This is the Mathematics Grade 10 self-learning module (Alternative Delivery Mode) for Quarter 1, Module 6: Finding the nth Term of a Geometric Sequence and Geometric Means. It is for Grade 10 learners who already know how to find the next terms of a geometric sequence by multiplying by the common ratio. After working through it, learners can use a formula to find any term of a geometric sequence and insert the geometric means between two given terms.
The module was written by Jess Mitts M. Belingon and published as a First Edition (2019) by the Department of Education, printed by DepEd Cordillera Administrative Region.
What learners will learn
After going through this module, the learner should be able to:
- become familiar with the formulas for finding the terms of a geometric sequence
- find the nth term of a geometric sequence
- determine the geometric mean or means of a geometric sequence
Lessons in this module
- Lesson 1: Finding the nth Term of a Geometric Sequence. Learners see why multiplying by the common ratio again and again is slow for far terms, such as the 10th, 20th, or 100th term, and derive the formula for the nth term.
- Lesson 2: Geometric Means. Learners learn what extremes and geometric means are and how to find the missing terms between two given terms of a geometric sequence.
Key idea 1: The nth term formula
The module starts with the sequence 5, 10, 20, …, where the common ratio is 2. Listing the terms one by one gives the 10th term as 2,560, but that takes ten steps. Writing each term as 5 times a power of 2 shows a pattern:
| Term | Pattern |
|---|---|
| a₁ = 5 | 5 × 2⁰ |
| a₂ = 10 | 5 × 2¹ |
| a₃ = 20 | 5 × 2² |
| a₄ = 40 | 5 × 2³ |
| a₅ = 80 | 5 × 2⁴ |
The exponent of the common ratio is always one less than the term’s position. This gives the general formula:
aₙ = a₁ · r^(n − 1)
- aₙ is the nth term
- a₁ is the first term
- r is the common ratio
- n is the position (number) of the term
Checking with the formula: a₁₀ = 5 · 2⁹ = 5 · 512 = 2,560, the same answer as listing the terms.
The module’s worked examples:
- Finding a term directly. For ½, 2, 8, …, the common ratio is 2 ÷ ½ = 4. So a₆ = (½)(4⁵) = (½)(1,024) = 512.
- When the first term is missing. If the fourth term is 128 and r = 4, first solve 128 = a₁ · 4³, so a₁ = 128 ÷ 64 = 2. Then a₇ = 2 · 4⁶ = 2 · 4,096 = 8,192.
Key idea 2: Extremes and geometric means
In a sequence such as 12, ___, 3, the given first and last terms are called the extremes. The terms in between are the geometric means. For example, in 2, 4, 8, 16, the numbers 4 and 8 are the geometric means of the extremes 2 and 16.
One geometric mean. Because the ratio between consecutive terms is constant, a₃ ÷ a₂ = a₂ ÷ a₁. For 12, ___, 3:
- 3 ÷ a₂ = a₂ ÷ 12, so (a₂)² = 36
- a₂ = 6 or −6
The same method gives the mean of 3 and 45 as 3√15 or −3√15.
Two or more geometric means. First find the common ratio with r = (n − k)th root of (aₙ ÷ aₖ), where n is the number of terms and k = 1 for the first term. For 2, ___, ___, 250, there are 4 terms, so r = cube root of (250 ÷ 2) = cube root of 125 = 5. The sequence is 2, 10, 50, 250.
Activities and real-life tasks
Each lesson has guided activities followed by short assessments:
- Lesson 1: “The nth Time” and “Find Me!” (worked examples), two “Find Out!” assessments, a fill-in table of the 5th, 7th, and 9th terms, and a word problem about money deposited in a bank at 5% interest compounded annually.
- Lesson 2: “What’s That Term?” (review), “Meet Me Halfway” (worked examples), the “Halfway” and “In Between” assessments, “Shade That Term of Mine,” and a “Salary Scheme” task where learners build a quarterly salary plan for a clerk with a basic salary of P10,000.00 and a constant rate of increase every three months.
