Geometric Progression
Objectives
The presentation intends to:
 teach students how to solve problems related
to geometric progressions;
 help students understand more about the given
topic;
What are geometric progressions?
 It is a sequence in which each term is
obtained by multiplying the preceding term
by a constant
 A sequence t1,t2, . . . . tn, . . . is called a
geometric sequence ( or more commonly a
geometric progression) if there exists a
number r such that
tn/tn-1 = r for n>1
Also known as a geometric sequence, is a
sequence of numbers where each term
after the first is found by multiplying the
previous one by a fixed non-zero number
called the common ratio.
Symbols
tn= last term
n= number of terms
r= common ratio
a= first term
Sn= sum of terms
Formulas
n-1
ar
• tn=
• a= tn/rn-1
n
• Sn= a(1-r )/1-r
• r= tn/tn-1
Examples
Find the common ratio and general term of
Geometric Progression
1) 1, 2/3, 4/9,…
2) 2,-4, 8,…
r= 2/3
r= -4/ 2
1
= -2
=2/3
tn= 2(-2)n-1
tn= (2/3)n-1
3) √2, √6, 3√2,… 4) If x+3, 4x, 2x+18, form
geometric progression, find x.
r= √6
4x = 2x+18
√2
x+3
4x
=√3
2= 2x2+6x+18x+54
n-1
16x
tn=√2(√3)
14x2-24x-54=0
2
7x2-12x-27=0
(7x+9)(x-3)=0
x= -9 x=3
7
5) The fifth term of a geometric progression is 81 and
the ninth term is 16. Find the first term, common
ratio, and the nth term.
t5= 81, t9=16, a=?, r=?, nth term= ?
tn=arn-1
t5= ar5-1
t9= t5r9-5
81=a(2/3)4
16= 81r4
6561/16
81
tn=arn-1
4√16 = 2
tn = 6561/16(2/3)n-1
4√81
3
Insert three positive geometric means between
a)1/2 and 128
½, t2, t3, t4, 128 t5= ar5-1
128= ½ r4
½
256= r4
4 √r4=4 √256
r=4
t2=1/2(4)=2
t3=2(4)= 8
t4=8(4)= 32
b)1 and 4
1, t2, t3,t4,
t5= ar5-1
4= 1r4
4 √r4=4 √4
r= 4√22
r= 2 2/4
r= 2 ½
r= √2
t2= 1 √2
t3= (√2) (√2) = 2
t4= 2 (√2) = 2√2
Find the sum of the first ten terms:
2+4+8+16+…
Sn= a(1-rn)
1-r
S10= 2[1-(2)10]
1-2
S10= 2(1-1624)
-1
S10= 2046
Find the sum of the geometric progression
10, -2, 2/5,… to 8 terms
Sn= a(1-rn)
1-r
S8= 10[1-(-1/5)8]
1-(-1/5)
S8= 10(1-0.00000256)
1.2
S8= 8.33
Find the t5of the geometric progression whose first
element is -3 and whose common ratio is -2.
tn= arn-1
t5= (-3)(-2)5-1
t5= (-3)(-2)4
t5= (-3)(16)
t5= -48
Infinite Arithmetic and Geometric
Progression
Definition
The sum of an infinite geometric progression a, ar,
ar2, … with r < 1 is given by,
S=a/1-r
whereas, -1<r<1
S is also called the sum to infinity of the G.P. a, ar, ar2
… In symbols, we write this as,
S=lim Sn
and read “S is the limit of Sn as n increases without
bound”
The rational number 1/3 has repeating decimal
.33333…
which we can consider to be as infinite sum:
1/3 = .3+.03+.003+.0003+…
This is an example of an infinite geometric
series with ratio .1. Thus, .03 = .3x.1, .003 =
.03x.1,.0003 = .0003x.1, and so on. We define an
infinite geometric series in the following way.
Find the sum of the following terms.
1) 1+2+3+4+5… 3) 9+3+1/3…
=infinite
r=1/3
S= a
2) 2+4+6+8…
1-r
=infinite
S= 9
1-1/3
S= 9
2/3
S= 27/2
4) 50+25+25/2…
r=1/2
S= a
1-r
S= 50
1-1/2
S= 50
1/2
S= 100
Exercises:
A. Determine whether the sequence is a geometric
progression or not. If the sequence is a geometric
progression give the common ratio (r) and the next
three terms.
