SPM Practice Question 1

Posted on April 22, 2020 by user

Question 1:
Diagram below 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 7 cm.

Rahman 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.
(b) the total price of the bricks used if the price of one brick is 50 sen.

Solution:
100, 98, 96, …, 4 is an arithmetic progression
a = 100 and d = –2

(a)
T_n = 4
a + (n – 1) d = 4
100 + (n – 1)(–2) = 4
100 – 2n + 2 = 4
2n = 98
n = 49
Height of the wall = 49 × 7 = 343 cm

(b)
Total number of bricks used
= S₄₉

= \frac{49}{2} (100 + 4) \leftarrow S_{n} = \frac{n}{2} (a + l)

= 2548
Total price = 2548 × RM0.50
= RM1,274

2. Geometric progression

Posted on April 22, 2020 by user

2. Geometric Progression

(A) Characteristics of Geometric Progressions

Geometric progression is a progression in which the ratio of any term to the immediate term before is a constant. The constant is called common ratio, r.

Example:

Determine whether or not each of the following number sequences is a geometric progression (GP).

(a) 1, 4, 16, 64, …..

(b) 10, –5, 2.5, –1.25, …..

(c) 2, 4, 12, 48, …..

(d) –6, 1, 8, 15, …..

Solution:

[Smart Tips: For GP, times a fixed number every time to get the next time.]

(a)

\begin{array}{l} Common ratio, r = \frac{T_{n}}{T_{n - 1}} \\ \frac{T_{3}}{T_{2}} = \frac{16}{4} = 4, \frac{T_{2}}{T_{1}} = \frac{4}{1} = 4 \\ \frac{T_{3}}{T_{2}} = \frac{T_{2}}{T_{1}} \end{array}

This is a GP, a =1, r = 4.

(b)

\begin{array}{l} Common ratio, \\ r = \frac{T_{2}}{T_{1}} = - \frac{5}{10} = - \frac{1}{2} \end{array}

This is a GP, a =1, r = – ½.

(c)

This is NOT a GP.

This is because the ratio of each term to its preceding term is not a similar constant.

(B) The steps to prove whether a given number sequence is a geometric progression.

Step 1: List down any three consecutive terms. [Example: T₁, T₂, T₃.]

Step 2: Calculate the values of

\frac{T_{3}}{T_{2}} and \frac{T_{2}}{T_{1}} .

Step 3: If

\frac{T_{3}}{T_{2}} = \frac{T_{2}}{T_{1}} = r

, then the number sequence is a geometric
progression.

Step 4: If

\frac{T_{3}}{T_{2}} \neq \frac{T_{2}}{T_{1}}

, then the number sequence is not a geometric progression.

SPM Practice Question 1

Posted on April 22, 2020 by user

Question 1:
Diagram below shows part of an arrangement of bricks of equal size.

= \frac{49}{2} (100 + 4) \leftarrow S_{n} = \frac{n}{2} (a + l)

= 2548
Total price = 2548 × RM0.50
= RM1,274

SPM Practice Question 10 – 12

Posted on April 22, 2020 by user

Question 10:
In a geometric progression, the first term is 27 and the fourth term is 1. Calculate
(a) the positive value of common ratio, r,
(b) the sum of the first n terms where n is sufficiently large till

r^{n} \approx 0

Solution:

Question 11:
Express the recurring decimal 0.187187187 … as a fraction in its simplest form.

Solution:

Question 12:

\begin{array}{l} Given that \frac{1}{p} = 0.1666666..... \\ = q + a + b + c + ..... \end{array}

where p is a positive integer. If q = 0.1 and a + b + c are the first three terms of a geometric progression, state the value of a, b and c in decimal form. Hence find the value of p.

Solution:

(F) Sum of the First n Terms of Geometric Progressions

Posted on April 22, 2020 by user

1.4.3 Sum of the First n Terms of Geometric Progressions

(F) Sum of the First n Terms of Geometric Progressions

\begin{array}{l} S_{n} = \frac{a (r^{n} - 1)}{r - 1}, r > 1 \\ S_{n} = \frac{a (1 - r^{n})}{1 - r}, r < 1 \end{array}

a = first term

r = common ratio

n = number of term

S_n = sum of the first n terms

Example 1:

Find the sum of each of the following geometric progressions.

(a) 1, 2, 4, ... up to the first 7 terms

(b) 9, 3, 1, ⅓, ... up to the first 6 terms
(c) 12, 3, ....,

\frac{3}{64}

[Smart TIPS: You can find n if you know the last term]

Solution:

SPM Practice Question 7 – 9

Posted on April 22, 2020 by user

Question 7:
Given a geometric progression

\frac{2}{z}, 3, \frac{9 z}{2}, q, ....

express q in terms of z.