How to use this module
- Take the pretest (What I Know). Its 15 items cover the common ratio, finding specified terms, and inserting geometric means. Your score shows whether you can move on or need help from your teacher or facilitator.
- Work through each lesson in order: What’s In (review), What’s New, What Is It (the discussion), What’s More (activities and assessments), What I Have Learned, and What I Can Do.
- Show complete solutions. Several assessments ask for full solutions, not just answers.
- Take the post-test (Assessment) at the end to check your learning.
- Check your answers with the answer key at the end of the module, honestly and only after you finish.
Learners should write all answers on a separate sheet of paper and not mark the module. Parents can help by making sure their child reads each worked example before trying the activities. Teachers can use the pretest and post-test, which have matching item types, to measure progress, and the two What I Can Do tasks as performance tasks.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 10 Mathematics, Quarter 1. It is Module 6 of the quarter.
What is the formula for the nth term of a geometric sequence?
The module derives aₙ = a₁ · r^(n − 1), where a₁ is the first term, r is the common ratio, and n is the position of the term. For 5, 10, 20, …, the 10th term is 5 · 2⁹ = 2,560.
What are geometric means?
Geometric means are the terms between two given terms, called the extremes, of a geometric sequence. In 2, 4, 8, 16, the numbers 4 and 8 are the geometric means of 2 and 16.
Does the module have an answer key?
Yes. Answer keys are provided at the end for the pretest, the post-test, and the lesson activities and assessments.
How many lessons are in this module?
There are two lessons: Lesson 1 on finding the nth term of a geometric sequence and Lesson 2 on geometric means.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1 – Module 6: Finding the nth Term of a Geometric Sequence and Geometric Means
10
[illustration: teachers and learners at a tree, at desks with a globe, and a learner writing by a lamp]
CO_Q1_Mathematics 10_ Module 6
Mathematics – Grade 10 Alternative Delivery Mode Quarter 1 – Module 6. Finding the nth Term of a Geometric Sequence and Geometric Means First Edition, 2019
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed material (i.e. songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this book are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials form their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writer’s Name: Jess Mitts M. Belingon
Editor’s Name: Vic Jomar M. Laderas; Jim D. Alberto
Reviewer’s Name: Bryan A. Hidalgo, Heather G. Banagui, Laila B. Kiw-isen, Selalyn Maguilao
Management Team: May B. Eclar, Benedicta B. Gamatero, Carmel F. Meris, Marciana M. Aydinan, Ethielyn E. Taqued, Edgar H. Madlaing, Lydia I. Belingon
Printed in the Philippines by:
Department of Education – Cordillera Administrative Region
Office Address: Wangal, La Trinidad, Benguet Telefax: (074) 422-4074 E-mail Address: car@deped.gov.ph
10
Mathematics
Quarter 1 – Module 6: Finding the nth Term of a Geometric Sequence and Geometric Means
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.
Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.
Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.
In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.
Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
What I Need to Know
This module aims to provide the material necessary to introduce the mathematical concept of geometric sequences to Grade – 10 students. This module will discuss the procedures in finding the nth term of geometric sequences and identifying geometric means. It also includes interesting activities which will help learners understand well the derivation of formulas for nth term of geometric sequences and geometric means.
After going through this module, the learner should be able to:
a. familiarize the formulas in finding terms of geometric sequence, b. find the nth term of a geometric sequence, and c. determine the geometric mean/s of a geometric sequence.
What I Know
A. Multiple Choice. Read and analyze the following items and determine the letter of the correct answer from the given choices. Write your answer on a separate sheet of paper.
What expression is same as (a ∙ a ∙ a ∙ a)(b ∙ b)?
A. (4a)(2b)
B. a2 b2
C. a4 b2
D. 8abWhat is the common ratio of the sequence 36, 12, 4, …?
A. 3
B. –3
C. 1/3
D. –1/3The first term of a geometric sequence is 5, and the second term is 10, what is the sixth term?
A. 160
B. 80
C. 320
D. 640The common ratio of a geometric sequence is 1/2, and the fourth term is 1/4. What is the third term of the sequence?
