Sequences and Series — Class 11 Maths Notes

Sequences and Series · Class 11 Maths · 10 topics.

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Topics covered in Sequences and Series

1.Introduction of Sequences and Series
Sequences
Definition
A sequence is an ordered list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term.
Types of Sequences
1. Arithmetic Sequence (AP):
  - Definition: A sequence in which the difference between any two consecutive terms is constant.
  - Common Difference (d): The constant difference between consecutive terms.
  - General Term: If 𝑎a is the first term and 𝑑d is the common difference, the nth term (𝑎𝑛an) is given by:𝑎𝑛=𝑎+(𝑛−1)𝑑an=a+(n−1)d
  - Example:
    - Consider the sequence: 2, 5, 8, 11,...
    - Here, the first term 𝑎=2a=2 and the common difference 𝑑=3d=3.
    - To find the 5th term:𝑎5=2+(5−1)×3=2+12=14a5=2+(5−1)×3=2+12=14
    - So, the 5th term is 14.
2. Geometric Sequence (GP):
  - Definition: A sequence in which the ratio of any two consecutive terms is constant.
  - Common Ratio (r): The constant ratio between consecutive terms.
  - General Term: If 𝑎a is the first term and 𝑟r is the common ratio, the nth term (𝑎𝑛an) is given by:𝑎𝑛=𝑎⋅𝑟(𝑛−1)an=a⋅r(n−1)
  - Example:
    - Consider the sequence: 3, 6, 12, 24,...
    - Here, the first term 𝑎=3a=3 and the common ratio 𝑟=2r=2.
    - To find the 5th term:𝑎5=3⋅2(5−1)=3⋅24=3⋅16=48a5=3⋅2(5−1)=3⋅24=3⋅16=48
    - So, the 5th term is 48.
Series
A series is the sum of the terms of a sequence. If you add up the first n terms of a sequence, you get a series.
Types of Series
1. Arithmetic Series:
  - Sum of the First n Terms: The sum 𝑆𝑛Sn of the first n terms of an arithmetic sequence can be found using the formula:𝑆𝑛=𝑛2×(2𝑎+(𝑛−1)𝑑)Sn=2n×(2a+(n−1)d)
  - Example:
    - For the sequence 2, 5, 8, 11,... find the sum of the first 4 terms.
    - Here, 𝑎=2a=2, 𝑑=3d=3, and 𝑛=4n=4.
    - Using the formula:𝑆4=42×(2×2+(4−1)×3)=2×(4+9)=2×13=26S4=24×(2×2+(4−1)×3)=2×(4+9)=2×13=26
    - So, the sum of the first 4 terms is 26.
2. Geometric Series:
  - Sum of the First n Terms: The sum 𝑆𝑛Sn of the first n terms of a geometric sequence can be found using the formula:𝑆𝑛=𝑎(1−𝑟𝑛)1−𝑟if 𝑟≠1Sn=a1−r(1−rn)ifr=1
  - Example:
    - For the sequence 3, 6, 12, 24,... find the sum of the first 4 terms.
    - Here, 𝑎=3a=3, 𝑟=2r=2, and 𝑛=4n=4.
    - Using the formula:𝑆4=3(1−24)1−2=3(1−16)−1=3×15=45S4=31−2(1−24)=3−1(1−16)=3×15=45
    - So, the sum of the first 4 terms is 45.
Detailed Real-Life Example
Let's consider an example related to saving money, which is an arithmetic sequence.
Scenario: You decide to save money in a piggy bank. You start with ₹100 in the first month and plan to increase your savings by ₹50 every month.
- First Term (a): ₹100
- Common Difference (d): ₹50
To find the amount saved in the 6th month:
𝑎6=𝑎+(6−1)×𝑑=100+5×50=100+250=350a6=a+(6−1)×d=100+5×50=100+250=350
So, in the 6th month, you'll save ₹350.
To find the total savings after 6 months:
𝑆6=62×(2×100+(6−1)×50)=3×(200+250)=3×450=1350S6=26×(2×100+(6−1)×50)=3×(200+250)=3×450=1350
So, the total savings after 6 months will be ₹1350.
