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MathSining
Today I don’t feel like doing anything
So I’m just gonna do some sequences
Let me start with some arithmetic
And then I’ll jump in geometrics
Oh and we shouldn’t forget the series
I’m gonna start with finding the common
difference
𝒕 𝟐 minus 𝒕 𝟏is what we’re looking for
But 𝒕 𝟓 minus 𝒕 𝟒 is okay
𝑻 𝒏 equals a plus n minus 1 d
List everything that’s given that it’ll be easy
And this will help you solve the questions
Oh, yes the sequence, the series
Oh they are not so bad
The series is the sum of sequences
They have commas between the terms
𝒏
𝟐
times 𝒂 plus 𝒕 𝒏
That’s the formula for 𝑺 𝒏
But use it when you know the last term
Remember what I said,
Ohohohohohohohoh
Remember what I said
Now we’re moving on the geometrics
Basically the same thing
But you multiply
Find the common ratio, babe.
𝑻 𝒏 over 𝑻 𝒏 minus 1
Is the key
𝑻 𝒏 equals
𝒂 − 𝒓 − 𝒏 minus 1
Divide two consecutive terms
Oh yes, the sequence, the series,
Oh they are not so bad!
The last is the geometric series
It’s the sum of a geometric sequence
Some questions may give you the N,
Or some may give you the 𝑻 𝒏
So know the formula that you’ll use
So infinite here we come
We’ll find out the partial sum
Ohohohohohohohoh
Divergent is limitless
Convergent is limited
Yeah yeah yeah yeah
There’s a boundary fort the ratio
Bigger than neg one
Smaller than one
So I hope you learned the sequences
And I hope you loved our song
‘cause it takes to look to make it
Arithmetic Means
SEQUENCES
1. Choose two (2) different numbers.
2. Denote the smaller number as x and the larger number as y.
3. Find the mean of this two numbers. That is, add these two
number then divide the sum by 2. In symbols,
𝑥+𝑦
2
.
4. Denote the first mean as 𝑚2.
5. Now, find the mean of the smaller number 𝑥 and 𝑚2. In symbols,
𝑥 + 𝑚2
2
.
6. Denote the second mean as 𝑚1.
7. Then, find the mean of the larger number 𝑦 and 𝑚2. In symbols,
𝑦+𝑚2
2
.
8. Denote the third mean as 𝑚3.
9. Lastly, arrange all the numbers in the form 𝑥, 𝑚1, 𝑚2, 𝑚3, 𝑦.
10. Share your answer with your partner.
Illustrative Example 1:
Insert three arithmetic means between 3
and 11.
Solution 1:
We look three numbers 𝒎 𝟏, 𝒎 𝟐 and 𝒎 𝟑
such that 3, 𝒎 𝟏, 𝒎 𝟐, 𝒎 𝟑, 11 is an arithmetic
sequence. In this case, we have 𝒂 𝟏 = 𝟑, 𝒏 =
𝟓, 𝒂 𝟓 = 𝟏𝟏 .
𝒂 𝒏 = 𝒂 𝟏 + (𝒏 − 𝟏)𝒅
𝟏𝟏 = 𝟑 + 𝟓 − 𝟏 𝒅  solve for d
𝟏𝟏 = 𝟑 + 𝟒𝒅
𝟏𝟏 − 𝟑 = 𝟑 − 𝟑 + 𝟒𝒅
𝟖 = 𝟒𝒅
𝟖
𝟏
𝟒
= 𝟒𝒅
𝟏
𝟒
𝒅 = 𝟐
Since d = 2, so we have
𝑚1 = 𝑎1 + 𝑑
𝑚1 = 3 + 2 = 5
𝑚2 = 𝑚1 + 𝑑
𝑚2 = 5 + 2 = 7
𝑚3 = 𝑚2 + 𝑑
𝑚3 = 7 + 2 = 9
Solution 2: Still, we look three numbers
𝒎 𝟏, 𝒎 𝟐 ,and 𝒎 𝟑 such that 3, 𝒎 𝟏, 𝒎 𝟐 ,𝒎 𝟑, 11
is an arithmetic sequence. In this case, we
need to solve for 𝒎 𝟐, the mean of 𝒂 𝟏 = 𝟑
and 𝒂 𝟓 = 𝟏𝟏. That is,
𝒎 𝟐 =
𝒂 𝟏 + 𝒂 𝟓
𝟐
=
𝟑 + 𝟏𝟏
𝟐
=
𝟏𝟒
𝟐
= 𝟕
Now, solve for 𝒎 𝟏, the mean of 𝒂 𝟏 = 𝟑 and
𝒎 𝟐 = 𝟕.That is,
𝒎 𝟏 =
𝒂 𝟏 + 𝒎 𝟐
𝟐
=
𝟑 + 𝟕
𝟐
=
𝟏𝟎
𝟐
= 𝟓
Then, solve for 𝒎 𝟑, the mean
of 𝒂 𝟓 = 𝟑 and 𝒎 𝟐 = 𝟕. That is,
Forming the sequence
𝟑, 𝒎 𝟏, 𝒎 𝟐, 𝒎 𝟑, 𝟏𝟏, we have 3, 5, 7, 9, 11.