A. 1
B. 1/2
C. 1/4
D. 1/8The common ratio of a geometric sequence is 3/2. Find the first three terms if the fifth term is 1.
A. 16/81, 8/27, 4/9
B. 8/27, 4/9, 2/3
C. 32/243, 16/81, 8/27
D. 2, 1, 1/2
For items 6 – 10, find the specified term of the following geometric sequences.
6.) 3, 6, 12, … find a7 7.) 10, 5, 5/2, … find a6 8.) 1, 3, 9, … find a9 9.) 2, – 1, 1/2, … find a10 10.) 100, 20, 4, … find a8
For items 11 – 15, insert geometric means in each geometric sequence.
11.) 3, _____, 8 12.) 3/2, _____, 3/8 13.) 7, _____, _____, 56 14.) 36, _____, _____, _____, 9/64 15.) – 1, _____, – 9, _____, – 81
Lesson 1: Finding the nth term of a Geometric Sequence
What’s In
In the previous modules, you learned about geometric sequences and how to find the next terms of geometric sequences.
Look at the sequences below.
5, 10, 20,… 36, 12, 4,…
If you are asked to find the next three terms of the geometric sequences, what method are you going to use?
Basically, you are going to multiply the common ratio to obtain the next terms as taught in the previous modules.
The sequence 5, 10, 20… is a geometric sequence with a common ratio of 2. Therefore, the next three terms are 40, 80 and 160.
The same through as the sequence 36, 12, 4, …with a common ratio of 1/3. Therefore, the next three terms are 4/3, 4/9, and 4/27.
Essential Question: What if I am going to ask for the 10th term, 20th term and 100th respectively? Are you going to use the same method?
Using the same method (multiplying the common ratio) to identify the 10th term, 20th term and 100th is such a waste of time. Don’t worry! This module will solve this problem.
What’s New
THE RIGHT TERM
Use any method to find the nth term of the given geometric sequences.
- What is the 7th term of the sequence 5, 10, 20, …? _______________
Since you obtained already the common ratio of this sequence which is 2 and the next three terms are 40, 80, and 160, you can easily identify the 7th term. By multiplying 160 by 2, you will obtain the 7th term which is 320.
Now, what is the 10th term? 20th term?
Here is another representation:
First term: 5 Second term: 5 x 2 = 10 Third term: 10 x 2 = 20 Fourth term: 20 x 2 = 40 Fifth term: 40 x 2 = 80 Sixth term: 80 x 2 = 160 Seventh term: 160 x 2 = 320 Eighth term: 320 x 2 = 640 Ninth term: 640 x 2 = 1 280 Tenth term: 1 280 x 2 = 2 560
Therefore, the 10th term is 2 560.
This method seems to be time – consuming and impractical, so, let us try to derive a shorter and a more accurate formula.
What Is It
Using the geometric sequence: 5, 10, 20, … you are asked to find for the 10th term.
Let us now derive a formula which may help us find an unknown term of a geometric sequence.
We obtained the 10th term of the sequence by using the representation. It can be written as:
| Term | Computation | Written as |
|---|---|---|
| 1st term (a1) | 5 | 5 |
| 2nd term (a2) | 5 x 2 = 10 | 5 x 2 |
| 3rd Term (a3) | 10 x 2 = 20 | 5 x 2 x 2 |
| 4th Term (a4) | 20 x 2 = 40 | 5 x 2 x 2 x 2 |
| 5th Term (a5) | 40 x 2 = 80 | 5 x 2 x 2 x 2 x 2 |
| 6th Term (a6) | 80 x 2 = 160 | 5 x 2 x 2 x 2 x 2 x 2 |
| 7th Term (a7) | 160 x 2 = 320 | 5 x 2 x 2 x 2 x 2 x 2 x 2 |
| 8th Term (a8) | 320 x 2 = 640 | 5 x 2 x 2 x 2 x 2 x 2 x 2 x 2 |
| 9th Term (a9) | 640 x 2 = 1 280 | 5 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 |
| 10th Term (a10) | 1 280 x 2 = 2 560 | 5 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 |
Looking at the illustration, we can derive a pattern that is being followed by the sequence. Let us consider the first five terms.
