1.
Find the sum of the arithmetic series
17 + 27 + 37 +...+ 417.
(Total 4 marks)
R.
17 + 27 + 37 + ... + 417
17 + (n – 1)10 = 417
10(n – 1) = 400
n = 41
(M1)
(A1)
41
(2(17) + 40(10))
2
= 41(17 + 200)
= 8897
S41 =
(M1)
(A1)
OR
S41 =
41
(17 + 417)
2
(M1)
41
(434)
2
= 8897
=
(A1) (C4)
[4]
4.
Find the sum of the infinite geometric series
2  4  8  16  ...
3 9 27 81
(Total 4 marks)
R.
2
u1
3
S=

1 r
 2
1  
 3
2 3
= 
3 5
2
=
5
(M1)(A1)
(A1)
(A1) (C4)
[4]
1
5.
The Acme insurance company sells two savings plans, Plan A and Plan B.
For Plan A, an investor starts with an initial deposit of $1000 and increases this by $80 each
month, so that in the second month, the deposit is $1080, the next month it is $1160 and so on.
For Plan B, the investor again starts with $1000 and each month deposits 6% more than the
previous month.
(a)
Write down the amount of money invested under Plan B in the second and third months.
(2)
Give your answers to parts (b) and (c) correct to the nearest dollar.
(b)
Find the amount of the 12th deposit for each Plan.
(4)
(c)
Find the total amount of money invested during the first 12 months
(i)
under Plan A;
(2)
(ii)
under Plan B.
(2)
(Total 10 marks)
R.
(a)
(b)
Plan A: 1000, 1080, 1160...
Plan B: 1000, 1000(1.06), 1000(1.06)2…
2nd month: $1060, 3rd month: $1123.60
(A1)(A1)
For Plan A,
For Plan B,
(c)
(i)
(ii)
For Plan A,
For Plan B,
T12 = a + 11d
= 1000 + 11(80)
= $1880
(M1)
(A1)
T12 = 1000(1.06)11
= $1898 (to the nearest dollar)
(M1)
(A1)
12
[2000 + 11(80)]
2
= 6(2880)
= $17280 (to the nearest dollar)
S12 =
1000 (1.06 12  1)
1.06  1
= $16870 (to the nearest dollar)
S12 =
2
4
(M1)
(A1)
(M1)
(A1)
4
[10]
6.
Each day a runner trains for a 10 km race. On the first day she runs 1000 m, and then increases
the distance by 250 m on each subsequent day.
(a)
On which day does she run a distance of 10 km in training?
(b)
What is the total distance she will have run in training by the end of that day? Give your
answer exactly.
(Total 4 marks)
2
R.
(a)
(b)
a1 = 1000, an = 1000 + (n – 1)250 = 10000
10 000  1000
n=
+ 1 = 37.
250
She runs 10 km on the 37th day.
37
(1000 + 10000)
2
She has run a total of 203.5 km
S37 =
(M1)
(A1)
(M1)
(A1)
[4]
7.
Portable telephones are first sold in the country Cellmania in 1990. During 1990, the number of
units sold is 160. In 1991, the number of units sold is 240 and in 1992, the number of units sold
is 360.
In 1993 it was noticed that the annual sales formed a geometric sequence with first term 160,
the 2nd and 3rd terms being 240 and 360 respectively.
(a)
What is the common ratio of this sequence?
(1)
Assume that this trend in sales continues.
(b)
How many units will be sold during 2002?
(3)
(c)
In what year does the number of units sold first exceed 5000?
(4)
Between 1990 and 1992, the total number of units sold is 760.
(d)
What is the total number of units sold between 1990 and 2002?
(2)
During this period, the total population of Cellmania remains approximately 80 000.
(e)
Use this information to suggest a reason why the geometric growth in sales would not
continue.
(1)
(Total 11 marks)
R.
360 240 3

