Grade 10 Math Q1 Ep4: Computing Arithmetic Means
Summary
TLDRIn today's episode of 'Adapted TV,' host Sir Jason Flores, also known as Math Buddy, guides viewers through the concept of arithmetic means in sequences. The lesson focuses on defining arithmetic means, identifying them within a sequence, and calculating them using the formula for the common difference. Examples include finding means between given extremes and inserting specified numbers of means between extremes. The engaging tutorial aims to enhance logical reasoning and critical thinking skills, making math fun and accessible.
Takeaways
- 📘 Arithmetic means are the numbers that lie between the first and last terms of a finite arithmetic sequence.
- 🔢 The arithmetic extremes are the first and last terms of an arithmetic sequence, with the terms in between being the arithmetic means.
- 💡 To find the arithmetic mean between two numbers, sum them and divide by two, which is also known as the average.
- 📐 The formula to calculate the common difference D in an arithmetic sequence is D = (a_n - a_k) / (n - k), where a_n is the last term and a_k is the first term.
- 📈 The common difference can be used to find the missing terms in an arithmetic sequence by adding it to the preceding term.
- 📝 When inserting multiple arithmetic means between two extremes, the sequence's terms are calculated using the common difference.
- 📊 The arithmetic mean of an arithmetic sequence can be represented as the average of the first and last terms.
- 📌 In the context of the script, examples are provided to demonstrate how to find arithmetic means and insert terms into a sequence.
- 🎓 The lesson encourages students to practice and apply these concepts to solve problems involving arithmetic sequences.
- 🎉 The video concludes with a reminder that learning math can be fun and easy, emphasizing a positive learning attitude.
Q & A
What are the arithmetic extremes in a finite arithmetic sequence?
-The first and last terms of a finite arithmetic sequence are called arithmetic extremes.
What is the term used for the numbers that lie between the arithmetic extremes in a sequence?
-The numbers that lie between the arithmetic extremes in a sequence are referred to as arithmetic means.
How is the arithmetic mean between two numbers calculated?
-The arithmetic mean between two numbers is calculated by adding the two numbers together and dividing the sum by two.
What formula is used to find the common difference when inserting arithmetic means between two extremes?
-The formula used to find the common difference \( D \) is \( D = \frac{a_n - a_K}{N - K} \), where \( a_n \) is the last term, \( a_K \) is the first term, and \( N \) and \( K \) are the positions of the last and first terms, respectively.
How do you find the arithmetic mean of the sequence 4, 8, 12, 16, 20, and 24?
-The arithmetic means of the sequence 4, 8, 12, 16, 20, and 24 are the terms 8, 12, 16, and 20, as they lie between the first and last terms, 4 and 24.
What is the arithmetic mean between 10 and 24?
-The arithmetic mean between 10 and 24 is 17, which is calculated by adding 10 and 24 to get 34, then dividing by 2.
How many arithmetic means are there between 8 and 16 if three means are inserted?
-When three arithmetic means are inserted between 8 and 16, there are a total of five terms including the extremes, with the three arithmetic means being 10, 12, and 14.
What is the formula to find the common difference in an arithmetic sequence?
-The formula to find the common difference \( D \) in an arithmetic sequence is \( D = \frac{a_n - a_1}{n - 1} \), where \( a_n \) is the last term, \( a_1 \) is the first term, and \( n \) is the number of terms.
How do you determine the missing terms in an arithmetic sequence when the second and fifth terms are known?
-To determine the missing terms in an arithmetic sequence when the second and fifth terms are known, first find the common difference using the formula for \( D \), then add this difference to the preceding term to find the next term.
What is the arithmetic mean between the arithmetic extremes 5 and 19?
-The arithmetic mean between the arithmetic extremes 5 and 19 is 12, calculated by adding 5 and 19 to get 24, then dividing by 2.
How many arithmetic means should be inserted between -5 and 1 if two means are to be added?
-If two arithmetic means are to be inserted between -5 and 1, the common difference \( D \) is calculated to be 2, and the terms would be -3 and -1.
Outlines
📘 Introduction to Arithmetic Means
Sir Jason Flores, the host of 'Adapted TV,' introduces the topic of arithmetic means in the context of arithmetic sequences. He encourages viewers to prepare their learning materials and engage in developing logical reasoning and critical thinking skills. The lesson aims to define arithmetic means and determine them within a sequence. Sir Jason explains that in a finite arithmetic sequence, the first and last terms are called 'arithmetic extremes,' while the terms in between are 'arithmetic means.' He uses the sequence 4, 8, 12, 16, 20, 24 to illustrate the concept, identifying 8, 12, 16, and 20 as the arithmetic means. The lesson also covers the formula for finding the common difference (D) in a sequence, which is essential for determining the arithmetic means when more than two terms are involved.