Real-Life Applications and Careers
Finance and Economics:
- Interest Calculations: Understanding sequences and series helps in calculating simple and compound interest, essential for banking and finance.
- Investment Analysis: Analyzing stock market trends often involves geometric sequences to understand growth rates.
Engineering:
- Signal Processing: Engineers use sequences and series to process and analyze signals in telecommunications.
- Control Systems: Designing control systems often requires understanding of sequences for stability analysis.
Computer Science:
- Algorithm Efficiency: Many algorithms use sequences and series to determine time and space complexity, which is crucial for optimizing code.
- Cryptography: Sequences play a role in encoding and decoding information securely.
Detailed Step-by-Step Explanation of a Geometric Sequence
1. Identify the First Term (a) and Common Ratio (r):
  - Example: 3, 6, 12, 24,...
  - First term 𝑎=3a=3
  - Common ratio 𝑟=2r=2
2. Find the nth Term:
  𝑎𝑛=𝑎⋅𝑟(𝑛−1)an=a⋅r(n−1)
  - To find the 5th term:𝑎5=3⋅2(5−1)=3⋅24=3⋅16=48a5=3⋅2(5−1)=3⋅24=3⋅16=48
  - So, the 5th term is 48.
2.Sequences
Definition of a Sequence
A sequence is an ordered list of numbers that follow a specific rule or pattern. Each number in a sequence is called a term. The position of each term in the sequence is called its index.
Notation
A sequence is usually denoted by {𝑎𝑛}{an}, where 𝑎𝑛an represents the nth term of the sequence. For example, {𝑎1,𝑎2,𝑎3,…}{a1,a2,a3,…} denotes the first, second, third, and subsequent terms of the sequence.
Types of Sequences
1. Arithmetic Sequence (AP)
  - Definition: A sequence in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (𝑑d).
  - General Term: The nth term of an arithmetic sequence is given by:𝑎𝑛=𝑎+(𝑛−1)𝑑an=a+(n−1)dwhere 𝑎a is the first term and 𝑑d is the common difference.
  - Example:
    - Sequence: 3, 7, 11, 15,...
    - Here, the first term 𝑎=3a=3 and the common difference 𝑑=4d=4.
    - To find the 5th term (𝑎5a5):𝑎5=3+(5−1)×4=3+16=19a5=3+(5−1)×4=3+16=19
    - So, the 5th term is 19.
2. Geometric Sequence (GP)
  - Definition: A sequence in which the ratio between any two consecutive terms is constant. This constant ratio is called the common ratio (𝑟r).
  - General Term: The nth term of a geometric sequence is given by:𝑎𝑛=𝑎⋅𝑟(𝑛−1)an=a⋅r(n−1)where 𝑎a is the first term and 𝑟r is the common ratio.
  - Example:
    - Sequence: 2, 6, 18, 54,...
    - Here, the first term 𝑎=2a=2 and the common ratio 𝑟=3r=3.
    - To find the 4th term (𝑎4a4):𝑎4=2⋅3(4−1)=2⋅27=54a4=2⋅3(4−1)=2⋅27=54
    - So, the 4th term is 54.
3. Fibonacci Sequence
  - Definition: A sequence in which each term is the sum of the two preceding terms.
  - General Term: The nth term (𝐹𝑛Fn) is given by:𝐹𝑛=𝐹𝑛−1+𝐹𝑛−2Fn=Fn−1+Fn−2with initial terms 𝐹0=0F0=0 and 𝐹1=1F1=1.
  - Example:
    - Sequence: 0, 1, 1, 2, 3, 5, 8, 13,...
    - To find the 6th term (𝐹6F6):𝐹6=𝐹5+𝐹4=5+3=8F6=F5+F4=5+3=8
    - So, the 6th term is 8.
Properties of Sequences
1. Monotonic Sequences: A sequence that is either entirely non-increasing or non-decreasing.
  - Increasing Sequence: Each term is greater than or equal to the previous term.
