 46 759 2
  • 18. Partial Sums ⚫ Example: 30702 is a partial sum in the arithmetic series with first term 17 and common difference 3. Which partial sum is it? (What is n?)
  • 19. Partial Sums ⚫ Example: 30277 is a partial sum in the arithmetic series with first term 17 and common difference 3. Which partial sum is it? (What is n?) ( )( )= + −12 1 2 n n S t n d ( )( )( )= + −30277 2 17 1 3 2 n n ( )= +30277 31 3 2 n n = + 2 60554 31 3n n
  • 20. Partial Sums (cont.) So n = 137. + − =2 3 31 60554 0n n − + − =2 3 411 442 60554 0n n n ( ) ( )− + − =3 137 442 137 0n n n ( )( )+ − =3 442 137 0n n = − 442 , 137 3 n
  • 21. Classwork ⚫ College Algebra ⚫ Page 1014: 14-26; page 1004: 52-66; page 978: 38-46