Embed Size (px)
Citation preview
CHAPTER 3: SEQUENCE AND SERIES
LEARNING OUTCOME:
At the end of this topic, students are able to:
i. understand the terms involve in sequence and series.
ii. use the suitable formula in writing the pattern of sequence.
iii. perform the right idea on expansion of Binomial Series.
INTRODUCTION:
Many real world processes generate lists of numbers. For instance, the balance in a bank account
at the end of each month, the interest earned each month when you deposit a sum of money into
your account, the height of bouncing balls over certain time and so on.
NOTES AND EXAMPLES:
3.1 Sequence (Introduction)
Sequence is a set of things (usually numbers) that are in order.
If the sequence goes on forever it is called an infinite sequence, otherwise it is a finite sequence
Example:
{1, 2, 3, 4 ,...} is a very simple sequence (and it is an infinite sequence)
{1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence)
Notation
Sequences also use the same notation as sets:
list each element, separated by a comma,
and then put curly brackets around the whole thing.
Example: {3, 5, 7, ...}
Sequence term
𝑥𝑛 is the term
𝑛 is the term number
Example: to mention the "5th term" you just write: 𝑥5
3.1.1 Sequence (Type of Sequence)
1) Arithmetic Sequences
In an Arithmetic Sequence the difference between one term and the next is a constant.
In other words, you just add some value each time ... on to infinity.
Example: {1,2,3, … }, {1,3,5, … }, {−2,3,8, … }
The rule of writing the pattern of the sequence is below:
𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
where:
𝑎 is the first term, and
𝑑 is the difference between the terms (called the "common difference")
𝑑 = 𝑥𝑛+1 − 𝑥𝑛
In General you could write an arithmetic sequence like this:
{𝑎, 𝑎 + 𝑑, 𝑎 + 2𝑑, 𝑎 + 3𝑑, … , 𝑎 + (𝑛 − 1)𝑑}
Example A: Finding the pattern of the sequence
i. 1,4,7,10,13,16, …
ii. 1,3,5,7, …
Solution:
i. 𝑎 = 1, 𝑑 = 3
𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
𝑥𝑛 = 1 + 3(𝑛 − 1)
𝑥𝑛 = 1 + 3𝑛 − 3
𝑥𝑛 = 3𝑛 − 2 ⋕
ii. 𝑎 = 1, 𝑑 = 2
𝑥𝑛 = 𝑎 + 2(𝑛 − 1)
𝑥𝑛 = 1 + 2(𝑛 − 1)
𝑥𝑛 = 1 + 2𝑛 − 2
𝑥𝑛 = 2𝑛 − 1 ⋕
Example B: Finding the nth term of the sequence
i. Find the nth term of the arithmetic sequence whose initial term 𝑎 = 3, and common
difference 𝑑 = 4. Hence find the value of 6th term.
ii. Given the arithmetic sequence {4𝑛 + 3}, find the common difference, and the 8th term
and list down the first four terms.
Solution:
i. 𝑎 = 3, 𝑑 = 4
Formula for arithmetic sequence is
𝑥𝑛 = 𝑎 + (𝑛 − 1)𝑑
= 3 + (𝑛 − 1)4
= 3 + 4𝑛 − 4
= 4𝑛 − 1 ⋕
Hence find the value of 6th term.
𝑥6 = 4(6) − 1 = 24 − 1 = 23 ⋕
ii. The first four terms,
𝑥𝑛 = 4𝑛 + 3
𝑥1 = 4(1) + 3 = 7
𝑥2 = 4(2) + 3 = 11
𝑥3 = 4(3) + 3 = 15
𝑥4 = 4(4) + 3 = 19
The first four terms are {7,11,15,19} ⋕
The 8th term,
𝑥8 = 4(8) + 3
= 35 ⋕
The common difference,
𝑑 = 𝑥2 − 𝑥1
= 11 − 7
= 4 ⋕
Example C: Finding the 𝑎 and 𝑑
i. Find the initial term and the common difference if the 12th term is 53 and 16th term is 69.
ii. Find the pattern of the sequence if given that 𝑥10 = 5 and 𝑥14 = 7.