| Term | Value | Pattern |
|---|---|---|
| (a1) | 5 | 5 x 2⁰ |
| (a2) | 5 x 2 = 10 | 5 x 2¹ |
| (a3) | 10 x 2 = 20 | 5 x 2² |
| (a4) | 20 x 2 = 40 | 5 x 2³ |
| (a5) | 40 x 2 = 80 | 5 x 2⁴ |
[chart: annotated diagram of the first five terms of the sequence; labels: “First Term of the sequence denoted by a1” (points to 5), “Common Ratio denoted by r” (points to the exponents’ pattern in the 2ⁿ terms)]
Let’s try to change the numbers to variables.
$a_1 = a_1 \cdot 1 = a_1 \cdot r^0$
$a_2 = a_1 \cdot r = a_1 \cdot r^1$
$a_3 = a_1 \cdot r \cdot r = a_1 \cdot r^2$
$a_4 = a_1 \cdot r \cdot r \cdot r = a_1 \cdot r^3$
$a_5 = a_1 \cdot r \cdot r \cdot r \cdot r = a_1 \cdot r^4$
Recall: In the laws of exponents can be written as a³, and any non-zero number raised to zero (0) is always equal to 1. Hence, a⁰ = 1.
What can you observe from the exponent of r and the subscript of a?
The exponent of r is less than one of the subscript of a. The subscript of a refers to the placement of the term which is denoted by n. Therefore, using these variables, we derived the formula
GENERAL IDEA. The formula in finding the nth term of a geometric sequence is
$a_n = a_1 r^{n-1}$
Wherein:
- $a_n$ – nth term
- $a_1$ – first term
- r – common ratio
- n – number of terms
Using the sequence: 5, 10, 20, …, we identified the 7th term and the 10th term, which are 320 and 2,560 respectively. Now, let us use the derived formula to check if we are on the right track.
For the 7th term ($a_7$) Given: $a_1 = 5, r = 2, n = 7$ $a_n = a_1 r^{n-1}$ $a_7 = (5)(2^{7-1})$ $a_7 = (5)(2^6)$ $a_7 = (5)(64)$ $a_7 = 320$ ✓
For the 10th term ($a_{10}$) Given: $a_1 = 5, r = 2, n = 10$ $a_n = a_1 r^{n-1}$ $a_{10} = (5)(2^{10-1})$ $a_{10} = (5)(2^9)$ $a_{10} = (5)(512)$ $a_{10} = 2{,}560$ ✓
What’s More
ACTIVITY 1. THE nth TIME
Find the specified term of the given geometric sequence.
- $\frac{1}{2}$, 2, 8,…; find $a_6$
To solve for the unknown term, identify first the given values. The first term is ½, and the unknown term is the sixth term, so, the value of n is equal to 6. To find common ratio:
$r = \frac{a_n}{a_{n-1}} \longrightarrow \frac{a_2}{a_1} = \frac{2}{\frac{1}{2}} = 4$
The given values will then be substituted to the formula in finding the nth term of a geometric sequence.
Given: $a_1 = \frac{1}{2}$; $r = 4$; $n = 6$ Unknown = $a_6$
$a_n = a_1 r^{n-1}$ $a_6 = \left(\frac{1}{2}\right)(4)^{6-1}$ $a_6 = \left(\frac{1}{2}\right)(4)^5$ $a_6 = \left(\frac{1}{2}\right)(1,024)$ $a_6 = 512$
ASSESSMENT 1. FIND OUT!
Find the specified term of each of the given geometric sequences. Show your complete solutions.
1.) 3, 6, 12, … $a_7$ 2.) 18, 6, 2, … $a_9$ 3.) $1, \frac{1}{2}, \frac{1}{4}$, … $a_7$
ACTIVITY 2. FIND ME!
- What is the seventh term of a geometric sequence whose fourth term is 128 and common ratio equal to 4?
To begin with the problem, you must have to analyze carefully what does it ask for. The problem is asking for the 7th term but the first term was not given.
