A1=19
Sequence & Series 2. n=2
A2 = 5(2) + 14
Sequence = 10+14
A2=24
- it is a list of numbers or quantities written in a
specific order by commas. 3. n=3
Ex: 1, 4, 9, 16, 25, … A3=5(3) + 14
= 15+14
Series A3=29
4. n=4
- its is the sum of all term is a given sequence. A4= 5(4) + 14
Ex: 1+2+9+16+25 = 20+14
Arithmetic A4=34
Sequence Series 5. n=5
- a type of sequence in - the sum of the terms of an A5= 5(5) + 14
which each term after the arithmetic sequence. = 25+14
1st term is obtained by A5= 39
𝑛
adding a constant (a 𝑠𝑛 = [2𝑎1+ (𝑛 − 1)𝑑]
common difference). 2 Example 2
𝑛
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑠𝑛 = [𝑎1 + 𝑎𝑛 ] Fin the 8th term in the sequence 4, 16, 64, …
2
An = last term D= common difference Given:
A1 = 1st term Sn= sum of all terms A1=4
N = no. of terms An= last term A8=?
N=8
R=4
Geometric
Sequence Series Type of Problem: Geometric Sequence
- a type of sequence in which - the sum of the terms of
each term after the 1st term an arithmetic sequence. Solution:
is obtained by multiplying a
𝑎 −𝑎 𝑟𝑛
𝑎𝑛 = 𝑎1 𝑟 𝑛−1
constant (a common ratio). 𝑠𝑛 = 1 1−𝑟1 , 𝑟 ≠ 1
𝑆𝑛 = 𝑎1−𝑟𝑎𝑛 , 𝑟 ≠ 1 𝑎8 = 4(48−1 )
𝑛−1
𝑎𝑛 = 𝑎1 𝑟 1−𝑟 𝑎8 = 4(47 )
An= last term R = common ratio 𝑎8 = 4(16,384)
A1= 1st term Sn= sum of all terms 𝑎8 = 65,536
N= no. of terms
Example 3
Example 1 A theater has 26 seats in row 1, 29 seats in row 2,
Find a formula for any term an on a sequence; a1=19 and 32 seats in row 3 and so on. If the pattern
and d=5. Determine the first five terms of the continues, determine the total seating capacity of
sequence. the theater if it has 40 rows.
Given: Given:
A1=19 A1=26
D=5 N=40
D=3
Type of problem: Arithmetic Sequence
Solution: Type of Problem: Arithmetic Series
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑛
𝑎𝑛 = 19 + (𝑛 − 1)5 𝑠𝑛 = [2𝑎1+ (𝑛 − 1)𝑑]
2
𝑎𝑛 = 19 + 5𝑛 − 5
𝑎𝑛 = 5𝑛 + 19 − 5 40
𝑠40 = [2(26) + (40 − 1)3]
𝑎𝑛 = 5𝑛 + 14 (Formula for any term an) 2
𝑠40 = 20[52 + 117]
1. n=1 340 = 20(169)
A1 = 5(1) + 14 𝑠40 = 3,380 seats
= 5+14
Sequence & Series 2. n=2
A2 = 5(2) + 14
Sequence = 10+14
A2=24
- it is a list of numbers or quantities written in a
specific order by commas. 3. n=3
Ex: 1, 4, 9, 16, 25, … A3=5(3) + 14
= 15+14
Series A3=29
4. n=4
- its is the sum of all term is a given sequence. A4= 5(4) + 14
Ex: 1+2+9+16+25 = 20+14
Arithmetic A4=34
Sequence Series 5. n=5
- a type of sequence in - the sum of the terms of an A5= 5(5) + 14
which each term after the arithmetic sequence. = 25+14
1st term is obtained by A5= 39
𝑛
adding a constant (a 𝑠𝑛 = [2𝑎1+ (𝑛 − 1)𝑑]
common difference). 2 Example 2
𝑛
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑠𝑛 = [𝑎1 + 𝑎𝑛 ] Fin the 8th term in the sequence 4, 16, 64, …
2
An = last term D= common difference Given:
A1 = 1st term Sn= sum of all terms A1=4
N = no. of terms An= last term A8=?
N=8
R=4
Geometric
Sequence Series Type of Problem: Geometric Sequence
- a type of sequence in which - the sum of the terms of
each term after the 1st term an arithmetic sequence. Solution:
is obtained by multiplying a
𝑎 −𝑎 𝑟𝑛
𝑎𝑛 = 𝑎1 𝑟 𝑛−1
constant (a common ratio). 𝑠𝑛 = 1 1−𝑟1 , 𝑟 ≠ 1
𝑆𝑛 = 𝑎1−𝑟𝑎𝑛 , 𝑟 ≠ 1 𝑎8 = 4(48−1 )
𝑛−1
𝑎𝑛 = 𝑎1 𝑟 1−𝑟 𝑎8 = 4(47 )
An= last term R = common ratio 𝑎8 = 4(16,384)
A1= 1st term Sn= sum of all terms 𝑎8 = 65,536
N= no. of terms
Example 3
Example 1 A theater has 26 seats in row 1, 29 seats in row 2,
Find a formula for any term an on a sequence; a1=19 and 32 seats in row 3 and so on. If the pattern
and d=5. Determine the first five terms of the continues, determine the total seating capacity of
sequence. the theater if it has 40 rows.
Given: Given:
A1=19 A1=26
D=5 N=40
D=3
Type of problem: Arithmetic Sequence
Solution: Type of Problem: Arithmetic Series
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑛
𝑎𝑛 = 19 + (𝑛 − 1)5 𝑠𝑛 = [2𝑎1+ (𝑛 − 1)𝑑]
2
𝑎𝑛 = 19 + 5𝑛 − 5
𝑎𝑛 = 5𝑛 + 19 − 5 40
𝑠40 = [2(26) + (40 − 1)3]
𝑎𝑛 = 5𝑛 + 14 (Formula for any term an) 2
𝑠40 = 20[52 + 117]
1. n=1 340 = 20(169)
A1 = 5(1) + 14 𝑠40 = 3,380 seats
= 5+14