  - Decreasing Sequence: Each term is less than or equal to the previous term.
2. Bounded Sequences: A sequence that has an upper bound or a lower bound.
  - Upper Bounded: All terms are less than or equal to some fixed number.
  - Lower Bounded: All terms are greater than or equal to some fixed number.
Real-Life Examples of Sequences
Arithmetic Sequence Example: Imagine you start saving money, and you decide to increase your savings by a fixed amount each month. If you save ₹500 in the first month, ₹550 in the second month, ₹600 in the third month, and so on, this forms an arithmetic sequence with the first term 𝑎=500a=500 and the common difference 𝑑=50d=50.
Geometric Sequence Example: Consider a scenario where a population of bacteria doubles every hour. If you start with 1 bacterium, after 1 hour you'll have 2, after 2 hours you'll have 4, after 3 hours you'll have 8, and so on. This forms a geometric sequence with the first term 𝑎=1a=1 and the common ratio 𝑟=2r=2.
Practice Problems
1. Find the 10th term of the arithmetic sequence 4, 9, 14, 19,...
  - First term 𝑎=4a=4
  - Common difference 𝑑=5d=5
  - Using the formula:𝑎10=4+(10−1)×5=4+45=49a10=4+(10−1)×5=4+45=49
  - So, the 10th term is 49.
2. Find the 7th term of the geometric sequence 3, 9, 27, 81,...
  - First term 𝑎=3a=3
  - Common ratio 𝑟=3r=3
  - Using the formula:𝑎7=3⋅3(7−1)=3⋅729=2187a7=3⋅3(7−1)=3⋅729=2187
  - So, the 7th term is 2187.
Real-Life Applications and Careers
Finance and Economics:
- Interest Calculations: Sequences help in calculating compound interest and analyzing financial growth.
- Investment Analysis: Financial analysts use sequences to model and predict market trends.
Computer Science:
- Algorithm Design: Understanding sequences is crucial for designing efficient algorithms, particularly in sorting and searching.
- Data Structures: Sequences are fundamental in understanding arrays, linked lists, and other data structures.
Engineering:
- Signal Processing: Engineers use sequences to analyze and process signals in various applications like telecommunications.
- Control Systems: Sequences are used in designing and analyzing control systems for stability and performance.
3.Series
A series is the sum of the terms of a sequence. If {𝑎𝑛}{an} is a sequence, then the series is expressed as:
𝑆=𝑎1+𝑎2+𝑎3+…+𝑎𝑛+…S=a1+a2+a3+…+an+…
where 𝑆S represents the sum of the sequence's terms.
Types of Series
1. Arithmetic Series
2. Geometric Series
3. Harmonic Series
4. Fibonacci Series
Arithmetic Series
Definition
An arithmetic series is the sum of the terms of an arithmetic sequence.
Sum of the First 𝑛n Terms
The sum of the first 𝑛n terms of an arithmetic series can be found using the formula:
𝑆𝑛=𝑛2×(2𝑎+(𝑛−1)𝑑)Sn=2n×(2a+(n−1)d)
where:
- 𝑆𝑛Sn is the sum of the first 𝑛n terms.
- 𝑎a is the first term.
- 𝑑d is the common difference.
- 𝑛n is the number of terms.
Alternatively, it can also be written as:
𝑆𝑛=𝑛2×(𝑎+𝑙)Sn=2n×(a+l)
where 𝑙l is the last term of the series.
Example
Consider the arithmetic sequence: 2, 5, 8, 11,...
- First term (𝑎a): 2
- Common difference (𝑑d): 3
- Number of terms (𝑛n): 4
To find the sum of the first 4 terms:
𝑆4=42×(2×2+(4−1)×3)=2×(4+9)=2×13=26S4=24×(2×2+(4−1)×3)=2×(4+9)=2×13=26
So, the sum of the first 4 terms is 26.
Geometric Series
Definition
A geometric series is the sum of the terms of a geometric sequence.
