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= + −
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
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