Chapter 3

Practice questions

In questions 16, the first five terms of each sequence are

found:
1 s( n) = 2n 3: 1, 1, 3, 5, 7
2 g( k ) = 2k 3: 1, 1, 5, 13, 29
3 f ( n) = 3 2 n :

3 3 3 3 3
, , , ,
2 4 8 16 32

4 a1 = 5
an = an1 + 3: 5, 8, 11, 14 , 17
5 an = ( 1)n ( 2n ) + 3: 1, 7, 5, 19, 29
6 b1 = 3
bn = bn1 + 2n: 3, 7, 13, 21, 31
7 The sequence 52, 55, 58, 61, ... is arithmetic;
d = 61 58 = 58 55 = 55 52 = 3.
8 The sequence 1, 3, 9, 27, 81, ... is geometric;
81 27 9
3
r=
=
=
=
= 3.
27
9
3
1
9 The sequence 0.1, 0.2, 0.4 , 0.8, 1.6, 3.2, ... is geometric;
0.2 0.4
3.2
r=
=
= ... =
= 2.
0.1 0.2
1.6
10 The sequence 3, 6, 12, 18, 21, 27, ... is neither arithmetic
nor geometric.
11 The sequence 6, 14 , 20, 28, 34 , ... is neither arithmetic nor
geometric.
1.2 The sequence 2.4 , 3.7, 5, 6.3, 7.6, ... is arithmetic;
d = 3.7 2.4 = 5 3.7 = 6.3 5 = 7.6 6.3 = 1.3.
13 3, 2, 7, 12, ...
a a8 = 32

b an = 5n 8
c an = an 1 + 5, a1 = 3
14 19, 15, 11, 7, ...
a a8 = 9

b an = 23 4 n
c an = an 1 4 , a1 = 19
15 8, 3, 14 , 25, ...
a a88 = 69
b ann = 11n 19

c ann = ann 11 + 11, a11 = 8
16 10.05, 9.95, 9.85, 9.75, ...
a a8 = 9.35
b an = 10.15 0.1n
c an = an 1 0.1, a1 = 10.05

17 100, 99, 98, 97, ...

a a8 = 93
b an = 101 n
c an = an 1 1, a1 = 100

5
1
, 1, , ...
2
2
17
a a8 =
2
7 3n
b an =
2
3
c an = an 1 , a1 = 2
2

18 2,

19 3, 6, 12, 24 , ...
a a8 = 384

b an = 3 2n1
c an = 2an 1, a1 = 3
20 4 , 12, 36, 108, ...
a a8 = 8748

b an = 4 3n 1
c an = 3an 1, a1 = 4
21 5, 5, 5, 5, ...
a a8 = 5

b an = 5 ( 1)n 1
c an = an 1, a1 = 5
22 3, 6, 12, 24 , ...

a a8 = 384
b an = 3 ( 2)n 1
c an = 2an 1, a1 = 3

23 972, 324, 108, 361, ...

4
a a8 =
9
n 1

1
b an = 972
3
a1 = 972

c
1
an = 3 an1, n > 1

Chapter 3

24 a1 = 15
a7 = 21
a7 = a1 + 6d 15 + 6d = 21 d = 6

The arithmetic sequence is: 15, 9, 3, 23, 29, 215, 221.

25 a1 = 99
a5 = 100

3n 1 = 39 n = 10

1
a5 = a1 + 4 d 99 + 4 d = 100 d = = 0.25
4
The arithmetic sequence is: 99, 99.25, 99.5, 99.75, 100.

26 a3 = 11
a12 = 47

a1 + 2d = 11
9d = 36 d = 4 a1 = 3
a1 + 11d = 47
So
o, an = 3 + ( n 1) 4 = 4 n 1.