Solution:
i. When common difference needed, surely it comes from arithmetic sequence. Therefore,
the formula related is
𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
Substitute two terms given,
𝑥12 = 𝑎 + 𝑑(12 − 1) = 53 − − − − − (1)
𝑥16 = 𝑎 + 𝑑(16 − 1) = 69 − − − − − (2)
Simplify
𝑎 + 11𝑑 = 53 − − − − − (1)
𝑎 + 15𝑑 = 69 − − − − − (2)
Solve simultaneously
(2) − (1) ∶ 4𝑑 = 16 ⇒ 𝑑 = 4 ⋕
From (1): 𝑎 + 11(4) = 53 ⇒ 𝑎 = 9 ⋕
ii. 𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
Substitute two terms given,
𝑥10 = 𝑎 + 𝑑(10 − 1) = 5 − − − − − (1)
𝑥14 = 𝑎 + 𝑑(14 − 1) = 7 − − − − − (2)
Simplify
𝑎 + 9𝑑 = 5 − − − − − (1)
𝑎 + 13𝑑 = 7 − − − − − (2)
Solve simultaneously
(2) − (1) ∶ 4𝑑 = 2 ⇒ 𝑑 = 1/2
From (1): 𝑎 + 9(1/2) = 5 ⇒ 𝑎 = 1/2
𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
=1
2+
1
2(𝑛 − 1)
=1
2𝑛 ⋕
2) Geometric Sequences
In a Geometric Sequence each term is found by multiplying the previous term by a constant.
Example: {2,6,18,54 … }, {3,9,27,81, … }, {2,4,8,16 … }
The rule of writing the pattern of the sequence is below:
𝑥𝑛 = 𝑎𝑟𝑛−1 where:
𝑎 is the first term, and
𝑟 is the factor between the terms (called the ‘’common ratio’’)
𝑟 =𝑥𝑛+1
𝑥𝑛
In General you could write a geometric sequence like this:
{𝑎, 𝑎𝑟, 𝑎𝑟2, 𝑎𝑟3, … , 𝑎𝑟𝑛−1}
Example D: Finding the pattern of the sequence
i. 2, 4, 8, 16, …
ii. 1,3,9,27,81, …
Solution:
i. 𝑎 = 2, 𝑟 = 2
𝑥𝑛 = 𝑎𝑟𝑛−1
𝑥𝑛 = 2(2𝑛−1)
𝑥𝑛 = 21(2𝑛−1)
𝑥𝑛 = 21+𝑛−1
𝑥𝑛 = 2𝑛 ⋕
ii. 𝑎 = 1, 𝑑 = 3
𝑥𝑛 = 𝑎𝑟𝑛−1
𝑥𝑛 = 1(3𝑛−1)
𝑥𝑛 = 3𝑛−1 ⋕
Example F:
i. Find the nth term of the geometric sequence whose initial term 𝑎 = 3, and common ratio
𝑟 = 4. Hence find the value of 7th term.
ii. Given the geometric sequence {3𝑛}, find the common ratio, and the 8th term and list down
the first four terms.
Solution:
i. 𝑎 = 2, 𝑟 = 4
Formula for geometric sequence is
𝑥𝑛 = 𝑎𝑟𝑛−1
= 2(4𝑛−1)
= 2(22(𝑛−1))
= 21(22𝑛−2)
= 21+(2𝑛−2)
= 22𝑛−1 ⋕
Hence find the value of 7th term.