Illustrative Example 2:
The 4th term of an arithmetic sequence is 28 and
the 15th term is 105. Find the common difference
and the first term of the sequence.
Solution:
We know that 𝒂 𝟒 = 𝟐𝟖 and 𝒂 𝟏𝟓 = 𝟏𝟎𝟓. Thus
we have
Substituting the given values in the
equation, we have
𝟐𝟖 = 𝒂 𝟏 + (𝟒 − 𝟏)𝒅 (eq. 1)
𝟏𝟎𝟓 = 𝒂 𝟏 + (𝟏𝟓 − 𝟏)𝒅 (eq. 2)
Eliminating 𝒂 𝟏, we subtract (eq. 1) from
(eq. 2)
𝟕𝟕 = 𝟏𝟏𝒅
Therefore, 𝒅 = 𝟕.
Solving for 𝒂 𝟏, we substitute 𝒅 = 𝟕
to (eq. 1)
𝟐𝟖 = 𝒂 𝟏 + (𝟒 − 𝟏)𝟕
𝟐𝟖 = 𝒂 𝟏 + (𝟑)𝟕
𝟐𝟖 = 𝒂 𝟏 + 𝟐𝟏
𝒂 𝟏= 𝟕 Therefore, the common difference and the
first term of the given sequence are d = 7 and 𝒂 𝟏 = 7,
respectively.
1. How did you obtain the
missing term of the
arithmetic sequence?
2. Is the common
difference necessary to
obtain the missing term of
the sequence?
3. How did you obtain the
common difference?
4. If we cannot solve the
common difference by
subtracting two consecutive
terms, is there any other way
to solve for it?
5. What is arithmetic
mean?
START TIMERTIME’S UP! TIME LIMIT:
10 minutes
Criteria
Correct Answer  10
Presentation & Creativity
10
Group Cooperation 5
Fastest Group  5
Group 1
a. Insert two terms in the arithmetic
sequence 15, ___, ___, 36.
Given: 𝒂 𝟏 = ____ ; n = ____ ; 𝒂 𝟒 = ____
b. Insert three arithmetic means between 12
and 56.
Given: 𝒂 𝟏 = ____; 𝒂 𝟓 = ____
Group 2
1. Insert two arithmetic means
between 𝟐𝟎 and 𝟑𝟖.
2. Insert three arithmetic means
between 𝟓𝟐 and 𝟒𝟎.
Group 3
2. Insert three arithmetic means
between 𝟓𝟐 and 𝟒𝟎.
3. Find the missing terms of the
arithmetic sequence
𝟓, ___, ___, ___, ___, 𝟐𝟓.
Group 4
3. Find the missing terms of the
arithmetic sequence
𝟓, ___, ___, ___, ___, 𝟐𝟓.
4. Find the missing terms of the
arithmetic sequence
𝟎, ___, ___, ___, ___, ___, 𝟏𝟓.
Group 5
4. Find the missing terms of the
arithmetic sequence
𝟎, ___, ___, ___, ___, ___, 𝟏𝟓.
5. The fifteenth term of an
arithmetic sequence is – 𝟑 and the
first term is 𝟐𝟓. Find the common
difference and the tenth term.
𝑨𝒓𝒊𝒕𝒉𝒎𝒆𝒕𝒊𝒄 𝑴𝒆𝒂𝒏𝒔 are the terms between any
two nonconsecutive terms of an arithmetic
sequence. It is necessary to solve the common
difference of an arithmetic sequence to insert
terms between two nonconsecutive terms of an
arithmetic sequence. The formula for the
general term of an arithmetic sequence,
𝒂 𝒏 = 𝒂 𝟏 + (𝒏 − 𝟏)𝒅 and the mid-point
between two numbers,
𝑥+𝑦
2
can also be used.
𝟓. 𝒅 = −𝟐, 𝒂 𝟏𝟎 = 𝟕
Output # 1.1
Answer the following problems.
