Solutions to Practice Problems for Module 9 Unit #3
Find the first four terms of the geometric sequence whose first term and
common ratio are given.
1.- a1  0.03 and r  2
Answer: 0.03, 0.03  2, 0.03  22 , 0.03  23  0.03, 0.06, 0.12, 0.24
1
2.- a1  2500 and r 
10
2
3
5
1
1
1
Answer: 2500, 2500    , 2500    , 2500     2500, 250, 25,
2
 10 
 10 
 10 
1
3.- a1  768 and r 
2
2
3
2
3
1
1
1
Answer: 768, 768    , 768    , 768     768, 384, 192, 96
2
2
2
1
4.- a1  400 and r 
5
16
1
1
1
Answer: 400, 400    , 400    , 400     400, 80, 16,
5
5
5
5
4
and r  3
5.- a1 
9
4 4
4
4
4 4
Answer: ,  3,  32 ,  33  , , 4, 12
9 9
9
9
9 3
Decide if the given sequence is a geometric sequence, and if so, give the
“common ratio” (r).
1 1 1
, , ,...
2 6 18
Answer: Perform the quotient between each term (starting with the second
one) and the previous one and check if that renders a constant number:
1 1 1
1 1 1
  ;
  . The ratio is common, so this is a geometric sequence
6 2 3 18 6 3
1
and its common ratio is: r  .
3
6.-
7.- 3,1,3,...
1
Answer: Perform the quotients: 1   3   ; 3  1  3 . The ratios are
3
different, so this is not a geometric sequence.
1 1
8.-  ,  , 2,...
8 2
1
1
1
Answer: Perform the quotients:     4;  2    4 . The ratios are the
2
8
8
same, so this is a geometric sequence with common ratio r  4 .
2 1 2
9.-  , ,  ,...
5 5 5
1  2
1
2 1
      ;    2 . The ratios are
5  5
2
5 5
different, so this is not a geometric sequence.
Answer: Perform the quotients:
10.-
1 1 1
,  , ,...
4 8 64
1 1
1 1  1
1
      . The ratios are
Answer: Perform the quotients:     ;
8 4
2 64  8 
8
different, so this is not a geometric sequence.
11.- 2,
10 50
, ,...
3 9
10
5 50 10 5
2  ;
  . The ratios are the same, so
3
3 9 3 3
5
this is a geometric sequence with common ratio r  .
3
Perform the quotients:
1 3 1
12.-  , ,  ,...
5 2 10
3  1
15
1 3
3
      ;     . The ratios are
2  5
2
10 2
5
different, so this is not a geometric sequence.
Answer: Perform the quotients:
13.-
2 3 2
, , ,...
3 2 3
3 2 9 2 3 4
  ;   . The ratios are different,
2 3 4 3 2 9
so this is not a geometric sequence.
Answer: Perform the quotients:
5 15 45
14.-  , ,  ,...
2 8 32
15
5
3
45 15
3
    ;     . The ratios are the same,
8
2
4
32 8
4
3
so this is a geometric sequence with common ratio r   .
4
Perform the quotients:
15.-
2 4 8
, , ,...
3 9 27
4 2 2 8 4 2
  ;
  . The ratios are the same, so this
9 3 3 27 9 3
2
is a geometric sequence with common ratio r  .
3
Perform the quotients:
Find the indicated term for each of the given geometric sequence:
16.- a6 for the sequence with a1  2 and r 
1
2
5
Answer: Use the
nth
1
1
term formula: a6  2    
16
2
17.- a8 for the sequence with a1  1.3 and r  2
Answer: Use the nth term formula: a8  1.3  2  166.4
7
18.- a10 for the sequence with a1  0.1 and r  3
Answer: Use the nth term formula: a10  0.1  3  1968.3
9
19.- a12 for the sequence with a1  500 and r  0.8
Answer: Use the nth term formula: a12  500  0.8  42.94967296
11
20.- a6 for the sequence with a1  8748 and r  
1
3
6
Answer: Use the
nth
1
term formula: a7  8748    12
3
Find the first term of each geometric sequence with the given characteristics:
21.- a3  12.6 and common ratio r  3
Answer: Use the nth term formula to solve for the first term:
a
12.6 12.6
 1.4
an  a1 r n1  a1  nn1 . In this case: a1  31 
9
r
 3
22.- a4  6 2 and common ratio r  2
Answer: Use the nth term formula to solve for the first term:
6 2
6 2
a1 