𝑥7 = 22(7)−1 = 8192 ⋕
ii. The first four terms,
𝑥𝑛 = 3𝑛
𝑥1 = 31 = 3
𝑥2 = 32 = 9
𝑥3 = 33 = 27
𝑥4 = 34 = 81
The first four terms are {3,9,27,81} ⋕
The 8th term,
𝑥8 = 38
= 6561 ⋕
The common difference,
𝑑 =𝑥2
𝑥1
=9
3
= 3 ⋕
Example G:
i. Find the initial term and the common ratio if the 12th term is 53 and 16the term is 69.
ii. Find 𝑎 and 𝑟 if 𝑥1 = 4 and 𝑥4 = 108
Solution:
i. When common ratio needed, surely it comes from geometric sequence. Therefore, the
formula related is
𝑥𝑛 = 𝑎𝑟𝑛−1
Substitute two terms given,
𝑥12 = 𝑎𝑟12−1 = 512 − − − − − (1)
𝑥14 = 𝑎𝑟16−1 = 2048 − − − − − (2)
Simplify
𝑎𝑟12 = 512 − − − − − (1)
𝑎𝑟14 = 2048 − − − − − (2)
Solve simultaneously
(2) ÷ (1) ∶
𝑎𝑟14
𝑎𝑟12=
2048
512
𝑎1−1𝑟14−12 = 4
𝑟2 = 4
𝑟 = √42
⇒ 𝑟 = 2 ⋕
From (1):
𝑎(212−1) = 512
𝑎 =512
2048 ⇒ 𝑎 =
1
4 ⋕
ii. 𝑥1 = 𝑎 = 4 ⋕
𝑥4 = 4𝑟4−1 = 108
4𝑟3 = 108
𝑟3 = 27
𝑟 = 3 ⋕
3) Other
(a) Sequence including Radical
Example H: Find the pattern of the following sequence
i. 1, √2, √3, √4, …
ii. 2,4
√3,
8
√5,
16
√7, …
Solution:
i. 1 = √1
1, √2, √3, √4, … = √1, √2, √3, √4, …
In this example, the value in the radical is 1,2,3,4,…
Which is arithmetic sequence. Therefore,
𝑥𝑛 = √𝑎 + 𝑑(𝑛 − 1)
𝑎 = 1, 𝑑 = 1
𝑥𝑛 = √1 + 1(𝑛 − 1)
𝑥𝑛 = √1 + 𝑛 − 1
𝑥𝑛 = √𝑛 ⋕
ii. 2,4
√3,
8
√5,
16
√7, … =
2
√1,
4
√3,
8
√5,
16
√7, …
Numerator: geometric sequence
Denominator: radical with arithmetic sequence
𝑥𝑛 =𝑦𝑛
√𝑧𝑛
Numerator:
𝑎 = 2, 𝑟 = 2
𝑦𝑛 = 2(2𝑛−1) = 21(2𝑛−1) = 2𝑛
Denominator:
𝑎 = 1, 𝑑 = 2
√𝑧𝑛 = √1 + 2(𝑛 − 1) = √1 + 2𝑛 − 2 = √2𝑛 − 1
𝑥𝑛 =2𝑛
√2𝑛 − 1⋕
The rule of writing the pattern of the sequence is below:
𝑥𝑛 = √𝑎 + 𝑑(𝑛 − 1) or 𝑥𝑛 = √𝑎𝑟𝑛−1
(b) Different sign sequence
The rule of writing the pattern of the sequence is below:
𝑥𝑛 = (−1)𝑛+1
Example I: Find the pattern of the following sequence
i. 2, −2,2, −2, …
ii. 2, −5,8, −11, …
Solution:
i. Clearly see sequence of +ve and –ve among all number 2.
𝑥𝑛 = (−1)𝑛+1 ∙ 2
𝑥𝑛 = 2(−1)𝑛+1 ⋕
ii. 𝑥𝑛 = (−1)𝑛+1{2,5,8,11, … }
{2,5,8,11, … } is arithmetic sequence.