🔢 Calculating Arithmetic Means
This section delves into the process of calculating arithmetic means between two given numbers. Sir Jason demonstrates how to find the arithmetic mean between 10 and 24, using the average formula, which results in 17. He then guides viewers through an example of inserting three arithmetic means between 8 and 16. The formula for the common difference (D = (a_n - a_k) / (n - k)) is applied to determine the unknown terms. The common difference is calculated as 2, and using this, the missing terms are found to be 10, 12, and 14, respectively. The process involves adding the common difference to the preceding term to find the next term in the sequence.
🧩 Solving for Missing Terms in a Sequence
The focus of this segment is on solving for missing terms in an arithmetic sequence. Given an incomplete sequence with the second term as 6 and the fifth term as 30, the task is to find the first, third, and fourth terms. Sir Jason uses the formula for the common difference to determine that the difference (D) is 8. With this, he calculates the missing terms: the first term is -2, the third term is 14, and the fourth term is 22. This part of the lesson reinforces the concept of using the common difference to solve for unknown terms in an arithmetic sequence.
📐 More Examples of Arithmetic Means
This section presents additional examples to further illustrate the concept of arithmetic means. Sir Jason first calculates the arithmetic mean between the numbers 5 and 19, which is 12. He then moves on to a more complex example involving algebraic expressions: finding the arithmetic mean between 3x squared plus 8 and x squared minus 6. By aligning similar terms and dividing the sum by 2, he simplifies the expression to 2x squared plus 1. The lesson continues with an example of inserting three arithmetic means between 2 and 22, where the common difference is found to be 5, leading to the determination of the missing terms: 7, 12, and 17.
🎓 Conclusion and Further Practice
In the final part of the lesson, Sir Jason summarizes the key points and provides guidance on how to find the arithmetic mean between two arithmetic extremes using the averaging method. He also explains how to compute arithmetic means when more than two are inserted between extremes, using the common difference formula. The lesson concludes with a call to action for students to practice the concepts learned by answering additional questions. Sir Jason emphasizes that learning math can be fun and easy, encouraging students to continue their mathematical journey.
Mindmap
Keywords
💡Arithmetic Means
💡Arithmetic Extremes
💡Common Difference
💡Average
💡Sequence
💡Critical Thinking
💡Logical Reasoning
💡nth Term
💡Algebraic Expressions
💡Problem-Solving
Highlights
Introduction to the concept of arithmetic means in sequences
Definition of arithmetic extremes and means in a finite arithmetic sequence
Explanation of how to find arithmetic means between two numbers
Example of finding arithmetic means in the sequence 4, 8, 12, 16, 20, 24
Formula for the common difference D in an arithmetic sequence
Calculation of the arithmetic mean between 10 and 24
Insertion of 3 arithmetic means between 8 and 16 with step-by-step calculation
Method to find missing terms in an arithmetic sequence when some are known
Solving for the first, third, and fourth terms in a sequence with given second and fifth terms
Explanation of how to find the arithmetic mean of algebraic expressions
Example of finding the arithmetic mean between 5 and 19
Solving for the arithmetic mean of algebraic expressions 3x^2 + 8 and x^2 - 6
Guidance on inserting a specified number of arithmetic means between given extremes
Inserting three arithmetic means between 2 and 22 with detailed calculation
Summary of methods to find arithmetic means and insert them between extremes
Interactive problem-solving session with the audience for practice
Conclusion and encouragement for students to apply the learned concepts
Transcripts
[Music]
thank you
[Music]
foreign
[Music]
hi good day welcome in today's episode
of adapted TV I am sir Jason Flores also
your math buddy and I will be here to
help you in developing your logical
reasoning and critical thinking skills
is yourself learning module ready
what about your pen and paper
great let's begin a fun and exciting
lesson for this lesson you are expected
to First Define arithmetic means and
second determine arithmetic means of a
sequence
previously you learned how to determine
the nth term of an arithmetic sequence
what's new
well we will focus on finding the
arithmetic means
for example in the sequence for
eight
twelve sixteen
twenty and 24
find its arithmetic means
well
8 12 16 and 20 are the arithmetic means
of the sequence because these terms are
between 4 and 24 which are the first and
last terms respectively
but what are arithmetic means
the first and last terms of a finite
arithmetic sequence are called
arithmetic extremes
and the numbers in between are called
arithmetic means
again the first and last terms of a
finite arithmetic sequence are called
arithmetic extremes
and the terms in between are called
arithmetic means
in the sequence four eight
12 16 20 and 24 the terms 4 and 24 are
the arithmetic extremes while eight
12 16 and 20 are the arithmetic beans
also
eight is the arithmetic mean of the
arithmetic extremes 4 and 12.
