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WELCOME
Students!
DRILL
Subtraction
of Integers
(-5) - ( -2)
(8) - ( -3)
(-9) - (8)
(4) - (10)
(-3) - (11)
WHO AM I?
FINITE OR INFINITE ?
ARITHMETIC OR NOT?
82, 86, 90, 94, 98
2, 4, 6, 8, 10, 12, ...
10, 20, 30, 40, ....
5, 25, 125, 625
Quarter 1
Week 2
a. Determine the nth term of a given
arithmetic sequence;
b. Define an arithmetic mean;
c. Find the arithmetic mean between the
terms of an arithmetic sequence;
d. Describe an arithmetic series;
e. Determine the sum of the first n terms of
an arithmetic sequence and;
f. Confidently solve different problems involving
arithmetic sequence, arithmetic mean, and
arithmetic series.
O
B
J
E
C
T
I
V
E
S
Lesson 1
DETERMINING the
Nth of an ARITHMETIC
SEQUENCE
ARITHMETIC
SEQUENCE
is a sequence where every
term after the first term is
obtained by adding a
constant.
common difference
is the constant number added to
the preceding term of the
arithmetic sequence.
It can be calculated by subtracting
any 2 consecutive term in the
arithmetic sequence
ARITHMETIC
SEQUENCE
FORMULA:
an = a1 + ( n - 1 )d
FORMULA:
an = a1 + ( n - 1 )d
nth
term
first
term
number
of
terms
common
difference
3, 7, 11, 15,19
3
7
11
15
19
= 3
= 3+4
4 4 4 4
= 3+4+4
= 3+0(4)
= 3+1(4)
= 3+2(4)
= 3+3(4)
=3+4(4)
= a1 = a1+0d
= 3+4+4+4
= 3+4+4+4+4
= a2 = a1+1d
= a3 = a1+2d
= a4 = a1+3d
= a5 = a1+4d
d
an = a1+(n-1)d
ARITHMETIC
SEQUENCE
FORMULA:
an = a1 + ( n - 1 )d
Find the 16th term of
the arithmetic
sequence
15, 21, 27, 33, 39, …
Problem 1
Step 1: Identify the given and the
unknown in the problem.
an = a1 + ( n - 1 )d
Problem: Find the 16th term of the arithmetic
sequence 15, 21, 27, 33, 39, …
an = ?
a1
n
d
= 15
= 16
= 6
Step 2: Substitute the known
quantities to the formula.
an = a1 + ( n - 1 )d
a16 = 15 + ( 16 - 1 )6
a1 n d
= 15 = 16 = 6
Step 3: Simplify
a16 = 15 + ( 16 - 1 )
= 15 + ( 15 )6
= 15 + 90
a16 = 105
a16 = 15 + ( 16 - 1 )6
= 15 + ( 15 )6
= 15 + 90
Therefore, the 16th term is 105.
In the
arithmetic sequence
4, 10, 16, 22, 28,…
which term is 130?
Problem 2
Step 1: Identify the given and the
unknown in the problem.
an = a1 + ( n - 1 )d
an
= 130
a1
n
d
= 4
= ?
= 6
Problem: In the arithmetic sequence
4, 10, 16, 22, 28,… which term is 112?
Step 2: Substitute the known
quantities to the formula.
an = a1 + ( n - 1 )d
130 = 4 + ( n - 1 )6
an a1 d
= 130 = 4 = 6
Step 3: Simplify.
130 = 4 + ( n - 1 )6
130 = 4 + 6n - 6
130 = -2 + 6n
Step 3: Simplify.
130 = -2 + 6n
130 + 2 = 6n
132 = 6n
22 = n
Therefore, 130 is the 22nd term.
Find the 16th term of
the arithmetic sequence
whose first term is 11
and whose seventh term
is 59.
Problem 3
Step 1: Identify the given and the
unknown in the problem.
a1 = 11
a7 = 59
n =7
d= ?
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
a16 =?
Step 2: Solve for d
an = a1 + ( n - 1 )d
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
a7 = a1 + ( n - 1 )d
59 = 11 + ( 7 - 1 )d
Step 2: Solve for d
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
59 = 11 + ( 6 )d
59 - 11 = 6d
48 = 6d
8 = d
shortcut
in finding
d
a1 = 11 a7 = 59
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term
is 59.