First, identify the given values and the unknown variable/s.
Given: Fourth Term ($a_4$) = 128 Common Ratio ($r$) = 4
There are two unknowns, we have $a_1$ and $a_7$.
Let’s have the formula $a_n = a_1 r^{n-1}$
There are two unknowns in the problem and to solve $a_7$, we need to solve $a_1$ first.
Since the given term is the fourth term, which is 128, we can use it to solve for the value of $a_1$. Substitute the value of $a_4$, which is 128, n which is 4, and r, which is 4 in the formula $a_n = a_1 r^{n-1}$.
$a_n = a_1 r^{n-1}$ $128 = a_1(4)^{4-1}$ $128 = a_1(4)^3$ $\frac{128}{64} = \frac{64a_1}{64}$ $a_1 = 2$
Afterwards, identify the unknown term which is $a_7$.
$a_n = a_1 r^{n-1}$ $a_7 = (2)(4)^{7-1}$ $a_7 = (2)(4)^6$ $a_7 = 2(4,096)$ $a_7 = 8{,}192$
ASSESSMENT 2: FIND OUT!
Answer the problem completely.
- The third and eighth term of a geometric sequence are 8 and $\frac{1}{4}$ respectively. What is the first term and the common ratio of the sequence?
What I Have Learned
FILL IN THE TABLE
Directions. Find the specified term of the geometric sequences given the first term and the common ratio.
| Given | $a_5$ | $a_7$ | $a_9$ |
|---|---|---|---|
| 1. $a_1 = 3$ r = 3 | |||
| 2. $a_1 = 2$ r = 2 | |||
| 3. $a_1 = -\frac{1}{2}$ r = −6 |
What I Can Do
Solve for what is asked in the problem. Show your solutions.
Warren deposited his P12, 000. 00 in a local bank in 2019.
- If the bank offers 5% interest rate compounded annually, how much will be in his bank account in 2028?
- Suppose he will withdraw the principal in 2028 and will leave the interest in his account, how much will the interest earn after 5 years?
Note: A – final amount P – initial principal balance r – interest rate n – number of times interest applied per time period t – number of time periods elapsed
Lesson 2 Geometric Means
What’s In
In the previous lesson in this module, you were able to derive the formula in finding the nth term of geometric sequences.
To review what you have already learned in finding the nth term of a geometric sequence, accomplish the following activity.
WHAT’S THAT TERM?
Find the specified term in each of the following geometric sequence.
_____ 1. 3, 9, 27, … 6th term _____ 2. 1, 2, 4, … 7th term _____ 3. 12, $a_2$, 3 _____ 4. 2, $a_2$, $a_3$, 54
Were you able to answer items 3 and 4? How did you come up with your answers?
Using the formula in finding the nth term of a geometric sequence to supply the unknown term in Items 3 and 4, will really let you experience difficulties.
Through this module, let us discover a shorter way to identify the unknown term/s in between terms of geometric sequences.
What’s New
Look at the problem below:
What must be placed in the blank to form a geometric sequence?
12, _____, 3
You might use the formula in finding the nth term of geometric sequence to supply the middle term.
$a_n = a_1 r^{n-1}$ $a_2 = 12r^{2-1}$ $a_2 = 12r$
We’re stuck in here!
Let’s discover how to answer his problem.
What is It
How can we solve this problem?
[chart: diagram; “12, ____, 3” with arrows: “extremes” points to 12 and 3; “Geometric mean” points to the blank]
The given terms are the first and last terms. These terms are called the extremes, and the term/s in between the extremes are called geometric mean/s. In the geometric sequence, 2, 4, 8, 16, the numbers 4 and 8 are the geometric means of the extremes, 2 and 16.
Let us derive the formula:
The common ratio refers to the ratio of two consecutive terms,
$r = \frac{a_n}{a_{n-1}}$ same through as $\frac{a_3}{a_2}$ and $\frac{a_2}{a_1}$, therefore we can equate $\frac{a_3}{a_2} = \frac{a_2}{a_1}$,
This can be our basis in finding geometric means.
