Problem Solving
Arithmetic Progression
and Geometric Progression
Last Updated: October 11, 2005
Review -- Arithmetic
nth term
Tn  a  d(n  1)
Tn  Sn  Sn 1
Jeff Bivin -- LZHS
Sum of n terms
n(a1  an )
Sn 
2
n
Sn  [ 2a  ( n  1)d ]
2
Review -- Geometric
nth term
an =
(n-1)
a1·r
Sum of n terms
a1  a1r
Sn 
1 r
n
a1  an r
Sn 
1 r
Jeff Bivin -- LZHS
Example 1
The sum of the first n terms of a progression
is given by Sn = n2 + 3n. Find, in terms of n
the nth term.
Sn = n2 + 3n
Sn-1 = (n-1)2 + 3(n-1)
= n2 – 2n + 1 + 3n – 3
= n2 + n – 2
Tn = Sn – Sn – 1
= n2 + 3n – (n2 + n – 2)
= 2n + 2
Jeff Bivin -- LZHS
SPM ANALYSIS
YEAR PROGRESSION TOTAL
MARKS
2004 GP
8 MARKS
2005
AP
6 MARKS
2006
AP
7 MARKS
Year 2004,
SPM PAPER 2
y cm
x cm
The diagram shows the arrangement of the first 3 of an
infinite series of similar triangles. The first triangle has a
base of x cm and a height of y cm. The measurements of the
base and the height of each subsequent triangle are half of
the measurement of its previous one.
(a) Show that the areas of the triangles form a geometric
progression and state the common ratio.
[3 marks]
(b) Given that x = 80 and y = 40
[5 marks]
(i) determine which triangle has an area of 6.25 cm2
(ii) find the sum to infinity of the areas in cm2 of the triangles
Jeff Bivin -- LZHS
SOLUTION (a)
Area of 1st triangle: A1 =
1 xy
2
Area of 2nd triangle: A2=
1  1 x  1 y   1 xy

8
2  2 
2


Area of 3rd triangle: A3 =
1  1 x  1 y 
1

xy



2  4  4  32
1 xy
1 xy
8
 32
1
1 xy
1 xy
4
2
8
A2 A3 1


A1 A2 4
The area of triangles form a GP with common ratio  1
4
Jeff Bivin -- LZHS
SOLUTION (b)(i)
Area of nth triangle: An = A1r n – 1
n 1
1 xy  1 


2 4
n 1
 1 ( 80 )( 40 )  1 
2
4
6.25 =
n 1
6.25   1 
1600  4 
n 1
1 1
256  4 
4
1 1
 4
 4
 
 
n 1
n 1  4
n5
Jeff Bivin -- LZHS
The fifth triangle has area of 6.25 cm2
SOLUTION (b)(ii)
Sum to infinity
a

1 r
1 ( 80 )( 40 )
2
1 1
4
1600

3
4
 2133 1
3
Jeff Bivin -- LZHS
cm2
Year 2005, SPM PAPER 2
Diagram above shows part of an arrangement of bricks of
equal size.
The number of bricks in the lowest row is 100. For each of
the other rows, the number of bricks is 2 less than in the row
below. The height of each brick is 6 cm.
Ali builds a wall by arranging bricks in this way.
The number of bricks in the highest row is 4. Calculate
(a) The height, in cm, of the wall
[3 marks]
(b) The total price of the bricks used if the price of
of one brick is 40 sen.
[3 marks]
Jeff Bivin -- LZHS
Year 2005, SPM PAPER 2
The arrangement of bricks is in arithmetic progression:
100, 98, 96, 94, 92, ….8, 6, 4.
a = 100, d = -2
Let Tn = 4
100 + (n – 1)(-2) = 4
96 = 2(n – 1)
48 = n - 1
n = 49
(a) The height of the wall = 49  6 = 294 cm.
(b) Total bricks used =
49 (100  4 )  2 , 548
2
The total price of bricks used = 2, 548  0.40
= RM1, 019.20
Jeff Bivin -- LZHS
Year 2006, SPM PAPER 2
Two companies, Delta and Omega, start to sell cars at the same
time.
(a) Delta sells k cars in the first month and its sales increase
constantly by m cars every subsequent month. It sells 240 cars
in the 8th month and the total sales for the first 10 months are
1900 cars.
Find the value of k and of m.
[5 marks]
(a) Omega sells 80 cars in the first month and its sales increase
constantly by 22 cars every subsequent month.
If both companies sell the same number of cars in the nth
month, find the value of n.
[2 marks]
Jeff Bivin -- LZHS
k, k+m, k+2m, k+ 3m, k+4m,…..
240 = k + 7m……..(1)
1900 = 5(2k+9m)
380 = 2k + 9m…….(2)
(1)  2 – (2): 480 – 380 = 14m – 9m
100 = 5m
m = 20
Substitute m = 20 into (1):
240 = k + 7(20)
k = 240 – 140
k = 100
Delta: 100, 120, 140, 160, 180, ….
Tn = 100 + 20(n – 1)
Omega: 80, 102, 124, 146, 168, …
Tn = 80 + 22(n – 1)
100 + 20(n – 1) = 80 + 22(n – 1)
100 + 20n – 20 = 80 + 22n – 22
80 + 20n = 58 + 22n
22 = 2n
n = 11