2. 7 - 1 = 6
1. 59 - 11 = 48
3. 48 ÷ 6 = 8 d
Step 1: Identify the given and the
unknown in the problem.
a1 = 11
a7 = 59
n =7
d= 8
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
a16 =?
Step 2: Solve for a16
an = a1 + ( n - 1 )d
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
a16 = a1 + ( n - 1 )d
a16 = 11 + ( 16 - 1 )8
Step 2: Solve for a16
Find the 16th term of the arithmetic sequence
whose first term is 11 and whose seventh term is 59.
a16 = 11 + ( 16 - 1 )8
= 11 + ( 15 )8
= 11 + 120
a16 = 131
Therefore, the 16th term is 131.
Lesson 2
Arithmetic
Mean
Arithmetic Means
- are the terms between any two
nonconsecutive terms of an arithmetic
sequence
Steps in solving the arithmetic mean
2. Using the formula of arithmetic sequence solve for
the value of the common difference
3. To get the value of the missing term add
the common difference and the value before
the missing term.
1. Identify the no. of terms in the arithmetic sequence.
Problem 1:
What is the arithmetic
mean between 50 and
120?
formula:
a + b
2
Solution
a = 50 b = 120
a + b = 50+120
2 2
= 170
2
= 85
AM =
Problem 2:
Insert 6 arithmetic
mean between 5 and 61.
Insert six arithmetic means between 5 and 61.
5,
Given:
2
1 4
3 6
5 8
7
61,
___,
___, ___,
___, ___, ___,
Formula:
Solution:
13, 21, 29, 37, 45, 53,
a1= 5 a8= 61 n= 8
a2 = a1 + d
Solve for d.
1. 61 - 5 = 56
2. 8 - 1 = 7
3. 56 ÷ 7 = 8 = d
= 5 + 8
a2 = 13
a3 = a2 + d
a3 = 13 + 8 = 21
d= ?
Problem 3:
If the 3rd term of an AP is18
and the 9th term is 60,
what is the first term
and the 6th term of the
sequence?
__, __,18,__,__,__,__,__,60
a3 a9
a4 a5 a6 a7
a8
a1 a2
a1 =18 a7 =60
1. 60 – 18 = 42
Solve for d.
2. 7 – 1 = 6
3. 42/6 = 7
a2 = 18 – 7 = 11
a1 = 11 – 7 = 4
a4 = 18 + 7 = 25
a5 = 25 + 7 = 32
a6 = 32 + 7 = 39
4 39
a3 = 18
Lesson 3
Arithmetic
Series
Arithmetic Series
- It is the sum of the terms of a
given Arithmetic Sequence.
Formula:
Sn = n ( a1 + an )
2
Sn – sum of n terms
an – nth terms
a1 – first terms
n – no. of terms
Formula:
Sn = n [2a1 + (n-1)d]
2
Sn – sum of n terms
an – nth terms
a1 – first terms
n – no. of terms
d – common difference
Find the sum of the first
twenty terms of the arithmetic
sequence 18, 23, 28, 33, 38, ...
Problem 1
Step 1 : Identify the known
and unknown quantities.
S20 = ? n = 20 a1 = 18
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
Formula #1
Sn = n ( a1 + an )
2
S20=? n=20 a1=18 an=?
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
Step 2: Solve for a20
an = a1 + ( n - 1 )d
a20 = a1 + ( n - 1 )d
a20 = 18 + ( 20 - 1 )5
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
a20 = 18 + ( 20 - 1 )5
= 18 + ( 19 )5
= 18 + 95
a20 = 113
Therefore, the 20th term is 113.
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
Sn = n ( a1 + an )
2
S20 = ? n = 20
a1 = 18 a20 = 113
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
S = n ( a1 + an )
2
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
S20 = 20 ( 18 + 113 )
2
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
S20 = 20 ( 18 + 170 )
2
S20 = 10 ( 131 )
S20 = 1 310 )
Formula #2:
Sn = n [2a1 + (n-1)d]
2
Sn – sum of n terms
an – nth terms
a1 – first terms
n – no. of terms
d – common difference
Step 1 : Identify the known
and unknown quantities.
Find the sum of the first twenty terms of the
arithmetic sequence 18, 23, 28, 33, 38, ...
S20=? n=20 a1=18
d = 5 an=?
