User:TeacherAbigail/sandbox

= Number Sequence: Geometric Sequence =

Learning Competencies:

 * Determines geometric means, nth term of a geometric sequence and sum of the terms of a given finite or infinite geometric sequence.
 * Solves problems involving sequences.

Learning Objectives:
At the end of the lesson, you are expected to:


 * determine the nth term of a geometric sequence.
 * find the geometric means of a geometric sequence.
 * calculate the sum of a finite and infinite geometric sequence.
 * Solve problems involving sequences.

General Instruction:
Working with the whole class, answer the 'What have you learned' part after each topic discussion. Collaboratively discuss and completely show your solutions to one another as a way to assess your understanding and progress on the topics being discussed. You are free to write your questions for any clarifications, and you must share your individual learnings or takeaways in the last part.

What to learn...
The nth term of a geometric sequence is $$a_n=a_1r^n-1$$.

Example 1:  Find the tenth term of the geometric sequence 1, -4, 16, -64, …

Solution:                       a1 = 1;               r = -4;               n = 10

$$a_n=a_1r^{n-1}$$

$$a_{10}=1(-4)^{10-1}$$                     Substitute with the given information

$$a_{10}=1(-4)^{9}$$                          Simplify the exponent

$$a_{10}=1(-262,144)$$                  Simplify

$$a_{10}=(-262,144)$$                    Simplify

Therefore the 10th term is -262, 144.

Example 2: A certain substance decomposes and loses 20% of its weight each hour. If the original quantity of the substance is 500 grams, how much remains after 8 hours?

Solution:  In this geometric sequence, a1 = 500 and r = 0.80 (100% - 20% = 80% or 0.80). We are asked to find the amount remaining after 8 hours. Thus, we must find the amount remaining at the beginning of the 9th hour.

a1 = 500,                      r = 0.80,                      n = 9

$$a_n=a_1r^{n-1}$$.

$$a_9=(500)(0.80)^{9-1}$$. Substitute with the given information

$$a_9=(500)(0.80)^8$$.

$$a_9=83.88 grams$$. Simplify using PEMDAS rule

What have you learned...
Answer the following:

1) Find the 7th term of the geometric sequence with a1 = 3 and r = 5

2) A machine costing P1,000,000 depreciates in value 20 percent each year.  How much will it be worth at the end of 4 years?

What to learn...
A)  Geometric means are the terms between any two nonconsecutive terms in a geometric sequence.  If the numbers a1, a2, a3…., an form a geometric sequence, the a2, a3…., an-1 are called geometric means between a1 and an. Thus, the formula to be used is

$$a_n=a_1r^{n-1}$$

B) Let m be the geometric mean between two numbers a and b, so that a, m and b form a geometric sequence, so that

$$m/a = m/b $$ or $$m^2 = ab$$ or  $$m = \pm\sqrt{ab} $$

Example 1:  Insert three geometric means between 5 and 3125.

Solution:  Let a1 = 5 and a5 = 3125. We will insert a2, a3, and a4.

$$a_n=a_1r^{n-1}$$

$$a_5=5r^{5-1}$$

$$3125= 5r^{5-1}$$    Substitute

$$3125= 5r^{4}$$        Simplify the exponent

$$625= 5r^{4}$$          Divide by 5 on both sides

$$5^4= 5r^{4}$$             Rewrite the base 625 into a power of 4 (625= 5555 = 54)

$$r = \pm5$$               Therefore the ratio is 5 since both satisfy the equation.

If r = 5, the geometric means are:   a2 = 5(5) = 25,  a3 = 5(25) = 125,  a4 = 5(125) = 625.

Thus, the sequence is 5, 25, 125, 625, 3125.

If r = -5 then the geometric means are: a2= -5(5) = -25, a3 = -5(-25) =125, a4 = -5(125) = -625.

Thus, the sequence is 5, -25, 125, -625, 3125.

Example 2:       Find the geometric mean between 4 and 36.

Solution:             a = 4,               b = 36

$$m=\sqrt{(4)(36)}$$     Substitute

$$m=\sqrt{(144)}$$          Simplify

$$m=12$$                    Extract the root

Therefore, the geometric mean between 4 and 36 is $$\pm12$$.