1) 12,22,32
2) 1, -1/2, 1/3
3) 1/6, 1/2, 3/2
4) 4, 3, 9/4
5) 3, 3√3, 9
6) 2, √3, 2/3
7) 3.33, 2.22, 1.11
8) -36, -2, -1/9
B. Find the indicated term of the geometric
progression whose first element is a and whose
common ratio is r.
a= ½ , r= 2/3, find t4
a= 4 , r=-1/2 , find t5
a= 81, r= 1/3, find t7
a= 1/3 , r= 3/2, find t6
C. Find the sum to the infinity of the geometric
progression.
1.
2.
3.
4.
5.
16, 4, 1
½, 1/6, 1/18
-2, -1/2, -1/8
2/3, 1/9, 1/54
5.05, 1.212, 0.29088
1. Insert five geometric means between
a. 1/2 and 32
b. 2 and 54
E. For the specific value of n, find the nth term of
each geometric series which starts as follows.
1. (a) 2+6+18+…, n = 7 (b) 5+10+20+…, n = 9
2. (a) 4+2+1+…, n = 8
(b) 25-5+1+…, n = 6
Solutions and Answers
A.
1) 12,22,32
Not geometric progression
2) 1, -1/2, 1/3
Not geometric progression
3) 1/6, 1/2, 3/2
r=1/2 = 3
1/
6
3
( /2)(3)= 9/2
(9/2)(3)=27/2
(27/2)(3)= 81/2
next terms are 9/2,,27/2, 81/2
4) 4, 3, 9/4
r= 3/4
(3/4)(9/4)= 27/16
(3/4)(27/16)= 81/64
(3/4)(81/64)= 243/256
next terms are: 27/16,
81/64, 243/256
5) 3, 3√3, 9
r= √3
(√3)(9)= 9√3
(√3) (9√3)= 27
(√3)(27)=27√3
next terms are:
9√3, 27, 27√3
6) 2, √3, 2/3
Not geometric
progression
7) 3.33, 2.22, 1.11
Not geometric progression
8) -36, -2, -1/9
r= -2/-36= 1/18
(-1/9)(1/18)= -1/162
(-1/162)(1/18)= -1/2916
(-1/2916)(1/18)= -1/52488
next terms are:
-1/162, -1/2916, -1/52488
B.
1) a= ½ , r= 2/3, t4= ?
tn= arn-1
t4= (½)(2/3)4-1
t4= (½)(2/3)3
t4= (½)(8/27)
t4= 4/27
2) a= 4, r=-1/2 , t5= ?
tn= arn-1
t5= (4)(-1/2)5-1
t5= (4)(-1/2)4
t5= (4)(1/16)
t5= 4/16
t5= ¼
3) a= 81, r= 1/3, t7=?
tn= arn-1
t7= (81)(1/3)7-1
t7= (81)(1/3)6
t7= (81)(1/729)
t7= 1/9
4) a= 1/3 , r= 3/2, t6=?
tn= arn-1
t6= (1/3)(3/2)6-1
t6= (1/3)(3/2)5
t6= (1/3)(7 19/32)
t6= 2 17/32
C.
1) 16, 4, 1
r=1/4
S= a
1-r
S= 16
1-1/4
S= 16
3/
4
S= 64/3
2) ½, 1/6, 1/18
r= 1/3
S= a
1-r
S= ½
1-1/3
S= ½
2/3
S= 3/4
3) -2, -1/2, -1/8
r= ¼
S= a
1-r
S= -2
1-1/4
S= -2
3/
4
S= -8/3
4) 2/3, 1/9, 1/54
r= 1/6
S= a
1-r
S= 2/3
1-1/6
S= 2/3
5/6
S= 4/5
5) 5.05, 1.212,
0.29088
r= 0.24
S= a
1-r
S= 5.05
1-0.24
S= 5.05
0.76
S= 6 49/76
a. 1/2 and 32
t7=t1r6
32= ½ r6
32= ½ r6
½
64= r6
6√64=r
r=2
t2 =1
t3=2
t4=4
t5=8
t6=16
b. 2 and 54
t7= t1r6
54=2r6
2 2
6 √27=
6 √ r6
r= 6 √27
r= 6 √33
r= √3
t2= 2 √3
t3= 6
t4= 6 √3
t5= 18
t6= 18 √3
E.
1. (a) 2+6+18+…,,n=7
tn= arn-1
t7= (2)(3)6
=1458
(b) 5+10+20+…, n=9
tn= arn-1
t9= (5)(2)8
= 1280
2.
(a)4+2+1+…,n=8
tn=arn-1
t7=(4)(1/2)7
=1/32
(b) 25-5+1…,n=6
tn= arn-1
t6=(25)(-1/5)5
= -1/125
Prepared by:
Princess Fernandez
Karen Calanasan
Kyla Villegas