SPM Practice Question 4 – 6

Posted on May 16, 2020 by Myhometuition

Question 4:
The sum of the first n terms of the geometric progression 5, 15, 75, … is 5465.
Find the value of n.

Solution:

Question 5:
The first three terms of a geometric progression are 5k + 6, 2k, k – 2.
Find
(a) the positive value of k,
(b) the sum from the third term to the sixth term, using the value of k obtained in (a)

Solution:

Question 6:
The first three terms of a sequence are 2, x, 18. Find the positive value of x so that the sequence is
(a) an arithmetic progression,
(b) a geometric progression.

Solution:

SPM Practice Question 4

Posted on May 16, 2020 by Myhometuition

Question 4:
Mr Choong started working for a company on 1 January 2008 with an initial annual salary of RM28800. Every January, the company increased his salary by 5% of the previous year’s salary.

Calculate
(a) his annual salary, to the nearest RM on 1 January 2013,
(b) the minimum value of n such that his annual salary in the nth year will exceed RM40 000,
(c) the total salary, to the nearest RM, paid to him by the company, for the years 2008 to 2013.

Solution:

SPM Practice Question 3

Posted on May 16, 2020 by Myhometuition

Question 3:
An arithmetic progression has 16 terms. The sum of the 16 terms is 188, and the sum of the even terms is 96. Find
(a) the first term and the common difference,
(b) the last term.

Solution:
(a) Let the first term = a
Common difference = d

\begin{array}{l} Given S_{16} = 188 \\ Thus, \frac{16}{2} [2 a + 15 d] = 188 \\ 8 [2 a + 15 d] = 188 \\ 2 a + 15 d = \frac{188}{8} \\ 2 a + 15 d = 23.5 - - - (1) \end{array}

Given the sum of the even terms = 96

\begin{array}{l} T_{2} + T_{4} + T_{6} + ..... + T_{16} = 96 \\ (a + d) + (a + 3 d) + (a + 5 d) + ..... + (a + 15 d) = 96 \\ \frac{8}{2} [(a + d) + (a + 15 d)] = 96 \\ 4 [2 a + 16 d] = 96 \\ 2 a + 16 d = 24 - - - (2) \end{array}

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5
Substitute d = 0.5 into (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Therefore, first term = 8 and common difference = 0.5.

(b) Last term
= T₂
= 8 + 15 (0.5)
= 8 +7.5
= 15.5

SPM Practice Question 10 – 12

Posted on April 22, 2020 by user

Question 10:
The sequence –11, –5, 1,… is an arithmetic progression. State the three consecutive terms of this arithmetic progression where the sum of these three terms is 93.

Solution:

Question 11:
An arithmetic series, with the first term 12 and common difference d, consists of 23 terms. Given that the sum of the last 3 terms is 5 times the sum of the first 3 terms, find
(a) the value of d,
(b) the sum of the first 19 terms.

Solution:

Question 12:
The sum of n terms of an arithmetic progression is given by the formula

S_{n} = \frac{n}{2} (5 - 3 n)

. Find
(a) the first term,
(b) the common difference,
(c) the tenth term.

Solution:

SPM Practice Question 7 – 9

Posted on April 22, 2020 by user

Question 7:
Find the sum of all the multiples of 7 between 100 and 500.

Solution:

Question 8:
If

\log_{10} p, \log_{10} p q and \log_{10} p q^{2}

are the first three terms of a progression, show that it forms an arithmetic progression.

Solution:

Question 9:

Show that the volumes of the cylinders in the above diagram form an arithmetic progression and state its common difference.

Solution:

SPM Practice Question 1 – 3

Posted on April 22, 2020 by user

Question 1:
The third and eighth terms of an arithmetic progression are –5 and 15 respectively. Find
(a) The first term and the common difference
(b) The sum of the first 10 terms.