the arithmetic mean between two numbers
is sometimes called the average of two
numbers
again the arithmetic mean between two
numbers is sometimes called the average
of two numbers
however if more than one arithmetic
means will be inserted between two
arithmetic extremes the formula for the
common difference D which is D is equal
to a sub n minus a sub K All Over N
minus K can be used
where a sub n is equal to the last term
and a sub K is the first term of the
sequence
let's try what is the arithmetic mean
between 10 and 24.
using the average formula get the
arithmetic mean of 10 and 24. that is
adding 10 and 24
that's 34 divided by 2
17 is the arithmetic mean
let's go to the next example
insert 3 arithmetic means between 8 and
16.
if 3 arithmetic means will be inserted
between 8 and 16 then our a sub 1 or the
first term will be equal to 8 and a sub
5 or the last term will be equal to 16.
using the formula for D compute for the
common difference that's D is equal to a
sub n minus a sub K All Over N minus k
we will substitute this in our given
problem our a sub n will be the last
term
so that will be equal to a sub 5
minus our a sub K we will use the first
term that is a sub 1
all over and we will use five
and for K we will use one
use using the values for a sub 5 and a
sub 1 we will have 16
minus
8
all over
5 minus 1.
16 minus 8 will give us 8
and 5 minus 1 will give us four
our difference now is equal to two
using the common difference two we will
solve now for a sub 2 a sub 3 and a sub
4.
a sub 2
is equal to the sum of the first term
plus the difference our a sub 1 is equal
to
eight
plus the difference which is 2
hour a sub 2
is equal to 10.
for a sub 3 that is the sum
of the second term
plus the difference
our a sub 2
is 10
plus the difference which is 2
our a sub 3
will be equal to 12.
and for the fourth term
that is a sub 4 is equal to the sum of
the third term plus the difference
our a sub 3 is 12.
plus the difference which is
two
power a sub 4 will now be equal to 14.
thus the three unknown terms
a sub 2 is 10
a sub 3 is 12.
and a sub 4
is 14.
for the next example let's try to solve
for the missing terms
get your pen and paper
ready
find the missing terms of the arithmetic
sequence
unknown for a sub 1
6
the third and the fourth terms are also
unknown and 30.
the arrangement of the terms tells us
that the second term or a sub 2 is equal
to 6 and a sub 5 are the fifth term is
equal to 30. we are supposed to find the
first term for a sub 1
the third term a sub 3 and the last
fourth term or a sub 4.
to find for the unknown let us first
determine the common difference D that
is D is equal to a sub n minus a sub K
All Over N minus k
from the given values we will substitute
you will have d
is equal to our a sub n
is a sub 5
minus our a sub K which is the second
term
that is a sub 2
all over we will use n here as 5.
and our K as 2.
next use the values for a sub 5 which is
30
minus the value for a sub 2 which is 6.
over 5 minus 2.
then you'll have D is equal to 30 minus
6 that is 24
divided by 5 minus 2 that is 3.
our difference is now equal to 8.
since we have our common difference of
eight we can now solve for the missing
terms for a sub 1
or the first term
that is equal to
the second term
minus the common difference
which is eight
a sub 2 which is six
minus the difference which is eight our
first term now is equal to
negative 2.
for the third term we will use a sub 3
is equal to the sum of a sub 2 plus the
common difference
you will use
the value for a sub 2 which is
6
plus the difference which is eight
our third term is now equal to
14.
thank you
and for the fourth term you'll have a
sub 4 is equal to the sum of the third
term
Plus
the common difference
using the value for the third term that
is equal to a sub 3 is 14.
plus the difference which is 8
our a sub 4 is now equal to 22.
therefore the three unknown terms for
this sequence are a sub 1 negative two
our a sub 3
is 14.
and our a sub 4
is equal to
22.