𝑎 = 2, 𝑑 = 3 {2,5,8,11, … } = 2 + 3(𝑛 − 1) {2,5,8,11, … } = 2 + 3𝑛 − 3 {2,5,8,11, … } = 3𝑛 − 1
𝑥𝑛 = (−1)𝑛+1(3𝑛 − 1) ⋕
3.1.2 Properties of Sequence
Theorem: Properties of Sequence
If {𝑎𝑛} and {𝑏𝑛} are two sequences and 𝑐 is a real number, then
1. ∑ 𝑐
𝑛
𝑘=1
= ∑ 𝑐
𝑛
𝑘=1
2. ∑(𝑐𝑎𝑘) = 𝑐 ∑ 𝑎𝑘
𝑛
𝑘=1
𝑛
𝑘=1
3. ∑(𝑎𝑘 + 𝑏𝑘) = ∑ 𝑎𝑘
𝑛
𝑘=1
𝑛
𝑘=1
+ ∑ 𝑏𝑘
𝑛
𝑘=1
4. ∑(𝑎𝑘 − 𝑏𝑘) = ∑ 𝑎𝑘
𝑛
𝑘=1
𝑛
𝑘=1
− ∑ 𝑏𝑘
𝑛
𝑘=1
5. ∑ 𝑎𝑘 = ∑ 𝑎𝑘
𝑗
𝑘=1
𝑛
𝑘=1
+ ∑ 𝑎𝑘
𝑛
𝑘=𝑗+1
where 1 < 𝑗 < 𝑛
Example J: Write the following summation of sequence following the properties given above
i. ∑ 7
4
𝑘=1
ii. ∑(4𝑘 + 3)
5
𝑘=1
iii. ∑(5𝑘2 − 4𝑘 + 6)
6
𝑘=1
iv. ∑ (𝑘3 −1
2𝑘2 − 7𝑘 − 3)
7
𝑘=1
Solution:
i. ∑ 7
4
𝑘=1
= ∑ 7
4
𝑘=1
⋕
ii. ∑(4𝑘 + 3) =
5
𝑘=1
∑ 4𝑘 + ∑ 3
5
𝑘=1
5
𝑘=1
= 4 ∑ 𝑘 + ∑ 3
5
𝑘=1
5
𝑘=1
⋕
iii. ∑(5𝑘2 − 4𝑘 + 6) = ∑ 5𝑘2 −
6
𝑘=1
∑ 4𝑘 + ∑ 6
6
𝑘=1
6
𝑘=1
= 5 ∑ 𝑘2 −
6
𝑘=1
4 ∑ 𝑘 + ∑ 3
6
𝑘=1
6
𝑘=1
6
𝑘=1
⋕
iv. ∑ (𝑘3 −1
2𝑘2 − 7𝑘 − 3)
7
𝑘=1
= ∑ 𝑘3 −
7
𝑘=1
∑1
2𝑘2 −
7
𝑘=1
∑ 7𝑘 − ∑ 3
7
𝑘=1
7
𝑘=1
= ∑ 𝑘3 −
7
𝑘=1
1
2∑ 𝑘2 −
7
𝑘=1
7 ∑ 𝑘 − ∑ 3
7
𝑘=1
7
𝑘=1
⋕
Formula: Properties of Sequence
If {𝑎𝑛} and {𝑏𝑛} are two sequences and 𝑐 is a real number, then
1. ∑ 𝑐 = 𝑐𝑛
𝑛
𝑘=1
2. ∑ 𝑘 =𝑛(𝑛 + 1)
2
𝑛
𝑘=1
3. ∑ 𝑘2 =𝑛(𝑛 + 1)(2𝑛 + 1)
6
𝑛
𝑘=1
4. ∑ 𝑘3 =
𝑛
𝑘=1
(𝑛(𝑛 + 1)
2)2
Example K: Find/Solve the sum of all sequence below
i. ∑ 5𝑘
3
𝑘=1
ii. ∑(3𝑘 + 2)
4
𝑘=1
iii. ∑(𝑘2 − 𝑘 + 2)
4
𝑘=1
iv. ∑(−2𝑘3 + 𝑘2 + 5𝑘 − 6)
5
𝑘=1
Solution:
i. ∑ 5𝑘
3
𝑘=1
= 5 ∑ 𝑘
3
𝑘=1
= 5 (𝑛(𝑛 + 1)
2)
= 5 (3(3 + 1)
2)
= 5 (3(4)
2)
= 5 (12
2)
= 5(6)
= 30 ⋕
ii. ∑(3𝑘 + 2)
4
𝑘=1
= ∑ 3𝑘 + ∑ 2
4
𝑘=1
4
𝑘=1
= 3 ∑ 𝑘
4
𝑘=1
+ ∑ 2
4
𝑘=1
= 3 (𝑛(𝑛 + 1)
2) + 2𝑛
= 3 (4(4 + 1)
2) + 2(4)
= 3 (4(5)
2) + 8
= 3 (20
2) + 8
= 3(10) + 8 = 38 ⋕