Sum of the First 𝑛n Terms
The sum of the first 𝑛n terms of a geometric series can be found using the formula:
𝑆𝑛=𝑎(1−𝑟𝑛)1−𝑟if 𝑟≠1Sn=a1−r(1−rn)ifr=1
where:
- 𝑆𝑛Sn is the sum of the first 𝑛n terms.
- 𝑎a is the first term.
- 𝑟r is the common ratio.
- 𝑛n is the number of terms.
Sum of an Infinite Geometric Series
If the absolute value of the common ratio 𝑟r is less than 1 (∣𝑟∣<1∣r∣<1), the sum of an infinite geometric series is given by:
𝑆=𝑎1−𝑟S=1−ra
Example
Consider the geometric sequence: 3, 6, 12, 24,...
- First term (𝑎a): 3
- Common ratio (𝑟r): 2
- Number of terms (𝑛n): 4
To find the sum of the first 4 terms:
𝑆4=3(1−24)1−2=3(1−16)−1=3×15=45S4=31−2(1−24)=3−1(1−16)=3×15=45
So, the sum of the first 4 terms is 45.
Harmonic Series
Definition
A harmonic series is the sum of the reciprocals of the positive integers:
𝐻𝑛=1+12+13+…+1𝑛Hn=1+21+31+…+n1
Properties
- The harmonic series diverges, meaning it does not converge to a finite limit as 𝑛n approaches infinity.
Fibonacci Series
Definition
The Fibonacci series is a sequence where each term is the sum of the two preceding ones, starting from 0 and 1:
0,1,1,2,3,5,8,13,…0,1,1,2,3,5,8,13,…
Example
The sum of the first 5 terms of the Fibonacci series is:
0+1+1+2+3=70+1+1+2+3=7
Real-Life Examples of Series
Arithmetic Series Example: Consider saving money every month. You save ₹100 in the first month, ₹150 in the second month, ₹200 in the third month, and so on. This forms an arithmetic series where each term increases by ₹50.
- First term (𝑎a): ₹100
- Common difference (𝑑d): ₹50
To find the total savings after 4 months:
𝑆4=42×(2×100+(4−1)×50)=2×(200+150)=2×350=700S4=24×(2×100+(4−1)×50)=2×(200+150)=2×350=700
So, the total savings after 4 months is ₹700.
Geometric Series Example: Consider a scenario where a population of bacteria triples every hour. Starting with 1 bacterium, the population after 1 hour, 2 hours, and 3 hours would be 3, 9, 27, and so on.
- First term (𝑎a): 1
- Common ratio (𝑟r): 3
To find the sum of the population after 4 hours:
𝑆4=1(1−34)1−3=1(1−81)−2=1×40=40S4=11−3(1−34)=1−2(1−81)=1×40=40
So, the population sum after 4 hours is 40.
Real-Life Applications and Careers
Finance and Economics:
- Interest Calculations: Series are used to calculate compound interest and annuities.
- Investment Analysis: Analysts use series to model and predict financial growth.
Engineering:
- Signal Processing: Engineers use series to analyze and process signals in telecommunications.
- Control Systems: Series are used in designing and analyzing control systems for stability and performance.
Computer Science:
- Algorithm Design: Series are crucial in analyzing the complexity and performance of algorithms.
- Data Compression: Series are used in encoding and decoding information efficiently.