Solution:

Question 8:
The second and the fourth term of a geometry progression are 10 and

\frac{2}{5}

respectively. Find
(a) The first term and the common ratio where r > 0,
(b) The sum to infinity of the geometry progression.

Solution:

Question 9:
In a geometric progression, the first term is 18 and the common ratio is r.
Given that the sum to infinity of this progression is 21.6, find the value of r.

Solution:

(G) Sum to Infinity of Geometric Progressions (Part 1)

Posted on April 22, 2020 by user

1.4.4 Sum to Infinity of Geometric Progressions

(G) Sum to Infinity of Geometric Progressions

S \infty = \frac{a}{1 - r}, - 1 < r < 1

a = first term

r = common ratio

S∞ = sum to infinity

Example:

Find the sum to infinity of each of the following geometric progressions.

(a) 8, 4, 2, ...

(b)

\frac{2}{3}, \frac{2}{9}, \frac{2}{27}, .....

(c) 3, 1, ⅓, ….

Solution:

(a)

8, 4, 2, ….

a = 2, r = 4/8 = ½
S∞ = 8 + 4 + 2 + 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + …..

S \infty = \frac{a}{1 - r} = \frac{2}{1 - \frac{1}{2}} = 4

(b)

\begin{array}{l} \frac{2}{3}, \frac{2}{9}, \frac{2}{27}, ..... \\ a = \frac{2}{3}, r = \frac{2 / 9}{2 / 3} = \frac{1}{3} \\ S \infty = \frac{a}{1 - r} \\ S \infty = \frac{\frac{2}{3}}{1 - \frac{1}{3}} = 1 \end{array}

(c)

\begin{array}{l} 3, 1, \frac{1}{3}, ..... \\ a = 3, r = \frac{1}{3} \\ S \infty = \frac{a}{1 - r} \\ S \infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{2 / 3} = \frac{9}{2} \end{array}

Question 2:

Table shows the prices, the price indices and weightage for four types of stationery A, B, C and D.

(a) Find the value of

(i) x, (ii) y, (iii) z.

(b) The composite index for the price of the stationery in the year 2009 based on the year 2008 is 118.5. Calculate the value of m.

(c) The total expenditure for the stationery in the year 2008 is RM350. Calculate the corresponding total expenditure in the year 2009.

(d) The price index for B in the year 2010 based on the year 2008 is 132. Calculate the price index for B in the year 2010 based on the year 2009.

Solution:
(a)(i)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ x = \frac{P_{2009}}{P_{2008}} \times 100 \\ x = \frac{2}{2.5} \times 100 \\ x = 80 \end{array}

(a)(ii)

\begin{array}{l} 135 = \frac{P_{2009}}{2} \times 100 \\ 135 = \frac{y}{2} \times 100 \\ y = R M 2.70 \end{array}

(a)(iii)

\begin{array}{l} 116 = \frac{5.80}{P_{2008}} \times 100 \\ 116 = \frac{5.80}{z} \times 100 \\ z = RM5 \end{array}

(b)

\begin{array}{l} Composite index for items in the year 2009 based on the year 2008, \bar{I} = 118.5 \\ \bar{I} = \frac{\sum I W}{\sum W} \\ 118.5 = \frac{(80) (2) + (120) (3) + (135) (m) + (116) (4)}{2 + 3 + m + 4} \\ 118.5 = \frac{984 + 135 m}{9 + m} \\ 1066.5 + 118.5 m = 984 + 135 m \\ 16.5 m = 82.5 \\ m = 5 \end{array}

(c)

\begin{array}{l} \bar{I} = \frac{P_{2009}}{P_{2008}} \times 100 \\ 118.5 = \frac{P_{2009}}{350} \times 100 \\ ∴ Total expenditure in the year 2009, P_{2009} = R M 414.75. \end{array}

(d)

\begin{array}{l} For stationery B, \\ It is given \frac{P_{2010}}{P_{2008}} \times 100 = 132 \to \frac{P_{2010}}{P_{2008}} = \frac{132}{100} \\ and \frac{P_{2009}}{P_{2008}} \times 100 = 120 \to \frac{P_{2009}}{P_{2008}} = \frac{120}{100} \\ Price index for stationery B in the year 2010 based on the year 2009, \\ I_{10 / 09} = \frac{P_{2010}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{P_{2010}}{P_{2008}} \times \frac{P_{2008}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{132}{100} \times \frac{100}{120} \times 100 \\ I_{10 / 09} = 110 \end{array}

SPM Add Maths

SPM Practice Question 1

2. Geometric progression

SPM Practice Question 1

SPM Practice Question 10 – 12

(F) Sum of the First n Terms of Geometric Progressions

SPM Practice Question 7 – 9

(G) Sum to Infinity of Geometric Progressions (Part 1)

(Long Questions) – Question 4

(Long Questions) – Question 3

(Long Questions) – Question 2