iii. ∑(𝑘2 − 𝑘 + 2)
4
𝑘=1
= ∑ 𝑘2
4
𝑘=1
− ∑ 𝑘 + ∑ 2
4
𝑘=1
4
𝑘=1
=𝑛(𝑛 + 1)(2𝑛 + 1)
6−
𝑛(𝑛 + 1)
2+ 2𝑛
=4(4 + 1)(2(4) + 1)
6−
4(4 + 1)
2+ 2(4)
=4(5)(8 + 1)
6−
4(5)
2+ 8
= 30 − 10 + 8 = 28 ⋕
iv. ∑(−2𝑘3 + 𝑘2 + 5𝑘 − 6)
5
𝑘=1
= −2 ∑ 𝑘3 + ∑ 𝑘2
5
𝑘=1
+ 5 ∑ 𝑘
5
𝑘=1
– ∑ 6
5
𝑘=1
5
𝑘=1
= −2(𝑛(𝑛 + 1)
2)2 +
𝑛(𝑛 + 1)(2𝑛 + 1)
6+ 5
𝑛(𝑛 + 1)
2) − 6𝑛
= −2(5(5 + 1)
2)2 +
5(5 + 1)(2(5) + 1)
6+ 5(
5(5 + 1)
2) − 6(5)
= −2(5(6)
2)2 +
5(6)(11)
6+ 5(
5(6)
2) − 30
= −450 + 55 + 75 − 30
= −350 ⋕
3.2 Series (Introduction)
Series is the sum of all the terms in an infinite sequence.
{2,4,6,8,10, … } ⇒ Sequence
2 + 4 + 6 + 8 + 10 + ⋯ ⇒ Series
The sum is also written as 𝑆𝑛 . It gives 𝑆𝑛 = 2 + 4 + 6 + 8 + 10 + ⋯
3.2.1 Arithmetic Series
Arithmetic series is the sum of all terms of an arithmetic sequence.
Recall that the general arithmetic sequence is:
{𝑎, 𝑎 + 𝑑, 𝑎 + 2𝑑, 𝑎 + 3𝑑, … , 𝑎 + (𝑛 − 1)𝑑}
Therefore the general arithmetic series is:
𝑆𝑛 = 𝑎 + (𝑎 + 𝑑) + (𝑎 + 2𝑑) + (𝑎 + 3𝑑) + ⋯ + (𝑎 + (𝑛 − 1)𝑑)
The formula for the nth term and the sum for arithmetic series are:
𝑥𝑛 = 𝑎 + 𝑑(𝑛 − 1)
𝑆𝑛 =𝑛
2[2𝑎 + (𝑛 − 1)𝑑]
Example L: Find the sum of
i. 1 + 4 + 7 + 10 + ⋯ + (3𝑛 − 2)
ii. 3 + 5 + 7 + 9 + ⋯ + (2𝑛 + 1)
Solution:
i. 𝑎 = 1, 𝑑 = 4 − 1 = 3
𝑆𝑛 = 1 + 4 + 7 + 10 + ⋯ + (3𝑛 − 2) =𝑛
2[2𝑎 + (𝑛 − 1)𝑑]
=𝑛
2[2(1) + (𝑛 − 1)3]
=𝑛
2[2 + 3𝑛 − 3]
𝑆𝑛 =𝑛
2[3𝑛 − 1] ⋕
ii. 𝑎 = 3, 𝑑 = 5 − 3 = 2
𝑆𝑛 = 3 + 5 + 7 + 9 + ⋯ + (2𝑛 + 1) =𝑛
2[2𝑎 + (𝑛 − 1)𝑑]
𝑆𝑛 = 3 + 5 + 7 + 9 + ⋯ + (2𝑛 + 1) =𝑛
2[2(3) + (𝑛 − 1)2]
𝑆𝑛 = 3 + 5 + 7 + 9 + ⋯ + (2𝑛 + 1) =𝑛
2[6 + 2𝑛 − 2]
𝑆𝑛 =𝑛
2[2𝑛 + 4] ⋕
3.2.2 Geometric Series
Geometric series is the sum of all terms of a geometric sequence.