27 a7 = 48
a13 = 10

32 a1 = 7
r=3
a = 137 781
n
an = a1 r n 1 137 781 = 7 3n 1 3n 1 = 19 683
So,
is the 10th term of the sequence.
a3 =137781
18
33 a = 243
6
4
19 683
an =
64
a1 r 2 = 18
a r 5 = 243
1
4
243
27
3
3
r =
: 18 r 3 =
r=
4
8
2
2
3
2
a1 = 18 : r = 18 : = 8
2

a1 + 6d = 48
19
a1 = 86
6d = 38 d =
a1 + 12d = 10
3
So, an = 86 + ( n 1)

19 19
277
.
=
n
3
3
3

28 a1 = 7
a6 = 1701

n1

3
19 683
= 8

2
64
n1
9
3
3
9 =
n = 100
2
2

an = a1 r n 1

19 683
is the 10th term of the sequence.
64

a6 = a1 r 7r = 1701 r = 243 r = 3 r = 3 34 P = 2500, n = 2, t = 10 , r = 0.04

nt
20
The geometric sequence is: 7, 21, 63, 189, 567, 1701.

r
0.04
A = P 1 + = 2500 1 +
3714.87

2
29 a1 = 9
Tims account will hold 3714.87.
a3 = 64
5

a3 = a1 r 2 9r 2 = 64 r 2 =
a2 = 24

64
8
r1, 2 =
9
3

The geometric mean is positive, so the answer is 24.

30 a1 = 24
a3 = 6

a3 = a1 r 2 24 r 2 = 6 r 2 =
3

1
a4 = a1 = 3
2
n1

1
an = 24
2
3
31 r =
7
14
a4 =
3
14 3 98
a3 = a4 : r =
: =
3 7
9

1
1
r1, 2 =
4
2

35 P = 1000, n = 4 , t = 18, r = 0.06

nt

418

r
0.06
A = P 1 + = 1000 1 +
2921.16
n

4
William will have 2921.16 on his 18th birthday.

36 A = 3000, n = 4 , t = 6, r = 0.06
nt
46

r
0.06
A = P 1 + P = 3000 : 1 +
2098.63

4
You should invest 2098.63 now.
37 a1 = 13
d=6
an = 367
an = a1 + ( n 1)d 367 = 13 + ( n 1)6 n = 60
60
(13 + 367 ) = 11400
S60 =
2

38 The series is geometric.

a1 = 2
2
r=
3
4096
an =
177 147
2
an = a1r n 1 2
3

39

2

3

n1

c 3.2129 = 3.21 + 0.002 929 29... = 3.21 + 0.0029 + 0.000 029 + ...

n1

4096

177 147

11

2
n = 12
3
12
2 1

rn 1
3
S12 = a1
=2
1.190 75
2 1
r 1

3
=

11

(3 + 0.2k ) = 3 + 3.2 + ... + 5.2 =

k=0

12
( 3 + 5.2) = 49.2
2

40 This is an infinite geometric series with

2
a1 = 2 and r = .
3
S =

a1
6
2
=
=
1 r 1 2 5
+
3

41 This is an infinite geometric series with

1
2
a1 = and r =
.
2
3
1
a1
3
3 + 2 3+ 6
2
S =
=
=

=
1 r
2
2 2( 3 2 ) 3 + 2
1
3

42 In each case we have an infinite geometric series.


a 0.7 = 0.7777... = 0.7 + 0.07 + 0.007 + ...
1

= 7 10 + 7 10 + 7 10 + ...

7
1
a1 =
,r =
10
10


7

0.7 = S = 10 = 7

1
9

1
10


0.345 = 0.3 + 0.045 4545... = 0.3 + 0.045 + 0.000 45 + ...
b


= 0.3 + 45 10 3 + 45 10 5 + 45 10 7 + ...

45
1
a1 =
,r =
1000
100

45
3
45 19
1000

=
+
=
0.345 = 0.3 + S = 0.3 +
1
10 990 55

1

100

= 3.21 + 29 10 4 + 29 10 6 + 29 10 8 + ...
1
29
,r =
a1 =
100
10 000
29
7952
10 000 321 29
+
=
3.2129 = 3.21 + S = 3.21 +
=
1
100 9900 2475
1
100

43 ( 2x 3)9 = 9 ( 2x )9 i ( 3)i
i=0 i

9

9
i = 3 ( 2x )9 3 ( 3)3 = 84 64 x 6 ( 27 ) = 145152x 6
3

The coefficient of x6 is 145152.

7
7
44 ( a x + b )7 = ( a x )7 i ( 3)i

i=0

7
i=4
4

74 4
3 3 4
( a x ) b = 35a x b

The coefficient of x 3b 4 is 35a3 .

15
15
15 2 15 i
2
( z )i
45 2 z =
z 2
z

i=0

15 2 15 10
32
( z )10 = 3003 10 z 10 = 96 096
i = 10
2

10
z
z

The constant term is 96096.