4.Exercise Questions
Exercises 1 to 6: Write the first five terms
Exercise 1
𝑎𝑛=𝑛(𝑛+2)an=n(n+2)
- For 𝑛=1n=1:
  𝑎1=1(1+2)=1×3=3a1=1(1+2)=1×3=3
- For 𝑛=2n=2:
  𝑎2=2(2+2)=2×4=8a2=2(2+2)=2×4=8
- For 𝑛=3n=3:
  𝑎3=3(3+2)=3×5=15a3=3(3+2)=3×5=15
- For 𝑛=4n=4:
  𝑎4=4(4+2)=4×6=24a4=4(4+2)=4×6=24
- For 𝑛=5n=5:
  𝑎5=5(5+2)=5×7=35a5=5(5+2)=5×7=35
First five terms: 3,8,15,24,353,8,15,24,35
Exercise 2
𝑎𝑛=𝑛𝑛+1an=n+1n
- For 𝑛=1n=1:
  𝑎1=11+1=12a1=1+11=21
- For 𝑛=2n=2:
  𝑎2=22+1=23a2=2+12=32
- For 𝑛=3n=3:
  𝑎3=33+1=34a3=3+13=43
- For 𝑛=4n=4:
  𝑎4=44+1=45a4=4+14=54
- For 𝑛=5n=5:
  𝑎5=55+1=56a5=5+15=65
First five terms: 12,23,34,45,5621,32,43,54,65
Exercise 3
𝑎𝑛=2𝑛an=2n
- For 𝑛=1n=1:
  𝑎1=21=2a1=21=2
- For 𝑛=2n=2:
  𝑎2=22=4a2=22=4
- For 𝑛=3n=3:
  𝑎3=23=8a3=23=8
- For 𝑛=4n=4:
  𝑎4=24=16a4=24=16
- For 𝑛=5n=5:
  𝑎5=25=32a5=25=32
First five terms: 2,4,8,16,322,4,8,16,32
Exercise 4
𝑎𝑛=2𝑛−36an=62n−3
- For 𝑛=1n=1:
  𝑎1=2(1)−36=−16a1=62(1)−3=6−1
- For 𝑛=2n=2:
  𝑎2=2(2)−36=16a2=62(2)−3=61
- For 𝑛=3n=3:
  𝑎3=2(3)−36=36=12a3=62(3)−3=63=21
- For 𝑛=4n=4:
  𝑎4=2(4)−36=56a4=62(4)−3=65
- For 𝑛=5n=5:
  𝑎5=2(5)−36=76a5=62(5)−3=67
First five terms: −16,16,12,56,766−1,61,21,65,67
Exercise 5
𝑎𝑛=(−1)𝑛−1⋅5𝑛+1an=(−1)n−1⋅5n+1
- For 𝑛=1n=1:
  𝑎1=(−1)1−1⋅51+1=1⋅52=25a1=(−1)1−1⋅51+1=1⋅52=25
- For 𝑛=2n=2:
  𝑎2=(−1)2−1⋅52+1=−1⋅53=−125a2=(−1)2−1⋅52+1=−1⋅53=−125
- For 𝑛=3n=3:
  𝑎3=(−1)3−1⋅53+1=1⋅54=625a3=(−1)3−1⋅53+1=1⋅54=625
- For 𝑛=4n=4:
  𝑎4=(−1)4−1⋅54+1=−1⋅55=−3125a4=(−1)4−1⋅54+1=−1⋅55=−3125
- For 𝑛=5n=5:
  𝑎5=(−1)5−1⋅55+1=1⋅56=15625a5=(−1)5−1⋅55+1=1⋅56=15625
First five terms: 25,−125,625,−3125,1562525,−125,625,−3125,15625
Exercise 6
𝑎𝑛=𝑛2+54an=4n2+5
- For 𝑛=1n=1:
  𝑎1=12+54=64=32a1=412+5=46=23
- For 𝑛=2n=2:
  𝑎2=22+54=94a2=422+5=49
- For 𝑛=3n=3:
  𝑎3=32+54=144=72a3=432+5=414=27
- For 𝑛=4n=4:
  𝑎4=42+54=214a4=442+5=421
- For 𝑛=5n=5:
  𝑎5=52+54=304=152a5=452+5=430=215
First five terms: 32,94,72,214,15223,49,27,421,215
Exercises 7 to 10: Find the indicated terms
Exercise 7
𝑎𝑛=4𝑛−3an=4n−3
- For 𝑛=17n=17:
  𝑎17=4(17)−3=68−3=65a17=4(17)−3=68−3=65
- For 𝑛=24n=24:
  𝑎24=4(24)−3=96−3=93a24=4(24)−3=96−3=93
So, 𝑎17=65a17=65 and 𝑎24=93a24=93.