you did great on that part dear students
now let's have more activities for you
let's solve this problem together
what is the arithmetic mean between the
two given arithmetic extremes
item number one 5 and 19.
and item number two three x squared plus
eight and x squared minus six
let's deal first with the first item
five and nineteen we will use the
average method that is getting the sum
of five
and nineteen
and we will divide it by two since you
only have two numbers
five plus nineteen will give us 24
divided by 2
. therefore the arithmetic mean of these
two numbers
is 12.
item number two you will use the same
method we will use the average method
for this item so that is the sum of 3x
squared
plus eight
Plus
x squared
minus 6.
remember remember to align
similar terms and your constant getting
the sum you will have
positive two
and that is three plus one you have four
x squared
is this your final answer
no we will divide this into two
dividing this into two
you will get
4 divided by 2 that is two simply copy
this variable in the exponent 2 divided
by 2 is 1.
for item number two the arithmetic mean
is 2x squared plus one
let's have another example
insert the specified number of
arithmetic means between the two given
arithmetic extremes
insert three arithmetic means between 2
and 22.
the arrangement of terms tells us that
the first term
or the a sub 1 is equal to 2 and the
fifth term or a sub 5 is equal to 22.
therefore we are looking for the unknown
terms a sub 2 a sub 3 and a sub 4.
before that let us first determine the
common difference D again you will be
using the formula D is equal to a sub n
minus a sub K All Over N minus k
you will have D is equal to a sub n we
will use the last term or a sub 5.
minus
a sub 1
divided by for n we will use 5.
and for K we will use one
next use the values for a sub 5 and a
sub 1 you have D is equal to a sub 5 is
22.
minus a sub 1 which is 2
divided by 5
minus 1.
our D is equal to 222 minus 2 is 20.
divided by 5 minus 1 that is for our D
our difference is equal to 5.
now we will use the common difference D
to solve for the unknown terms for a sub
2.
we will get the sum of the first term
plus the common difference
that is a sub 1 is equal to 2
plus the difference which is 5
our a sub 2 is equal to 7.
for a sub 3 we will have
the sum of the second term
plus the difference
that is a sub 2 that is 7
plus the difference 5
our a sub 3
is equal to 12.
and for the fourth term we will have
a sub 4
is equal to the sum of a sub 3 plus the
difference our a sub 3 is 12.
plus the difference which is 5 our a sub
4 is equal to 17.
therefore the second term
is 7
third term is 12. and the fourth term
is 17.
easy right
congratulations for a job well done I
hope you learned something new
especially in Computing the arithmetic
means
for the next part of this lesson try to
answer the following questions
number one
how do we find the arithmetic mean of
two arithmetic extremes
we can find the arithmic mean of two
arithmetic extremes by using the
averaging method which means getting the
sum of the two numbers and dividing it
by two
question number two when two or more
arithmetic means are inserted between
two arithmetic extremes
how are they computed
[Music]
when two or more are arithmetic means
are inserted between two arithmetic
extremes they are computed using the
formula in finding the common difference
which is D is equal to a sub n minus a
sub K Over N minus K and adding the
difference to the preceding term to get
the next term
amazing for the next part show what you
can do by answering the following
questions
what is the arithmetic mean between the
given arithmetic extremes
X
second term is a known
9x
for this problem we are going to use the
average method that is adding X and 9x
and dividing it by two
X Plus 9x will give us 10x divided by 2.
the answer is 5X
next problem
insert the specified number of
arithmetic means between the given
arithmetic extremes
insert two arithmetic means between
negative 5 and 1.
for this problem we will use
the formula in finding the difference
which is D is equal to a sub n minus a
sub K Over N minus k
using the values for a sub n we'll use 1
minus a sub K is negative 5
divided by n which is 4
and K which is 1.
1 minus negative 5 will give us 6
divided by 4 minus one is three the
answer is 2.
using the value of the common difference
we can now solve for the unknown terms
for a sub 2 that is the sum of the first
term and the common difference a sub 1
which is negative 5 plus the common
difference 2 a sub 2 is equal to
negative 3.
for the third term we will have the sum
of a sub 2 plus the difference our a sub
2 is negative 3 plus the difference of
two our third term is negative one
great job dear students
and that concludes our lesson for today
see you on the next episode this has
been your teacher Jason also bear in
mind that learning math will always be
fun and easy be awesome be awesome
only here on deputv
[Music]
thank you
[Music]
thank you
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