46 ( 3n 2m )5 = 5 ( 3n)5 ( 2m )0 + 5 ( 3n)4 ( 2m )1
0
1
5
5
+ ( 3n)3 ( 2m )2 + ( 3n)2 ( 2m )3
2
3

5
5
+ ( 3n)1( 2m )4 + ( 3n)0 ( 2m )5
4
5

= 243n5 810n4 m + 1080n3m2 720n2 m3 + 240nm 4 32m5

9

9
i
47 ( 4 + 3r 2 ) = 9 4 9i (3r 2 )
i

i=0

9 4 2 5
i = 5 4 (3r ) = 126 256 243r 10 = 7 838 208r 10
5
The coefficient of r10 is 7838208.

48 a1 = 4
a4 = 19
an = 99
4 + 3d = 19

d = 5 and n = 20
4 + ( n 1)d = 99

Chapter 3

49 Nicks studying hours form an arithmetic sequence

with first term a1 = 12 and common difference d = 2.
Charlottes studying hours form a geometric sequence
with first term b1 = 12 and common ratio r = 1.1.
a a5 = a1 + 4 d = 12 + 4 2 = 20


b5 = b1r 4 = 12 1.14 17.57
In week 5, Nick studied for 20 hours and Charlotte
studied for 17.57 hours.

b Sarithmetic15

Sgeometric15

15
= [2 12 + (15 1)2] = 390
2
1.115 1
381.27
= 12
1.1 1

For the 15 weeks, Nick studied for a total of 390 hours

and Charlotte studied for a total of 381.27 hours.

c bn > 40 12 1.1n 1

40
12
> 40 n >
+ 1 n > 13.6
log 1.1
log

Charlotte will exceed 40 hours of study per week in

the 14th week.
d 
We need to determine n so that bn an . The easiest
way of doing this is by entering n (as X), an (as Y1),
and bn (as Y2) into Table in a GDC.

51 a The initial amount forms a geometric sequence with

a1 = 500 and common ratio r = 1.06 (fixed rate 6% per
annum).
After 10 years it will be worth: a11 = a1r10 = 500 . 1.0610
= 895.42 = 895 to the nearest euro.
b 
The future value is a partial sum of a geometric
sequence:
1.0611 1
r 11 1
1 = 6985.82
1 = 500
FV = a1
1.06 1

r 1

= 6986 to the nearest euro.

52 6, 9.5, 13, ... is an arithmetic sequence with a1 = 6 and

d = 3.5.
a a40 = a1 + 39d = 6 + 39 3.5 = 142.5
103
b S103 =
[2a1 + (103 1)d ]
2
103
=
(2 6 + 102 3.5) = 19 003.5
2
53 The training programme forms an arithmetic sequence
with a1 = 2 and d = 0.5.
a an = 20 2 + ( n 1) 0.5 = 20 n = 37
She first runs a distance of 20 km on the 37th day of
her training.
37
b
S37 =
(2 2 + 36 0.5) = 407
2
The total distance run during 37 days of training
would be 407 km.
54 a r =

2400 3600 3
=
=
1600 2400 2

b 
The number of new participants in 2012 is the 13th
term in the sequence.
12

We see that in week 12 Nick will study for 34 hours

while Charlotte studies for 34.24 hours. So, Charlotte
will catch up with Nick in week 12.
50 Plan A forms an arithmetic sequence with first term
a1 = 1000 and common difference d = 80. Plan B forms
a geometric sequence with first term b1 = 1000 and
common ratio r = 1.06.
a b2 = 1000 1.06 = 1060 g


b3 = 1000 1.062 = 1123.6 g

b a12 = a1 + 11d = 1000 + 11 80 = 1880 g

b12 = b1 1.0611 1898.33 g
12
c i SA12 = (2a1 + 11d ) = 6 (2 1000 + 11 80) = 17 280 g
2

ii SB12 = b1

1.0612 1
rn 1
16 869.9 g
= 1000
1.06 1
r 1

3
a13 = a1r 12 = 1600 = 207 594
2


n1
3
> 50 000
c an > 50 000 1600
2

n1
log 31.25
3
+ 1 n > 9.489
> 31.25 n >

2
log 1.5
The 10th term of the sequence will be greater than
50000; therefore, the number of new participants will
first exceed 50000 in 2009.