3
41
2
2
2
 
1
1
and common ratio r  
25
2
Answer: Use the nth term formula to solve for the first term:
1
16 2
24
a1 


51
24
3
1


 
 2
23.- a5 
24.- a2  1.2 and a5  0.0096
Answer: Use this information and the nth term formula to find the common
ratio by creating two equations with two unknowns and solving them:

a2  1.2  a1 r 21  a1 r
a r
1.2
1

 1 4  3 . Solve for the common

51
4
0.0096 a1 r
r
a5  0.0096  a1 r  a1 r 

0.0096
 0.008  r  3 0.008  0.2 and then use it to solve for a1 in
ratio: r 3 
1.2
any of the first two equations a1  6
1
8
Answer: Use this information and the nth term formula to find the common
ratio by creating two equations with two unknowns and solving them:
a2  2  a1 r 21  a1 r 
a1 r
2
1



16


. Solve for the common ratio:

1
a1 r 3 r 2
a4    a1 r 41  a1 r 3   1
8

8
2
r  16  r  16  4 and then use it to solve for a1 in any of the first two
25.- a2  2 and a4  
equations a1  8
Evaluate each of the following sums of a geometric sequence:
26.- S4 for the sequence with a1  1.2 and r  4
1   4
1 rn
 S4  1.4
 119
1 r
1   4
4
Answer: Use the partial sum formula: Sn  a1
1
10
1 rn
Answer: Use the partial sum formula: Sn  a1

1 r
27.- S5 for the sequence with a1  20 and r 
5
1
1  
10
S5  20    22.222
1
1  
 10 
28.- S7 for the sequence with a1  5 and r  2
1   2 
1 rn
Answer: Use the partial sum formula: Sn  a1
 S7  5
 215
1 r
1   2 
7
29.- S5 for the sequence with a1  81 and r  
2
3
5
 2
1   
3
Answer: Use the partial sum formula S5  81 
 55
 2
1   
 3
30.- S10 for the sequence 2, 6,18,...
Answer: Find the common ratio to be able to apply the partial sum
1   3
formula r  3 ; then the sum S10  2
 59048
1   3
10
31.- S8 for the sequence 80, 40, 20,...
Answer: Find the common ratio to be able to apply the partial sum
8
1
1  
1
1275
2
formula r  ; then the sum S8  80   
2
8
1
1  
2
32.- S7 for the sequence 1215,810,540,...
Answer: Find the common ratio to be able to apply the partial sum
7
2
1  
2
10295
3
formula r  ; then the sum S7  1215   
3
3
2
1  
3
33.- S5 for the sequence with a3  28 and r  2
Answer: Use the given information to find the first term by using the nth term
28
28
formula a1 

 7 . Now use the partial sum formula
31
4
2
 
 S5   7 
1   2
 217
1   2
5
1
3
Answer: Use the given information to find the first term by using the n th term
1
 27 . Now use the partial sum formula
formula a1 
41
1


 
3


34.- S6 for the sequence with a4  1 and r  
6
 1
1   
182
3

 S6   27  
9
 1
1   
 3
5
2
and r 
2
5
Answer: Use the given information to find the first term by using the n th term
5
625
formula a1  241 
. Now use the partial sum formula
16
2
 
5
35.- S4 for the sequence with a4 
4
2
1  
1015
 625   5 

 S4  

16
 16  1   2 
 
5
36.- S5 for the sequence with a5  144 and a3  36
Answer: Use this information and the nth term formula to find the common
ratio by creating two equations with two unknowns and solving them:
a5  144  a1 r 51  a1 r 4 
a r4

 4  1 2  r 2 . Solve for the common ratio:

a1 r
a3  36  a1 r 31  a1 r 2 

r 2  4  r  4  2 and then use it to solve for a1 in any of the first two
equations a1  9 . Now we can use the partial sum formula
1   2
S5   9 
 279
1   2
5
1
1
and a2  
16
2
Answer: Use this information and the nth term formula to find the common
ratio by creating two equations with two unknowns and solving them:
1

a5   a1 r 51  a1 r 4 
1 a1 r 4

16
 r 3 . Solve for the common ratio:
 
1
8 a1 r
a2    a1 r 21  a1 r 

2
37.- S6 for the sequence with a5 
1
1
1
r 3    r  3    and then use it to solve for a1 in any of the first two
8
8
2
equations a1  1. Now we can use the partial sum formula
6
 1
1   
21
2
S6  1 