Solution:
(a)

\begin{array}{l} T_{3} = - 5 \to Use T_{n} = a + (n - 1) d \\ a + 2 d = - 5 \to (1) \\ T_{8} = 15 \\ a + 7 d = 15 \to (2) \\ (2) - (1), \\ 5 d = 20 \\ d = 4 \\ Substitute d = 4 into (1), \\ a + 2 (4) = - 5 \\ a = - 13 \end{array}

(b)

\begin{array}{l} S_{10} = \frac{10}{2} [2 (- 13) + 9 (4)] \to Use S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ = 50 \end{array}

Question 2:
The first three terms of an arithmetic progression are 2k, 3k + 3, 5k + 1. Find
(a) the value of k,
(b) the sum of the first 15 terms of the progression.

Solution:

Question 3:
Given an arithmetic progression p + 9, 2p + 10, 7p – 1,… where p is a constant. Find
(a) value of p,
(b) the sum of the next five terms.

Solution:

The nth Term of Arithmetic Progression (Example 1 & 2)

Posted on April 22, 2020 by user

The nth Term of Arithmetic Progression (Examples)

Example 1:

If the 20th term of an arithmetic progression is 14 and the 40th term is –6,
Find

(a) the first term and the common difference,

(b) the 10th term.

Solution:

(a)

T₂₀ = 14

a + 19d = 14 ----- (1) ← (T_n= a + (n– 1) d

T₄₀ = – 6

a + 39d = – 6 ----- (2)

(2) – (1),

20d = – 20

d = – 1

Substitute d = – 1 into (1),

a + 19 (– 1) = 14

a = 33

(b)

T₁₀ = a + 9d

T₁₀ = 33 + 9 (– 1)

T₁₀ = 24

Example 2:

The 3rd term and the 7th term of an arithmetic progression are 20 and 12 respectively.

(a) Calculate the 20th term.

(b) Find the term whose value is – 34.

Solution:

(a)

T₃ = 20

a + 2d = 20 ----- (1) ← (T_n= a + (n – 1) d

T₇ = 12

a + 6d = 12 ----- (2)

(2) – (1),

4d = – 8

d = – 2

Substitute d = – 2 into (1),

a + 2 (– 2) = 20

a = 24

T₂₀ = a + 19d

T₂₀ = 24 + 19 (– 2)

T₂₀ = –4

(b)
T_n = –34
a + (n – 1) d = –34

24 + (n – 1) (–2) = –34

(n – 1) (–2) = –58

n – 1 = 29

n = 30

(C) The nth term of an Arithmetic Progression

Posted on April 22, 2020 by user

1.2.2 The nth term of an Arithmetic Progression

(C) The nth term of an Arithmetic Progression

T_n= a + (n − 1) d

where

a = first term

d = common difference

n = the number of term

T_n = the nth term

(D) The Number of Terms in an Arithmetic Progression

Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 1:

Find the number of terms for each of the following arithmetic progressions.

(a) 5, 9, 13, 17... , 121

(b) 1, 1.25, 1.5, 1.75,..., 8

Solution:

(a)

5, 9, 13, 17... , 121

AP,

a = 5, d = 9 – 5 = 4

The last term, T_n = 121

a + (n – 1) d = 121

5 + (n – 1) (4) = 121

(n – 1) (4) = 116

(n – 1) =

\frac{116}{4}

= 29

n = 30

(b)

1, 1.25, 1.5, 1.75,..., 8

AP,

a = 1, d = 1.25 – 1 = 0.25

T_n = 8

a + (n – 1) d = 8

1 + (n – 1) (0.25) = 8

(n – 1) (0.25) = 7

(n – 1) = 28

n = 29

(E) The Consecutive Terms of an Arithmetic Progression

If a, b, c are three consecutive terms of an arithmetic progression, then

c – b = b – a

Example 2:

If x + 1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression, find the value of x and its common difference.