Formula #2:
Sn = n [2a1 + (n-1)d]
2
S20=? n=20 a1=18 d = 5 an=?
S20 = 20 [2(18) + (20-1)5]
2
S20 = 20 [2(18) + (20-1)5]
2
S20 = 10 [36 + (19)5]
S20 = 10 [36 + 95]
S20 = 10 [ 131 ]
S20 = 1 310
Thank you!

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ARITHMETIC SEQUENCE, MEAN AND SERIES WEEK 2 QUARTER 1

  • 3. (-5) - ( -2)
  • 4. (8) - ( -3)
  • 9. FINITE OR INFINITE ? ARITHMETIC OR NOT?
  • 10. 82, 86, 90, 94, 98
  • 11. 2, 4, 6, 8, 10, 12, ...
  • 12. 10, 20, 30, 40, ....
  • 13. 5, 25, 125, 625
  • 15. a. Determine the nth term of a given arithmetic sequence; b. Define an arithmetic mean; c. Find the arithmetic mean between the terms of an arithmetic sequence; d. Describe an arithmetic series; e. Determine the sum of the first n terms of an arithmetic sequence and; f. Confidently solve different problems involving arithmetic sequence, arithmetic mean, and arithmetic series. O B J E C T I V E S
  • 16. Lesson 1 DETERMINING the Nth of an ARITHMETIC SEQUENCE
  • 17. ARITHMETIC SEQUENCE is a sequence where every term after the first term is obtained by adding a constant.
  • 18. common difference is the constant number added to the preceding term of the arithmetic sequence. It can be calculated by subtracting any 2 consecutive term in the arithmetic sequence
  • 20. FORMULA: an = a1 + ( n - 1 )d nth term first term number of terms common difference
  • 21. 3, 7, 11, 15,19 3 7 11 15 19 = 3 = 3+4 4 4 4 4 = 3+4+4 = 3+0(4) = 3+1(4) = 3+2(4) = 3+3(4) =3+4(4) = a1 = a1+0d = 3+4+4+4 = 3+4+4+4+4 = a2 = a1+1d = a3 = a1+2d = a4 = a1+3d = a5 = a1+4d d an = a1+(n-1)d
  • 23. Find the 16th term of the arithmetic sequence 15, 21, 27, 33, 39, … Problem 1
  • 24. Step 1: Identify the given and the unknown in the problem. an = a1 + ( n - 1 )d Problem: Find the 16th term of the arithmetic sequence 15, 21, 27, 33, 39, … an = ? a1 n d = 15 = 16 = 6
  • 25. Step 2: Substitute the known quantities to the formula. an = a1 + ( n - 1 )d a16 = 15 + ( 16 - 1 )6 a1 n d = 15 = 16 = 6
  • 26. Step 3: Simplify a16 = 15 + ( 16 - 1 ) = 15 + ( 15 )6 = 15 + 90 a16 = 105 a16 = 15 + ( 16 - 1 )6 = 15 + ( 15 )6 = 15 + 90 Therefore, the 16th term is 105.
  • 27. In the arithmetic sequence 4, 10, 16, 22, 28,… which term is 130? Problem 2
  • 28. Step 1: Identify the given and the unknown in the problem. an = a1 + ( n - 1 )d an = 130 a1 n d = 4 = ? = 6 Problem: In the arithmetic sequence 4, 10, 16, 22, 28,… which term is 112?
  • 29. Step 2: Substitute the known quantities to the formula. an = a1 + ( n - 1 )d 130 = 4 + ( n - 1 )6 an a1 d = 130 = 4 = 6
  • 30. Step 3: Simplify. 130 = 4 + ( n - 1 )6 130 = 4 + 6n - 6 130 = -2 + 6n
  • 31. Step 3: Simplify. 130 = -2 + 6n 130 + 2 = 6n 132 = 6n 22 = n Therefore, 130 is the 22nd term.
  • 32. Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. Problem 3
  • 33. Step 1: Identify the given and the unknown in the problem. a1 = 11 a7 = 59 n =7 d= ? Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 =?