 1  32
1   
 2
38.- S for the sequence with a1  5 and r 
2
5
Answer since r  1 , this infinite sum will converge to the value: S 
S 
5
25

3
2
1  
5
a1

1 r
4
3
Answer since r  1 , this infinite sum will not converge. We say that the sum
39.- S for the sequence with a1  3 and r 
diverges, since we are adding infinite number terms each of them larger than
one.
40.- S for the sequence with a1  270 and r 
3
4
Answer since r  1 , this infinite sum will converge to the value: S 
S 
a1

1 r
270
 1080
3
1  
4
Express the following repeating decimal as a fraction by writing it as an infinite
geometric sum and using the formula for S  :
41.- 0.5555...  0.5
Answer: The decimal consists of repeating “5’s”. Write the decimal expression
5
5
5
5


 ... . This is
as the sum: 0.5  0.05  0.005  0.0005  ...  
10 100 1000 10000
5 1
1
and r 
the infinite sum of a geometric sequence with a1  
. We
10 2
10
1
1
a1
5
2
can use the formula: S 
 S 
 2 
1 r
1 9 9
1  
 10  10
42.- 0.0404...  0.04
Answer: The repeating part of the decimal is the pair “04”. We write the
decimal expression as the sum:
4
4
4
0.04  0.0004  0.000004  ... 


 ... . This is the infinite sum
100 10000 100000
4
1
1

and r 
of a geometric sequence with a1 
. We can use the
100 25
100
1
1
a1
4
25
 25 
formula: S 
 S 
1 r
 1  99 99
1 

 100  100
43.- 1.2222...  1.2
Answer: The repeating part of the decimal is “2”, and the non-repeating “1”
We write the decimal expression as the sum:
2
2
2
1  0.2  0.02  0.002  ...  1  

 ... . This is the addition of the
10 100 1000
number “1” plus the infinite sum of a geometric sequence with
2 1
1

and r  . We can use the formula:
10 5
10
1
1
a
2
5
S  1  S 
 5  to express the repeating decimal part.
9
1 r
9
1
1  
 10  10
2 11
The decimal number is therefore the addition 1  
9 9
a1 
44.- 4.16666...  4.16
Answer: The repeating part of the decimal is “6”, and the non-repeating “4.1”
We write the decimal expression as the sum:
41 6
6
6
4.1  0.06  0.006  0.0006  ...  


 ... . This is the addition
10 100 1000 10000
41
of the number “ ” plus the infinite sum of a geometric sequence with
10
6
3
1
a1 

and r  . We can use the formula:
100 50
10
3
3
a1
1
S 
 S  50  50 
to express the repeating decimal part.
1 r
 1  9 15
1  
 10  10
41 1 125 25

The decimal number is therefore the addition  
10 15 30
6
45.- 3.0484848...  3.048
Answer: The repeating part of the decimal is “48”, and the non-repeating “3.0”
We write the decimal expression as the sum:
48
48
48
3  0.048  0.00048  0.0000048  ...  3 


 ... . This is the
1000 100000 1000000
addition of the number “ 3 ” plus the infinite sum of a geometric sequence with
48
6
1
a1 

and r 
. We can use the formula:
1000 125
100
6
6
a1
8
S 
 S  125  125 
to express the repeating decimal part.
1 r
 1  99 165
1 

 100  100
8
503

The decimal number is therefore the addition 3 
165 165
Find the total amount accumulated in an annuity under each of the following
conditions:
46.- Annual deposit: $2500; interest rate: 3.5%; after 20 years.
P (1  rate)n  1
 
Answer: Use the annuity formula: Total  
rate
20
2500 1.035  1

  $70699.20
Total 
0.035
47.- Annual deposit: $1600; interest rate: 2.6%; after 18 years.
18
1600 1.026   1

  $36140.07
Answer: Use the annuity formula  Total 
0.026
48.- Annual deposit: $3250; interest rate: 4.1%; after 25 years.
25
3250 1.041  1

  $137186.77
Answer: Use the annuity formula  Total 
0.041
How much per year should be invested in an annuity that pays 2.5% per year if
one would like to accumulate each of the following amounts:
49.- $25000 after 20 years
Answer: Solve for “P” in the annuity formula: P 
P
25000  0.025
1.02520 1
50.- $46000
Total  rate
(1  rate) n  1



 $978.68
after 25 years
Answer: Solve for “P” in the annuity formula  P 
46000  0.025
1.025
25
1
 $1346.69