Solution:

x + 1, 2x + 3, 6

c – b = b – a

6 – (2x + 3) = (2x + 3) – (x + 1)

6 – 2x – 3 = 2x + 3 – x – 1

3 – 2x = x + 2

\begin{array}{l} x = \frac{1}{3} \\ \frac{1}{3} + 1, 2 (\frac{1}{3}) + 3, 6 \\ \frac{4}{3}, 3 \frac{2}{3}, 6 \\ d = 3 \frac{2}{3} - \frac{4}{3} = 2 \frac{1}{3} \end{array}

1. Arithmetic progression

Posted on April 22, 2020 by user

Progression

1.2.1 Arithmetic progression
(A) Characteristics of Arithmetic Progression

An arithmetic progression is a progression in which the difference between any term and the immediate term before is a constant. The constant is called the common difference, d.

d = T_n – T_n-1 or d = T_n+1 – T_n

Example 1:

Determine whether the following number sequences is an arithmetic progression (AP) or not.

(a) –5, –3, –1, 1, …

(b) 10, 7, 4, 1, -2, …

(c) 2, 8, 15, 23, …

(d) 3, 6, 12, 24, …

Smart TIPS: For an arithmetic progression, you always plus or minus a fixed number

Solution:

(B) The steps to prove whether a given number sequence is an arithmetic progression

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

Step 2: Calculate the values of T₃ − T₂ and T₂ − T₁ .

Step 3: If T₃ − T₂ = T₂ − T₁ = d, then the number sequence is an arithmetic progression.

[Try Question 8 and 9 in SPM Practice 1 (Arithmetic Progression)]

Example 2:

Prove whether the following number sequence is an arithmetic progression

(a) 7, 10, 13, …

(b) –20, –15, –9, …

Solution:

(a)

7, 10, 13 ← (Step 1: List down T₁ , T₂ , T₃)

T₃– T₂ = 13 – 10 = 3 ← (Step 2: Find T₃– T₂ and T₂ – T₁)

T₂ – T₁ = 10 – 7 = 3 ← (Step 2: Find T₃– T₂ and T₂ – T₁)

T₃– T₂ = T₂ – T₁

Therefore, this is an arithmetic progression.

(b)

–20, –15, –9

T₃– T₂ = –9 – (–15) = 6

T₂ – T₁ = –15 – (–20) = 5

T₃– T₂ ≠ T₂ – T₁

Therefore, this is not an arithmetic progression.

3. The nth Term of Geometric Progressions (Part 1)

Posted on April 22, 2020 by user

1.4.2 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions

T_{n} = a r^{n - 1}

a = first term

r = common ratio

n = the number of term

T_n = the nth term

Example 1:

Find the given term for each of the following geometric progressions.

(a) 8 ,4 ,2 ,...... T₈

(b) $\frac{16}{27}, \frac{8}{9}, \frac{4}{3}, .....$ , T₆

Solution:

\begin{array}{l} T_{n} = a r^{n - 1} \\ T_{1} = a r^{1 - 1} = a r^{0} = a \leftarrow (First term) \\ T_{2} = a r^{2 - 1} = a r^{1} = a r \leftarrow (S e c o n d term) \\ T_{3} = a r^{3 - 1} = a r^{2} \leftarrow (T h i r d term) \\ T_{4} = a r^{4 - 1} = a r^{3} \leftarrow (Fourth term) \end{array}

(a)

\begin{array}{l} 8, 4, 2, ..... \\ a = 8, r = \frac{4}{8} = \frac{1}{2} \\ T_{8} = a r^{7} \\ T_{8} = 8 {(\frac{1}{2})}^{7} = \frac{1}{16} \end{array}

(b)

\begin{array}{l} \frac{16}{27}, \frac{8}{9}, \frac{4}{3}, ..... \\ a = \frac{16}{27} \\ r = \frac{T_{2}}{T_{1}} = \frac{\frac{16}{27}}{\frac{8}{9}} = \frac{2}{3} \\ T_{6} = a r^{5} \\ = \frac{16}{27} {(\frac{2}{3})}^{5} = \frac{512}{6561} \end{array}