1. Flower farms in Tagaytay grew different variety of
flowers including anthurium. Monica, a flower arranger,
went to Tagaytay to buy anthurium. She plans to arrange
the flowers following an arithmetic sequence with four (4)
layers. If she put one (1) anthurium on the first layer and
seven (7) on the fourth layer, how many anthurium should
be placed on the second and third layer of the flower
arrangement?
2. St. Mary Magdalene Parish Church in Kawit, one
of the oldest churches in Cavite, established in 1624
by Jesuit Missionaries. The church is made of red
bricks preserved for more than a hundred years.
Suppose that church wall contains 4bricks on the
top and 16bricks on the bottom layer. Assuming an
arithmetic sequence, how many bricks are there in
the 2nd, 3rd and 4th layer of the wall?
3. In some of the Kiddie parties nowadays, Tower
Cupcakes were quite popular because it is appealing
and less expensive. In Juan Miguel’s 1st birthday
party, his mother ordered a six (6) layer tower
cupcakes. If the 1st and 4th layer of the tower
contains 6 and 21 cupcakes, respectively, how many
cupcakes are there in the 6th layer (bottom) of the
tower assuming arithmetic sequence in the number of
cupcakes?
ASSIGNMENT
1. Follow-up
a. Find the arithmetic mean of –23 and 7.
2. Study:
Sum of Arithmetic Sequence
a. How to find the sum of terms in an arithmetic sequence?
b. Find the sum of the following arithmetic sequence
1, 4, 7, 10, 13, 16, 19, 22, 25
4, 11, 18, 25, 32, 39, 46, 53, 60
2, 6, 10, 14, 18, 22, 26, 30, 34
7, 12, 17, 22, 27, 32, 37, 42, 47
9, 12, 15, 18, 21, 24, 27, 30, 33
Directions: Use the following numbers inside
the box to complete the arithmetic sequence
below. You may use a number more than once.
1. 2, ___, ___, 14
2. 4, ___, ___, ___, 10
3. 6, ___, ___, ___, 16
4. 9, ___, ___, ___, ___, 24
5. ___, 17, ___, ___, 11
 arithmetic means

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arithmetic means

  • 2. Today I don’t feel like doing anything So I’m just gonna do some sequences Let me start with some arithmetic And then I’ll jump in geometrics Oh and we shouldn’t forget the series
  • 3. I’m gonna start with finding the common difference 𝒕 𝟐 minus 𝒕 𝟏is what we’re looking for But 𝒕 𝟓 minus 𝒕 𝟒 is okay 𝑻 𝒏 equals a plus n minus 1 d List everything that’s given that it’ll be easy And this will help you solve the questions
  • 4. Oh, yes the sequence, the series Oh they are not so bad The series is the sum of sequences They have commas between the terms 𝒏 𝟐 times 𝒂 plus 𝒕 𝒏 That’s the formula for 𝑺 𝒏 But use it when you know the last term
  • 5. Remember what I said, Ohohohohohohohoh Remember what I said Now we’re moving on the geometrics Basically the same thing But you multiply Find the common ratio, babe.
  • 6. 𝑻 𝒏 over 𝑻 𝒏 minus 1 Is the key 𝑻 𝒏 equals 𝒂 − 𝒓 − 𝒏 minus 1 Divide two consecutive terms Oh yes, the sequence, the series, Oh they are not so bad!
  • 7. The last is the geometric series It’s the sum of a geometric sequence Some questions may give you the N, Or some may give you the 𝑻 𝒏
  • 8. So know the formula that you’ll use So infinite here we come We’ll find out the partial sum Ohohohohohohohoh Divergent is limitless Convergent is limited Yeah yeah yeah yeah
  • 9. There’s a boundary fort the ratio Bigger than neg one Smaller than one So I hope you learned the sequences And I hope you loved our song ‘cause it takes to look to make it
  • 11. 1. Choose two (2) different numbers. 2. Denote the smaller number as x and the larger number as y. 3. Find the mean of this two numbers. That is, add these two number then divide the sum by 2. In symbols, 𝑥+𝑦 2 . 4. Denote the first mean as 𝑚2. 5. Now, find the mean of the smaller number 𝑥 and 𝑚2. In symbols, 𝑥 + 𝑚2 2 . 6. Denote the second mean as 𝑚1. 7. Then, find the mean of the larger number 𝑦 and 𝑚2. In symbols, 𝑦+𝑚2 2 . 8. Denote the third mean as 𝑚3. 9. Lastly, arrange all the numbers in the form 𝑥, 𝑚1, 𝑚2, 𝑚3, 𝑦. 10. Share your answer with your partner.