Recall that the general geometric sequence is:
{𝑎, 𝑎𝑟, 𝑎𝑟2, 𝑎𝑟3, … , 𝑎𝑟𝑛−1}
Therefore the general geometric series is:
𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + 𝑎𝑟3 + ⋯ + 𝑎𝑟𝑛−1
The formula for the nth term and the sum for geometric series are:
𝑥𝑛 = 𝑎𝑟𝑛−1
𝑆𝑛 =𝑎(𝑟𝑛−1)
𝑟−1 for 𝑟 > 1 or 𝑆𝑛 =
𝑎(1−𝑟𝑛)
1−𝑟 for 𝑟 < 1
Example M: Find the sum of
i. 1 + 3 + 9 + 27 + 81 + ⋯ + 3𝑛−1
ii. 1 +1
2+
1
4+
1
8+ ⋯ + (
1
2)
𝑛−1
Solution:
i. 𝑎 = 1
𝑟 =3
1= 3 (𝑟 > 1)
𝑆𝑛 = 1 + 3 + 9 + 27 + 81 + ⋯ + 𝑛 =𝑎(𝑟𝑛 − 1)
𝑟 − 1=
1(3𝑛 − 1)
3 − 1=
3𝑛 − 1
2⋕
ii. 𝑎 = 1
𝑑 =
121
=1
2 (𝑟 < 1)
𝑆𝑛 = 1 +1
2+
1
4+
1
8+ ⋯ + (
1
2)
𝑛−1
=𝑎(1 − 𝑟𝑛)
1 − 𝑟=
1 (1 − (12)
𝑛
)
1 − (12)
=1 − (
12)
𝑛
(12)
𝑆𝑛 = 2 (1 − (1
2)
𝑛
) ⋕
Theorem for Geometric Series
a) If |𝑟| < 1, then the geometric series converge with 𝑆∞ =𝑎
1−𝑟
b) If |𝑟| > 1, then the geometric series diverge
Example N:
a) Find the sum of infinite series below
1
2+
1
4+
1
8+ ⋯
b) 2 +1
2+
1
8+
1
32+ ⋯
i. Find the sum of the first 10 terms
ii. If it is converge, find 𝑆∞.
Solution:
a) 𝑎 =1
2, 𝑟 =
1
41
2
=1
2 ⇒ 𝑟 < 1
Since 𝑟 < 1, series is converge
Therefore the sum of infinite series,
𝑆∞ =𝑎
1 − 𝑟=
12
1 −12
= 1 ⋕
b) i) 𝑎 = 2, 𝑟 =1
2
2=
1
4
Sum of first 10 terms,
𝑆10 =𝑎(1−𝑟𝑛)
1−𝑟=
2(1−(1
4)
10)
1−(1
4)
=2(1−(
1
4)
10)
3
4
=8
3(1 − (
1
4)
10
) =8
3(0.9999) = 2.6667 ⋕
ii) When 𝑟 =1
4< 1 ⇒ The series converge
Therefore we can find 𝑆∞
𝑆∞ =𝑎
1 − 𝑟=
2
1 − (14)
=2
34
=8
3⋕
Example O:Express the recurring decimal below as an infinite geometric series.
