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Chapter 13: Sequences and
Series
Section 13.1: Arithmetic and
Geometric Sequences
arithmetic sequence – a sequence of
numbers such that the difference of any two
consecutive terms is constant.
tn =
To get the
nth term
t1
+
start with
the first
term
(n – 1)d
and add the
difference
n – 1 times
geometric sequence – a sequence such
that the ratio (common ratio) of any two
consecutive terms is constant.
tn = t1 · r (n – 1)
To get the
nth term
start with
the first
term
and multiply
by the ratio
n – 1 times
sequence – defined to be a function whose
domain is the set of positive integers.
Example 1:
State whether the given sequence is
arithmetic, geometric, or neither.
a) 1 · 2, 2 · 3, 3 · 4, 4 · 5, …
Neither the difference of consecutive terms nor
the ratio of consecutive terms is constant.
Neither
b) 51/3 , 101/3 , 201/3, 401/3, …
The common ratio of the terms is 21/3.
geometric
c) 15, 11, 7, 3, …
The common difference is -4.
Arithmetic
Example 2:
State the first four terms of the specified
sequence. Then tell whether the sequence
is arithmetic, geometric, or neither.
a) tn = 5n + 2
= 5(1) + 2
= 5(2) + 2
= 5(3) + 2
= 5(4) + 2
7, 12, 17, 22, …
arithmetic
b) tn = n + 1
n+2
c) tn = 3n
d) tn = n3
b) 2/3, ¾, 4/5, 5/6, … (neither)
c) 3, 9, 27, 81, … (geometric)
d) 1, 8, 27, 64, …(neither)
Example 3:
Find a formula for tn.
a) 1, 4, 7, 10, …
b) 8, 4, 2, 1, …
HOMEWORK (Day 1)
pg. 476; 1 – 8, 13, 15
Example 4:
a. Arithmetic sequence
t1 = 114, t3 = 100, Find t20.
100 = 114 + (3 – 1)d
-14 = 2d
-7 = d
t20 = 114 + (20 – 1)(-7)
t20 = -19
b. Geometric sequence
t1 = 5/64, t4 = 5/8, find t26.
5/8 = 5/64(r)(4 – 1)
8 = r3
r=2
t26 = 5/64(2)(26-1)
t26 = 2,621,440
Practice
1) Arithmetic Sequence
t1 = 5, t7 = 29, find t53.
d=4
t53 = 213
2) Geometric Sequence
t1 = ½ , t5 = 8, find t10.
r=2
t10 = 256
HOMEWORK (Day 2)
pg. 477; 29, 30, 33, 34, 36
Example 5:
In a certain sequence, t2 = 2 and t5 = 16.
Find t10 if the sequence is:
Arithmetic sequence
t2 = 2
2 = t1 + (2 – 1)d
2 = t1 + d
t5 = 16
16 = t1 + (5 – 1)d
16 = t1 + 4d
16 = t1 + 4d
- (2 = t1 + d)
14 = 3d
d = 14/3
2 = t1 + d
t1 = 2 – d
t1 = 2 – 14/3
t1 = -8/3
t10 = t1 + (10 – 1)d
= -8/3 + (9)(14/3)
= 118
3
Geometric sequence
t2 = 2
2 = t 1 · r2 – 1
2 = t1 · r
t5 = 16
16 = t1 · r5 – 1
16 = t1 · r4
t1 = 2
r
t1 = 16
r4
2 = 16
r
r4
2r4 = 16r
r3 = 8
r=2
t1 = 2/r
= 2/2
=1
t10 = t1 · r10 – 1
= 1 · r10 – 1
= 1 · 29
= 512
HOMEWORK (Day 3)
pg. 477; 31, 32, 35
1) Arithmetic; t1 = 3, t5 = 23, find t20.
2) Geometric; t1 = 24, t6 = ¾ , find t10.
3) Arithmetic; t2 = 5, t8 = 23, find t30.
4) Geometric; t4 = 32, t10 = 2,048, find t25.