  • 12. Illustrative Example 1: Insert three arithmetic means between 3 and 11. Solution 1: We look three numbers 𝒎 𝟏, 𝒎 𝟐 and 𝒎 𝟑 such that 3, 𝒎 𝟏, 𝒎 𝟐, 𝒎 𝟑, 11 is an arithmetic sequence. In this case, we have 𝒂 𝟏 = 𝟑, 𝒏 = 𝟓, 𝒂 𝟓 = 𝟏𝟏 . 𝒂 𝒏 = 𝒂 𝟏 + (𝒏 − 𝟏)𝒅
  • 13. 𝟏𝟏 = 𝟑 + 𝟓 − 𝟏 𝒅  solve for d 𝟏𝟏 = 𝟑 + 𝟒𝒅 𝟏𝟏 − 𝟑 = 𝟑 − 𝟑 + 𝟒𝒅 𝟖 = 𝟒𝒅 𝟖 𝟏 𝟒 = 𝟒𝒅 𝟏 𝟒 𝒅 = 𝟐 Since d = 2, so we have 𝑚1 = 𝑎1 + 𝑑 𝑚1 = 3 + 2 = 5 𝑚2 = 𝑚1 + 𝑑 𝑚2 = 5 + 2 = 7 𝑚3 = 𝑚2 + 𝑑 𝑚3 = 7 + 2 = 9
  • 14. Solution 2: Still, we look three numbers 𝒎 𝟏, 𝒎 𝟐 ,and 𝒎 𝟑 such that 3, 𝒎 𝟏, 𝒎 𝟐 ,𝒎 𝟑, 11 is an arithmetic sequence. In this case, we need to solve for 𝒎 𝟐, the mean of 𝒂 𝟏 = 𝟑 and 𝒂 𝟓 = 𝟏𝟏. That is, 𝒎 𝟐 = 𝒂 𝟏 + 𝒂 𝟓 𝟐 = 𝟑 + 𝟏𝟏 𝟐 = 𝟏𝟒 𝟐 = 𝟕 Now, solve for 𝒎 𝟏, the mean of 𝒂 𝟏 = 𝟑 and 𝒎 𝟐 = 𝟕.That is, 𝒎 𝟏 = 𝒂 𝟏 + 𝒎 𝟐 𝟐 = 𝟑 + 𝟕 𝟐 = 𝟏𝟎 𝟐 = 𝟓
  • 15. Then, solve for 𝒎 𝟑, the mean of 𝒂 𝟓 = 𝟑 and 𝒎 𝟐 = 𝟕. That is, Forming the sequence 𝟑, 𝒎 𝟏, 𝒎 𝟐, 𝒎 𝟑, 𝟏𝟏, we have 3, 5, 7, 9, 11.
  • 16. Illustrative Example 2: The 4th term of an arithmetic sequence is 28 and the 15th term is 105. Find the common difference and the first term of the sequence. Solution: We know that 𝒂 𝟒 = 𝟐𝟖 and 𝒂 𝟏𝟓 = 𝟏𝟎𝟓. Thus we have
  • 17. Substituting the given values in the equation, we have 𝟐𝟖 = 𝒂 𝟏 + (𝟒 − 𝟏)𝒅 (eq. 1) 𝟏𝟎𝟓 = 𝒂 𝟏 + (𝟏𝟓 − 𝟏)𝒅 (eq. 2) Eliminating 𝒂 𝟏, we subtract (eq. 1) from (eq. 2) 𝟕𝟕 = 𝟏𝟏𝒅 Therefore, 𝒅 = 𝟕.
  • 18. Solving for 𝒂 𝟏, we substitute 𝒅 = 𝟕 to (eq. 1) 𝟐𝟖 = 𝒂 𝟏 + (𝟒 − 𝟏)𝟕 𝟐𝟖 = 𝒂 𝟏 + (𝟑)𝟕 𝟐𝟖 = 𝒂 𝟏 + 𝟐𝟏 𝒂 𝟏= 𝟕 Therefore, the common difference and the first term of the given sequence are d = 7 and 𝒂 𝟏 = 7, respectively.
  • 19. 1. How did you obtain the missing term of the arithmetic sequence?
  • 20. 2. Is the common difference necessary to obtain the missing term of the sequence?
  • 21. 3. How did you obtain the common difference?
  • 22. 4. If we cannot solve the common difference by subtracting two consecutive terms, is there any other way to solve for it?