a) 0.68686868 …
b) 0.201201201 …
Solution:
a) 0.68 68 68 68 …
= 0.68 + 0.00 68 + 0.00 00 68
From this,
𝑎 = 0.68, 𝑟 =0.00 68
0.68= 0.01
Thus by using
𝑆∞ =𝑎
1 − 𝑟=
0.68
1 − 0.01=
68
99⋕
b) 0.201 201 201 …
= 0.201 + 0.000 201 + 0.000 000 201
From this,
𝑎 = 0.201, 𝑟 =0.000 201
0.201= 0.001
Thus by using
𝑆∞ =𝑎
1 − 𝑟=
0.201
1 − 0.001=
67
333⋕
EXERCISE:
1. Write the first five terms of the arithmetic sequences.
(a) 35 nan
(b)
n
na
2
1
(c)
n
na
2
1
(d) 12
432
2
n
nnan
(e) n
na )1(1
2. Find the nth term of the arithmetic sequence if
(a) Initial term, 𝑎 = 15 and common difference, 𝑑 = 4
(b) Initial term, 𝑎 = 0 and common difference, 𝑑 = −2/3
(c) Initial term, 𝑎 = −𝑦 and common difference, 𝑑 = 5𝑦
3. Find the initial term and common difference if
(a) 943 a and 856 a
(b) 1905 a and 1156 a
(c) 268 a and 4212 a
4. Write the next three terms of the geometric sequences.
(a) ,....48,24,12,6
(b) ,...27
1,
9
1,
3
1,1
(c) ,...515,15,53,3
(d) ,...40,20,10,5 32 xxx
5. Find the nth term of the geometric sequence if
(a) Initial term, 𝑎 = 2 and common ratio, 𝑟 = 3
(b) Initial term, 𝑎 = 5 and common ratio, 𝑟 = −1/10
(c) Initial term, 𝑎 = 3 and common ratio, 𝑟 = √5
6. Find the initial term and common ratio if
(a) 32 a and 64
35 a
(b) 184 a and 3
27 a
(c) 3
163 a and
27
645 a
7. Write the pattern for the apparent 𝑛th terms of the sequences.
(a) ,...19,15,11,7,3
(b) ,...10,8,6,4,2
(c) ,...16
1,
8
1,
4
1,
2
1
(d) ,...,120
1,
24
1,
6
1,
2
1,1
(e) ,...81
2,
27
2,
6
2,
3
2,2
5432
(f) ,...32
311,
16
151,
8
71,
4
31,
2
11
8. Find the sum
(a)
6
1
)13(i
i
(b)
5
1
5k
(c)
5
1
22i
i
(d)
6
1
23 )2(k
kkk
(e)
5
1
24
1
3 )4()2(ji
ji
9. Find the sum of finite arithmetic series
(a) ..3023169 until th10 term.
(b) ...201482 until th9 term.
(c) ..949698100 until th15 term.
(d) ..22
31
2
1 until
th18 term.
(e) ..2
72
2
11 until
th11 term.
10. Find the sum of finite geometric series
(a) ..8421 until th9 term.
(b) ...2
33612 until
th10 term.
(c) ...4
27
2
932 until
th8 term.
(d) ...241263 until th8 term.
(e) ...16
1
8
1
4
1
2
1 until
th7 term.
11. Find the sum of infinite geometric series below.
(a) ...8
1
4
1
2
11
(b) ...81
32
27
16
9
8
3
42
(c) ...1000
1
100
1
10
11
(d) ...313
1
9
1
(e) ...656
25
36
125
12. Express the recurring decimal below as an infinite geometric series.
(a) 297.0
(b) 83.1
(c) 681.0
(d) 35.2
PROBLEM:
1. Object Drop
An object is dropped from an airplane and falls 30 feet during the first second. During each
successive second it falls 44 feet more than in the preceding second. How many feet does it
travel during first 10 seconds? How far does it fall during tenth second?
2. Bacterial Culture
A certain bacterial culture doubles in number every day. If there were 1000 bacteria at the end
of the first day, how many will be there after 8 days? How many after 𝑛 days?
3. Area of Circle
The largest circle has radius 11 BA . The next circle has BABA 122
1 , the one after that has
radius BABA 232
1 , and so on. If these circles continue endlessly in this manner, what is the
sum of the areas of all the circles?