 = 1.5
240 160 2
(a)
r=
(b)
2002 is the 13th year.
u13 = 160(1.5)13–1
= 20759 (Accept 20760 or 20800.)
(c)
5000 = 160(1.5)n–1
5000
= (1.5)n–1
160
 5000 
log 
 = (n – 1)log1.5
 160 
(A1)
1
(M1)
(M1)
(A1)
3
(M1)
(M1)
3
 5000 
log 

160 

n–1=
= 8.49
log 1.5
(A1)
 n = 9.49  10th year
 1999
(A1)
OR
Using a gdc with u1 = 160, uk+1 =
3
uk, u9 = 4100, u10 = 6150
2
1999
d)
(e)
(M2)
(G2)
1.513  1 
S13 = 160 

 1 .5  1 
= 61958 (Accept 61960 or 62000.)
4
(M1)
(A1)
Nearly everyone would have bought a portable telephone so there
would be fewer people left wanting to buy one.
2
(R1)
OR
Sales would saturate.
(R1)
1
[11]
8.
The diagram shows a square ABCD of side 4 cm. The midpoints P, Q, R, S of the sides are
joined to form a second square.
A
Q
P
D
(a)
(i)
Show that PQ = 2 2 cm.
(ii)
Find the area of PQRS.
B
R
S
C
(3)
The midpoints W, X, Y, Z of the sides of PQRS are now joined to form a third square as
shown.
4
A
B
Q
X
W
P
R
Y
Z
S
D
(b)
C
(i)
Write down the area of the third square, WXYZ.
(ii)
Show that the areas of ABCD, PQRS, and WXYZ form a geometric
sequence. Find the common ratio of this sequence.
(3)
The process of forming smaller and smaller squares (by joining the midpoints) is continued
indefinitely.
(c)
(i)
Find the area of the 11th square.
(ii)
Calculate the sum of the areas of all the squares.
(4)
(Total 10 marks)
R.
(a)
(i)
AP 2  AQ 2
PQ =
=
(b)
(M1)
22  22 =
42 = 2 2 cm
(ii)
Area of PQRS = (2 2 )(2 2 ) = 8 cm2
(i)
Side of third square =
 2   2
2
2
(A1)
Area of third square = 4 cm
(A1)
1st 16 2nd 8


2nd 8 3rd 4
(M1)
 Geometric progression, r =
3
= 4 = 2 cm
2
(ii)
(A1)(AG)
8 4 1
 
16 8 2
(A1)
3
10
(c)
(i)
16
1
u11 = u1r = 16   =
1024
2
10
(M1)
=
1
( = 0.015625 = 0.0156, 3 sf) (A1)
64
5
(ii)
S =
u1
16
=
1
1– r
1–
2
= 32
(M1)
(A1)
4
[10]
11.
(a)
Consider the geometric sequence −3, 6, −12, 24, ….
(i)
Write down the common ratio.
(ii)
Find the 15th term.
Consider the sequence x − 3, x +1, 2x + 8, ….
(3)
(b)
When x = 5, the sequence is geometric.
(i)
Write down the first three terms.
(ii)
Find the common ratio.
(2)
(c)
Find the other value of x for which the sequence is geometric.
(4)
(d)
For this value of x, find
(i)
the common ratio;
(ii)
the sum of the infinite sequence.
(3)
(Total 12 marks)
R.
(a)
(i)
r = 2
(ii)
u15 = 3 (2)14
A1
= 49152 (accept 49200)
(b)
(c)
N1
(A1)
A1
N2
(i)
2, 6, 18
A1
N1
(ii)
r=3
A1
N1
Setting up equation (or a sketch)
x 1 2 x  8
(or correct sketch with relevant information)

x  3 x 1
x2 + 2x + 1 = 2x2 + 2x  24
M1
A1
(A1)
x2 = 25
x = 5 or x = 5
x = 5
Notes: If “trial and error” is used, work must be
documented with several trials shown.
A1
N2
6
Award full marks for a correct answer with this
approach.
If the work is not documented, award N2 for a
correct answer.
(d)
1
2
(i)
r=
(ii)
For attempting to use infinite sum formula for a GP
S=
A1
N1
(M1)
8
1
1
2
S = 16
A1
N2
[12]
7