12.2A Arithmetic
Sequences
Algebra II
Arithmetic Sequence:
• The difference between consecutive
terms is constant (or the same).
• The constant difference is also known
as the common difference (d).
(It’s also that number that you are adding
everytime!)
Example 1: Decide whether each
sequence is arithmetic.
•
•
•
•
•
•
•
-10,-6,-2,0,2,6,10,…
-6--10=4
-2--6=4
0--2=2
2-0=2
6-2=4
10-6=4
Not arithmetic (because
the differences are
not the same)
•
•
•
•
•
5,11,17,23,29,…
11-5=6
17-11=6
23-17=6
29-23=6
• Arithmetic (common
difference is 6)
Rule for an Arithmetic Sequence
an=a1+(n-1)d
Example 2: Write a rule for the nth
term of the sequence 32,47,62,77,… .
Then, find a12.
• The is a common difference where d=15,
therefore the sequence is arithmetic.
• Use an=a1+(n-1)d
an=32+(n-1)(15)
an=32+15n-15
an=17+15n
a12=17+15(12)=197
Ex. 3: One term of an arithmetic sequence
is a8=50. The common difference is 0.25.
Write a rule for the nth term.
• Use an=a1+(n-1)d to find the 1st term!
a8=a1+(8-1)(.25)
50=a1+(7)(.25)
50=a1+1.75
48.25=a1
* Now, use an=a1+(n-1)d to find the rule.
an=48.25+(n-1)(.25)
an=48.25+.25n-.25
an=48+.25n
Ex. 4 Now graph an=4+.25n.
• Remember to graph the ordered pairs of the
form (n,an)
• So, graph the points (1,48.25), (2,48.5),
(3,48.75), (4,49), etc.
12-2B Arithmetic Series
•
Example 1: Two terms of an arithmetic sequence
are a5=10 and a30=110. Write a rule for the nth term.
• Begin by writing 2 equations; one for each term
given.
a5=a1+(5-1)d OR 10=a1+4d
And
a30=a1+(30-1)d OR 110=a1+29d
• Now use the 2 equations to solve for a1 & d.
10=a1+4d
110=a1+29d (subtract the equations to cancel a1)
-100= -25d
So, d=4 and a1=-6 (now find the rule)
an=a1+(n-1)d
an=-6+(n-1)(4) OR an=-10+4n
Example 1(part 2): using the rule an=-10+4n,
write the value of n for which an=-2.
-2=-10+4n
8=4n
2=n
Arithmetic Series
• The sum of the
terms in an
arithmetic sequence
1st Term
• The formula to find
the sum of a finite
arithmetic series is:
Last
Term
 a1  an 
S n  n

 2 
# of terms
Example 2: Consider the arithmetic
series 20+18+16+14+… .
• Find the sum of the 1st • Find n such that Sn=-760
25 terms.
 a1  an 
S n  n

• First find the rule for
 2 
the nth term.
• an=22-2n
 20  (22  2n) 
 760  n

2


• So, a25 = -28 (last term)
 a1  an 
S n  n

 2 
 20  28 
S 25  25
 S 25  25(4)  100
2


 20  (22  2n) 
 760  n

2


-1520=n(20+22-2n)
-1520=-2n2+42n
2n2-42n-1520=0
n2-21n-760=0
(n-40)(n+19)=0
n=40 or n=-19
Always choose the positive solution!
Ex. 3 Find the sum of arithmetic
series
Ex. 4 Find the value of n
Assignment