$\frac{a_3}{a_2} = \frac{a_2}{a_1}$,
Let’s go back to the problem, 12, ____, 3. The first term is 12 and the last term is 3. Now, let us substitute.
$\frac{a_3}{a_2} = \frac{a_2}{a_1}$,
$\frac{3}{a_2} = \frac{a_2}{12}$,
$(a_2)^2 = 36$
$\sqrt{a_2} = \pm\sqrt{36}$
$a_2 = 6$ or $-6$ ✓
To insert terms, let us identify first the common ratio by using the formula:
$r = \sqrt[n-k]{\frac{a_n}{a_k}}$
Where: n = number of terms k = 1, since it is the first term placement r = common ratio
What’s More
ACTIVITY 1. MEET ME HALFWAY
Find the missing term/s in the following geometric sequences.
- 3, _____, 45
To insert a geometric mean, let us first identify the extremes and number of terms. By substitution,
$\frac{a_3}{a_2} = \frac{a_2}{a_1}$, $\frac{45}{a_2} = \frac{a_2}{3}$ $(a_2)^2 = 135$ $\sqrt{a_2} = \pm\sqrt{135}$ or $\pm 3\sqrt{15}$
Therefore, $a_2 = 3\sqrt{15}$ or $-3\sqrt{15}$
- 2, _____, _____, 250
There are two geometric means needed in this problem. Let us identify first the extremes and the number of terms. The extremes are 2 and 250, and there are 4 terms in the sequence.
$r = \sqrt[n-k]{\frac{a_n}{a_k}}$ $r = \sqrt[4-1]{\frac{a_4}{a_1}}$ $r = \sqrt[3]{\frac{250}{2}}$ $r = \sqrt[3]{125}$ $r = 5$
Therefore, the sequence is 2, 10, 50, 250…
ASSESSMENT 1: HALFWAY
Find the geometric mean of the given extremes.
| Given | Geometric Mean |
|---|---|
| 1. 3 and 8 | |
| 2. 100 and 25 | |
| 3. $\frac{1}{2}$ and $\frac{1}{8}$ | |
| 4. 3 and $\frac{1}{3}$ | |
| 5. x and x⁷ |
ASSESSMENT 2. IN BETWEEN
Insert geometric means in each geometric sequence.
1.) 2, _____, _____, 686 2.) _____, 24, _____, _____, $\frac{3}{64}$ 3.) _____, _____, _____, 4, 8 4.) _____, $\frac{1}{4}$, $\frac{1}{2}$, _____ 5.) 81, _____, _____, _____, _____, $\frac{1}{3}$
What I Have Learned
SHADE THAT TERM OF MINE
The table contains terms of a geometric sequence. Identify whether the terms in the right columns are geometric means or term of the sequence. Use any shade to color the geometric means or terms
| Geometric Sequence | Terms | |||
|---|---|---|---|---|
| 1.) 1, ___, ___, ___, 81 | 30 | 9 | 27 | 3 |
| 2.) ___, 3, ___, ___, 1/9 | 1 | 6 | 9 | 1/3 |
| 3.) 24, ___, ___, ___, 3/2 | 1 | 2 | 3 | 4 |
| 4.) 2, ___, 1/2, ___, 1/8 | 1 | 4 | 1/4 | 2 |
| 5.) ___, 1/2, 1 | 2 | 4 | 1/4 | 1/3 |
What I Can Do
SALARY SCHEME
Angelie is working for a year as a clerk in a certain department. Her employer promised her that she will be having a salary increase every after 3 months. Assume that her basic salary is P10,000.00 and there is a constant rate of increase in her salary every after 3 months.
Your task is to construct/write a salary scheme for Angelie that determines her salary in the first up to the last quarter of the year.
| Cycle of Salary Increase | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Salary Received |
Assessment
A. Multiple Choice. Read and analyze the following items and determine the letter of the correct answer from the given choices. Write your answer on a separate sheet of paper.