  • 23. 5. What is arithmetic mean?
  • 24. START TIMERTIME’S UP! TIME LIMIT: 10 minutes Criteria Correct Answer  10 Presentation & Creativity 10 Group Cooperation 5 Fastest Group  5
  • 25. Group 1 a. Insert two terms in the arithmetic sequence 15, ___, ___, 36. Given: 𝒂 𝟏 = ____ ; n = ____ ; 𝒂 𝟒 = ____ b. Insert three arithmetic means between 12 and 56. Given: 𝒂 𝟏 = ____; 𝒂 𝟓 = ____
  • 26. Group 2 1. Insert two arithmetic means between 𝟐𝟎 and 𝟑𝟖. 2. Insert three arithmetic means between 𝟓𝟐 and 𝟒𝟎.
  • 27. Group 3 2. Insert three arithmetic means between 𝟓𝟐 and 𝟒𝟎. 3. Find the missing terms of the arithmetic sequence 𝟓, ___, ___, ___, ___, 𝟐𝟓.
  • 28. Group 4 3. Find the missing terms of the arithmetic sequence 𝟓, ___, ___, ___, ___, 𝟐𝟓. 4. Find the missing terms of the arithmetic sequence 𝟎, ___, ___, ___, ___, ___, 𝟏𝟓.
  • 29. Group 5 4. Find the missing terms of the arithmetic sequence 𝟎, ___, ___, ___, ___, ___, 𝟏𝟓. 5. The fifteenth term of an arithmetic sequence is – 𝟑 and the first term is 𝟐𝟓. Find the common difference and the tenth term.
  • 30. 𝑨𝒓𝒊𝒕𝒉𝒎𝒆𝒕𝒊𝒄 𝑴𝒆𝒂𝒏𝒔 are the terms between any two nonconsecutive terms of an arithmetic sequence. It is necessary to solve the common difference of an arithmetic sequence to insert terms between two nonconsecutive terms of an arithmetic sequence. The formula for the general term of an arithmetic sequence, 𝒂 𝒏 = 𝒂 𝟏 + (𝒏 − 𝟏)𝒅 and the mid-point between two numbers, 𝑥+𝑦 2 can also be used.
  • 31.
  • 32. 𝟓. 𝒅 = −𝟐, 𝒂 𝟏𝟎 = 𝟕
  • 33. Output # 1.1 Answer the following problems. 1. Flower farms in Tagaytay grew different variety of flowers including anthurium. Monica, a flower arranger, went to Tagaytay to buy anthurium. She plans to arrange the flowers following an arithmetic sequence with four (4) layers. If she put one (1) anthurium on the first layer and seven (7) on the fourth layer, how many anthurium should be placed on the second and third layer of the flower arrangement?
  • 34. 2. St. Mary Magdalene Parish Church in Kawit, one of the oldest churches in Cavite, established in 1624 by Jesuit Missionaries. The church is made of red bricks preserved for more than a hundred years. Suppose that church wall contains 4bricks on the top and 16bricks on the bottom layer. Assuming an arithmetic sequence, how many bricks are there in the 2nd, 3rd and 4th layer of the wall?
  • 35. 3. In some of the Kiddie parties nowadays, Tower Cupcakes were quite popular because it is appealing and less expensive. In Juan Miguel’s 1st birthday party, his mother ordered a six (6) layer tower cupcakes. If the 1st and 4th layer of the tower contains 6 and 21 cupcakes, respectively, how many cupcakes are there in the 6th layer (bottom) of the tower assuming arithmetic sequence in the number of cupcakes?
  • 36.
  • 37.
  • 38. ASSIGNMENT 1. Follow-up a. Find the arithmetic mean of –23 and 7. 2. Study: Sum of Arithmetic Sequence a. How to find the sum of terms in an arithmetic sequence? b. Find the sum of the following arithmetic sequence 1, 4, 7, 10, 13, 16, 19, 22, 25 4, 11, 18, 25, 32, 39, 46, 53, 60 2, 6, 10, 14, 18, 22, 26, 30, 34 7, 12, 17, 22, 27, 32, 37, 42, 47 9, 12, 15, 18, 21, 24, 27, 30, 33
  • 39. Directions: Use the following numbers inside the box to complete the arithmetic sequence below. You may use a number more than once. 1. 2, ___, ___, 14 2. 4, ___, ___, ___, 10 3. 6, ___, ___, ___, 16 4. 9, ___, ___, ___, ___, 24 5. ___, 17, ___, ___, 11