Note: The geometric mean between two numbers a and b is

$$\sqrt{ab}$$     if a and b positive

$$-\sqrt{ab}$$  if a and b are negative.

What have you learned...
Answer the following.

1)  Insert three geometric means between 16 and 1296.

2)  Find the geometric mean between 10 and  1/10.

What to learn...
Finite sequence if it has a first term and a last term. The formula for the sum of the first n-terms in a geometric sequence is

$$S_n = {a_1-a_1r^n \over 1- r }$$

Where Sn= sum; a1= the first term; r = the common ratio

Example 1:       Find the sum of the first 5 terms of 3, 6, 12, 24, …

Solution:           n = 5;              a1 = 3;              r = 2

$$S_n = {a_1-a_1r^n \over 1- r }$$

$$S_n = {3-3 (2)^5 \over 1-2 }$$        Substitute with the given information

$$S_n = {3-3 (32) \over-1 }$$        Simplify using PEMDAS rule

$$S_n = 93$$                       Therefore the sum of the first 5 terms of 3, 6, 12, 24, … is 93.

Example 2:  Find the sum of the sequence -5, -10, -20, -40, … until a6 ?

Solution:          n = 6                a1 = -5;             r = -2

$$S_n = {a_1-a_1r^n \over 1- r }$$

$$S_n = {-5-{(-5)(-2)^6 }\over 1-2 }$$     Substitute

$$S_n = {-5-{(-5)(64) }\over -1 }$$        Simplify using PEMDAS rule

$$S_n = -315 $$                           Therefore the sum is -315

The sum S of the terms of an infinite geometric sequence is

$$S_{ \infty} \displaystyle = {a_1 \over 1-r}$$

where: a1= the first term; r= the common ratio and  $$/r/ < 1$$

Example 1:       Find the sum of the first 5 terms of 3, 6, 12, 24, …

Solution:           n = 5;              a1 = 3;              r = 2

$$S_{ \infty} \displaystyle = {a_1 \over 1-r}$$

$$S_{ \infty} \displaystyle = {64 \over 1-{1 \over 2}}$$         Substitute with the given information

$$S_{ \infty} \displaystyle = {64 \over{1 \over 2}}$$               Simplify

$$S_{ \infty} \displaystyle = 128$$              Therefore the sum is 128.

Example 2:  Find the sum of  $${5\over 10}+{5\over 100}+{5\over 1000},...$$

Solution :             $$a_1 = {5 \over 10}$$          $$r = {1 \over 10}$$

$$S_{ \infty} \displaystyle = {a_1 \over 1-r}$$

$$S_{ \infty} \displaystyle ={10 \over 1 -{1 \over 10}}$$          Substitute with the given information

$$S_{ \infty} \displaystyle ={5/10 \over 9/10}$$               Simplify

$$S_{ \infty} \displaystyle = {5 \over 9} $$                     Therefore the sum is 5/9.

What have you learned...
A. Solve the following.


 * 1) 1) The sum of the first 5 terms of the geometric sequence 4, 12, 36, 108, …
 * 2) 2) The sum of the infinite geometric sequence 64, 16, 4, 1, …

B. Solve the following problems.

The World Health Organization (WHO) reported that about 16 million adolescent girls between 15 and 19 years of age give birth each year. Knowing the adverse effects of adolescent childbearing on the health of the mothers as well as their infants, a group of students from Magiting High School volunteered to help the government in its campaign for the prevention of early pregnancy by giving lectures to 7 barangays about the WHO Guidelines on teenage pregnancy. The group started in Barangay 1 and 4 girls attended the lecture. Girls from other barangays heard about it, so 8 girls attended from Barangay 2, 16 from Barangay 3, and so on.

a.  Complete the table below representing the number of adolescent girls who attended the lecture from Barangay 1 to Barangay 7 assuming that the number of attendees doubles at each barangay. b.  Analyze the data in the table and create a formula.

c.  Determine the total number of girls who will benefit from the lecture.

My Answer
Kindly put your answers here. Use the format "Surname: [Enter] LessonNumber.ItemLetter:Answer". Refer to the example below.