  • 34. Step 2: Solve for d an = a1 + ( n - 1 )d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a7 = a1 + ( n - 1 )d 59 = 11 + ( 7 - 1 )d
  • 35. Step 2: Solve for d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. 59 = 11 + ( 6 )d 59 - 11 = 6d 48 = 6d 8 = d
  • 37. a1 = 11 a7 = 59 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. 2. 7 - 1 = 6 1. 59 - 11 = 48 3. 48 ÷ 6 = 8 d
  • 38. Step 1: Identify the given and the unknown in the problem. a1 = 11 a7 = 59 n =7 d= 8 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 =?
  • 39. Step 2: Solve for a16 an = a1 + ( n - 1 )d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 = a1 + ( n - 1 )d a16 = 11 + ( 16 - 1 )8
  • 40. Step 2: Solve for a16 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 = 11 + ( 16 - 1 )8 = 11 + ( 15 )8 = 11 + 120 a16 = 131 Therefore, the 16th term is 131.
  • 42. Arithmetic Means - are the terms between any two nonconsecutive terms of an arithmetic sequence Steps in solving the arithmetic mean 2. Using the formula of arithmetic sequence solve for the value of the common difference 3. To get the value of the missing term add the common difference and the value before the missing term. 1. Identify the no. of terms in the arithmetic sequence.
  • 43. Problem 1: What is the arithmetic mean between 50 and 120?
  • 44. formula: a + b 2 Solution a = 50 b = 120 a + b = 50+120 2 2 = 170 2 = 85 AM =
  • 45. Problem 2: Insert 6 arithmetic mean between 5 and 61.
  • 46. Insert six arithmetic means between 5 and 61. 5, Given: 2 1 4 3 6 5 8 7 61, ___, ___, ___, ___, ___, ___, Formula: Solution: 13, 21, 29, 37, 45, 53, a1= 5 a8= 61 n= 8 a2 = a1 + d Solve for d. 1. 61 - 5 = 56 2. 8 - 1 = 7 3. 56 ÷ 7 = 8 = d = 5 + 8 a2 = 13 a3 = a2 + d a3 = 13 + 8 = 21 d= ?
  • 47. Problem 3: If the 3rd term of an AP is18 and the 9th term is 60, what is the first term and the 6th term of the sequence?
  • 48. __, __,18,__,__,__,__,__,60 a3 a9 a4 a5 a6 a7 a8 a1 a2 a1 =18 a7 =60 1. 60 – 18 = 42 Solve for d. 2. 7 – 1 = 6 3. 42/6 = 7 a2 = 18 – 7 = 11 a1 = 11 – 7 = 4 a4 = 18 + 7 = 25 a5 = 25 + 7 = 32 a6 = 32 + 7 = 39 4 39 a3 = 18
  • 50. Arithmetic Series - It is the sum of the terms of a given Arithmetic Sequence.
  • 51. Formula: Sn = n ( a1 + an ) 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms
  • 52. Formula: Sn = n [2a1 + (n-1)d] 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms d – common difference
  • 53. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... Problem 1
  • 54. Step 1 : Identify the known and unknown quantities. S20 = ? n = 20 a1 = 18 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
  • 55. Formula #1 Sn = n ( a1 + an ) 2 S20=? n=20 a1=18 an=? Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
  • 56. Step 2: Solve for a20 an = a1 + ( n - 1 )d a20 = a1 + ( n - 1 )d a20 = 18 + ( 20 - 1 )5 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
  • 57. a20 = 18 + ( 20 - 1 )5 = 18 + ( 19 )5 = 18 + 95 a20 = 113 Therefore, the 20th term is 113. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
  • 58. Sn = n ( a1 + an ) 2 S20 = ? n = 20 a1 = 18 a20 = 113 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
  • 59. S = n ( a1 + an ) 2 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20 = 20 ( 18 + 113 ) 2
  • 60. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20 = 20 ( 18 + 170 ) 2 S20 = 10 ( 131 ) S20 = 1 310 )
  • 61. Formula #2: Sn = n [2a1 + (n-1)d] 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms d – common difference
  • 62. Step 1 : Identify the known and unknown quantities. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20=? n=20 a1=18 d = 5 an=?
  • 63. Formula #2: Sn = n [2a1 + (n-1)d] 2 S20=? n=20 a1=18 d = 5 an=? S20 = 20 [2(18) + (20-1)5] 2
  • 64. S20 = 20 [2(18) + (20-1)5] 2 S20 = 10 [36 + (19)5] S20 = 10 [36 + 95] S20 = 10 [ 131 ] S20 = 1 310