_____ 1. What expression is same as (𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎)(𝑏 ∙ 𝑏 ∙ 𝑏)?
- A. 15ab
- B. a⁵ b³
- C. a³ b
- D. 5a³b
_____ 2. What is the common ratio of the sequence 1/7, 1, 7, …?
- A. 7
- B. – 7
- C. 1/7
- D. – 1/7
_____ 3. The first term of a geometric sequence is 2, and the second term is 10, what is the fifth term?
- A. 250
- B. 100
- C. 500
- D. 1 250
_____ 4. The common ratio of a geometric sequence is 1/4, and the fourth term is 1/2. What is the third term of the sequence?
- A. 1
- B. 1/2
- C. 1/4
- D. 2
_____ 5. Find the 12th term of the geometric sequence 1, 2, 4,…
- A. 512
- B. 1 024
- C. 2 048
- D. 1 112
For items 6 – 10, find the specified term of the following geometric sequences.
_____ 6.) 6, 12, 24, … find a7
_____ 7.) 20, 10, 5, … find a6
_____ 8.) 7, – 14, 28, … find a8
_____ 9.) a1 = 2, r = 3 find a10
_____ 10.) a1 = 0, r = 1/2 find a8
For items 11 – 15, insert geometric means in each geometric sequence.
11.) 3, ____, 27
12.) 1/2, _____, 1/8
13.) 5/2, _____, _____, 625/16
14.) 3,,,_____, 3
15.) – 1,, − 25 ,, – 625
Answer Key
[transcriber’s note: this page is printed upside down; the cards are transcribed upright]
What I Know
- C
- C
- A
- B
- A
- 192
- 5/16
- 6 561
- −1/256 [?]
- 3 125/4 [?]
- 2√6 or −2√6
- 3/4 or −3/4 [?]
- 14, 28
- 9, 4/9, 4/16 [?]
- 3, 27 or −3, −27
ASSESSMENT
- B
- A
- D
- D
- C
- 384
- 5/8
- −896
- 39,366
- 0
- 9 or −9
- 1/4 or −1/4
- 25/4, 125/8
- −3, 3, −3 or 3, 3, 3
- 5, 125
What I Have Learned (Lesson 1)
- 243, 2 187, 19,683
- 32, 128, 512
- −648, −23 328, −839 808
What’s More Lesson 1 Assessment 1
- 192
- 2/729
- 1/64
What’s In Lesson 2
- 729
- 64
- 6 or −6
- 6, 18
What’s More Lesson 2 Assessment 1
- 2√6 or 5 − 2√6
- 50 or −50
- 1/4 or −1/4
- 1 or −1
- x⁴
What’s More Lesson 1 Assessment 2
- r = 1/2
- a1 = 32
What’s More Lesson 2 Assessment 2
- 2, 14, 98, 686
- 192, 24, 3, 3/8, 3/64
- 1/2, 1, 2, 4, 8
- 1/8, 1/4, 1/2, 1
What I Have Learned (Lesson 2)
| Geometric Sequence | Terms | |||
|---|---|---|---|---|
| 1.) 1, ___, ___, ___, 81 | 30 | 9 | 27 | 3 |
| 2.) ___, 3, ___, ___, 1/9 | 1 | 6 | 9 | 1/3 |
| 3.) 24, ___, ___, ___, 3/2 | 1 | 2 | 3 | 4 |
| 4.) 2, ___, 1/2, ___, 1/8 | 1 | 4 | 1/4 | 2 |
| 5.) ___, 1/2, 1 | 2 | 4 | 1/4 | 1/3 |
References
Applications of Geometric Sequences and Series to Financial Maths. www.yumpu.com/en/document/read/35995428
Cabral, J., Bernabe, J., & Valencia, E. (2004). New Trend in Math Series, Algebra II. Quezon City: Vibal Publishing House Incorporated.
For inquiries or feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
Ground Floor, Bonifacio Bldg., DepEd Complex Meralco Avenue, Pasig City, Philippines 1600
Telefax: (632) 8634-1072; 8634-1054; 8631-4985
Email Address: blr.lrqad@deped.gov.ph * blr.lrpd@deped.gov.ph
Leave a Reply