DelaCruz

-Lesson 1a: 56

Cuamag

-Lesson 1: Number 1

An = A1rn-1

A7 = 3(5)7-1

A7 = 3(5)6

A7 = 46, 875

Therefore, the 7th term is 46,875.

Batac

-Lesson 1: Number 2

An = A1rn-1

A4 = 1000000(0.80)4-1

A4 = 1000000(0.80)3

A4 = 409,600

Therefore, it is worth 409,600 at the end of 4 years.

Penales

-Lesson 2: Number 1

Solution:  Let a1= 16 and a5 = 1296. We will insert a2, a3, and a4.

$$a_5=a_5r^ {n-1}$$

$$1296=16r^ {5-1}$$         Substitute

$$1296=16r^4$$            Simplify the exponent

$$1296=16r^4$$            Divide by 16 on both sides

$$81=r^4$$                     Rewrite the base 81 into a power of 4 ( 81=3x3x3x3=34)

$$r = \pm3$$                      Therefore the ratio is 3 since both satisfy the equation.

If r = 3, the geometric means are:   a2 = 3(16) = 48, a3 = 3(48)= 144, a4 = 3(144) = 432

Thus, the sequence is 16, 48, 144, 432, 1296.

If r = -3, then the geometric means are: a2=-3(16)=-48, a3 =-3(-48)=144, a4 =-3(144)= -432

Thus, the sequence is 16, -48, 144, -432, 1296.

Baguio

-Lesson 2: Number 2

Solution:             a = 10,                   b = 1/10

$$m=\sqrt{(10)*(1/10)}$$       Substitute

$$m=\sqrt{(1)}$$                          Simplify

$$m=1$$                                  Extract the root

Therefore, the geometric mean between 10 and 1/10 is 1.

Adanza

-Lesson 3A: Number 1

Solution:  n = 5 ; a1 = 4; r = 3

Sn = [a1-a1(r)n]/(1-r)

S5  =  [4-4(3)⁵]/(1-3)

S5 =  [4-4(243)]/(-2)

S5  =  484

Therefore, the sum of the first 5 terms of the geometric sequence 4, 12, 36, 108, … is 484.

-Lesson 3A: Number 2

Solution:  a1 = 64; r = 1/4

S∞ =   a1/(1-r)

S∞=  64/[1-(1/4)]

S∞ =  64/(3/4)

S∞ =  256/3 or 85 1/3

Therefore, the sum of the infinite geometric sequence 64, 16, 4, 1, … is 256/3 or 85 1/3.

-Lesson 3B: Letter a-c

a) b) Formula: an=4(2)n-1

c) The total number of girls who benefited the lecture is 16, 380.

My Takeaways
Kindly put your takeaways from the lesson here. Use the format "Surname:MyTakeaway".

'''Cuamag: Understanding geometric sequences involves recognizing the pattern of multiplication between successive terms, determining the common ratio, and applying formulas to find specific terms or the sum of the series. Real-world applications may include population growth, financial investments, and various exponential phenomena.'''

'''Batac: Geometric sequences find practical applications in various real-life situations, serving as mathematical models for phenomena characterized by consistent growth. For instance, the compound interest in financial investments follows a geometric sequence, where each period's interest is calculated based on the previous amount.'''

'''Baguio: Geometric sequences are sequences of numbers where each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. The common ratio determines whether the sequence grows or decays, and geometric sequences have many real-life applications just like in calculating compound interest, where the interest is based on a fixed percentage growth each year. Moreover, they are closely related to exponential functions and provide a useful tool for understanding growth and decay patterns.'''

'''Adanza: The geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant called the common ratio. It is a useful concept in mathematics and has applications in various fields such as finance and physics. Understanding the properties and formulas of geometric sequences can help in solving real-life problems involving exponential growth or decay.'''

'''Penales: Geometric sequences are mathematical patterns that offer valuable insights into the world of exponential growth and decay. By studying geometric sequences, we can uncover important properties, understand their applications in different fields, and enhance our problem-solving skills. The key takeaways and understanding derived from studying geometric sequences empower us to solve real-world problems, make informed decisions, and develop a solid mathematical foundation. Let us embrace the beauty and relevance of geometric sequences as we continue our mathematical journey.'''