1.513 1
r 13 1
= 619 582
= 1600
1.5 1
r 1
e 
This trend in growth would not continue due to
market saturation.

d
S13 = a1

55 a1 = 25
a4 = 13 a1 + 3d = 13 25 + 3d = 13 d = 4
an = 11995 a1 + (n 1) d = 11995

25 + (n 1) (4) = 11995 n = 3006

56 a MN = + =
2
2

60 The Kell scheme forms an arithmetic sequence with

a1 = 18000 and d = 400. The IBO scheme forms a
geometric sequence with b1 = 17000 and r = 1.07.

2
2
=
4
2

2
1
b
AreaMNPQ = =
2
2
2

a
All answers are in euros.
2

1 2
1 2
c i RS = 2 2 + 2 2 =

i Kell: a2 = 18 000 + 400 = 18 400, a3 = a2 + 400 = 18 800

IBO: b2 = 17 000 1.07 = 18 190, b3 = b2 1.07 = 19 463.3
10
ii Kell: S10 = ( 2 18 000 + 9 400 ) = 198 000
2
1.0710 1
= 234 879.62
IBO: S10 = 17 000
1.07 1

iii Kell: a10 = 18 000 + 9 400 = 21600

1 1 1
+ =
8 8 2

2
1
1
Area
=
=

RSTU
2
4
1
1
1, 1 , 1 , ... r = 2 = 4 = 1
ii
1 2
2 4
1
2
9
1
1
d i Area10 = 1 =
2
512
a1
1

=
=2
ii S =
1 r 1 1

2

IBO: b10 = 17 000 1.07 = 31253.81

b i 
From a ii we can see that b3 > a3, so Merijayne will
start earning more than Tim in the third year.
9

ii 
We can compare their total earnings with the help
of a GDC. If X denotes the year, Y1 represents
total earnings for Tim and Y2 for Merijayne, we
have:

57 Tims swimming programme forms an arithmetic

sequence with a1 = 200 and d = 20.
a
a52 = a1 + 51d = 200 + 51 20 = 1220
Tim will swim 1220 metres in the final week.
52
52
b S52 =
(2a1 + 51d ) = (2 200 + 51 20) = 36 920
2
2

3
58 a Area A = = 1
3

1
1
AreaB = =
3
9

In the fourth year, Merijaynes total earnings will

exceed those of Tim.

Altogether, Tim swims 36920 metres.

1
1
b AreaC = =
9
81

61 The number of seats in each row forms an arithmetic

sequence with a1 = 16 and d = 2.
a
a24 = a1 + 23d = 16 + 23 2 = 62
24
b
S24 = 16 + 18 + ... + 62 =
(16 + 62) = 936
2

8
1
c Shaded area2 = 1 + 8 = 1 +
9
9

8
8
8
= 1 + +

9
9
81
2
8
8
1
d Shaded area = 1 + + + ... =
=9
9
8
9

1
9
Unshaded area = 9 Shaded area = 0
Shaded area3 = Shaded area2 + 8

59 a i The series 2 + 22 + 222 + 2222 + ... is neither

arithmetic nor geometric.
4 8 16
ii 
The series 2 + + +
+ ... is geometric
3 9 27
2
with r = <1; thus converging.

62 The values of the investment after each year form a

geometric sequence with a1 = 7000, r = 1.0525, and an+1
represents the value of the investment after n years.
a
Value of investment = 7000 . 1.0525t
10
t
t
b 7000 1.0525 = 10 000 1.0525 = 7
10
log
7
t =
= 6.97
log(1.0525)

The minimum number of years is 7.

c 
If the rate of 5% is compounded quarterly, the value of
3
the investment over 7 years would be:
iii 
The series 0.8 + 0.78 + 0.76 + 0.74 + ... is arithmetic
74
5
with d = 0.02.
7000 1 +
= 9911.95.

8 32 128
4 100
iv The series 2 + +
+
+ ... is geometric
3 9
27
For 5.25% compounded annually, the value of the
4
investment would be 7000 . 1.05257 = 10015.04.
with r = > 1; thus diverging.
3
2
Therefore, the investment at 5.25% annually is better.
b
For series ii we have: S =
= 6.
2
1
3
5

Chapter 3

63 a S1 = 9 a1 = 9

S2 = 20 a1 + a2 = 20 9 + a2 = 20 a2 = 11
b
d = a2 a1 = 11 9 = 2
c
a4 = a1 + 3d = 9 + 3 2 = 15