Exercise 8
𝑎𝑛=𝑛22𝑛⋅𝑎7an=2nn2⋅a7
To find 𝑎7a7:
𝑎7=7227=49128a7=2772=12849
So, 𝑎7=49128a7=12849.
Exercise 9
𝑎𝑛=(−1)𝑛−1⋅𝑛3an=(−1)n−1⋅n3
- For 𝑛=9n=9:𝑎9=(−1)9−1⋅93=(−1)8⋅729=729a9=(−1)9−1⋅93=(−1)8⋅729=729
So, 𝑎9=729a9=729.
Exercise 10
𝑎𝑛=𝑛(𝑛−2)𝑛+3⋅𝑎20an=n+3n(n−2)⋅a20
To find 𝑎10a10 using 𝑎20a20:
𝑎10=10(10−2)10+3⋅𝑎20=8013⋅𝑎20a10=10+310(10−2)⋅a20=1380⋅a20
So, 𝑎10=8013⋅𝑎20a10=1380⋅a20.
Exercises 11 to 14: Write the first five terms and obtain the corresponding series
Exercise 11
𝑎1=3a1=3 𝑎𝑛=3𝑎𝑛−1+2an=3an−1+2 for all 𝑛>1n>1
- 𝑎1=3a1=3
- 𝑎2=3𝑎1+2=3(3)+2=9+2=11a2=3a1+2=3(3)+2=9+2=11
- 𝑎3=3𝑎2+2=3(11)+2=33+2=35a3=3a2+2=3(11)+2=33+2=35
- 𝑎4=3𝑎3+2=3(35)+2=105+2=107a4=3a3+2=3(35)+2=105+2=107
- 𝑎5=3𝑎4+2=3(107)+2=321+2=323a5=3a4+2=3(107)+2=321+2=323
First five terms: 3,11,35,107,3233,11,35,107,323
Series: 3+11+35+107+323=4793+11+35+107+323=479
Exercise 12
𝑎1=−1a1=−1 𝑎𝑛=𝑎𝑛−1𝑛an=nan−1 for 𝑛≥2n≥2
- 𝑎1=−1a1=−1
- 𝑎2=𝑎12=−12=−12a2=2a1=2−1=−21
- 𝑎3=𝑎23=−123=−16a3=3a2=3−21=−61
- 𝑎4=𝑎34=−164=−124a4=4a3=4−61=−241
- 𝑎5=𝑎45=−1245=−1120a5=5a4=5−241=−1201
First five terms: −1,−12,−16,−124,−1120−1,−21,−61,−241,−1201
Series: −1−12−16−124−1120=−1.825−1−21−61−241−1201=−1.825
Exercise 13
𝑎1=2a1=2 𝑎2=2a2=2 𝑎𝑛=𝑎𝑛−1−1an=an−1−1 for 𝑛>2n>2
- 𝑎1=2a1=2
- 𝑎2=2a2=2
- 𝑎3=𝑎2−1=2−1=1a3=a2−1=2−1=1
- 𝑎4=𝑎3−1=1−1=0a4=a3−1=1−1=0
- 𝑎5=𝑎4−1=0−1=−1a5=a4−1=0−1=−1
First five terms: 2,2,1,0,−12,2,1,0,−1
Series: 2+2+1+0−1=42+2+1+0−1=4
Exercise 14
Fibonacci sequence: 𝑎1=1a1=1 𝑎2=1a2=1 𝑎𝑛=𝑎𝑛−1+𝑎𝑛−2an=an−1+an−2 for 𝑛>2n>2
- 𝑎1=1a1=1
- 𝑎2=1a2=1
- 𝑎3=𝑎2+𝑎1=1+1=2a3=a2+a1=1+1=2
- 𝑎4=𝑎3+𝑎2=2+1=3a4=a3+a2=2+1=3
- 𝑎5=𝑎4+𝑎3=3+2=5a5=a4+a3=3+2=5
First five terms: 1,1,2,3,51,1,2,3,5
To find 𝑎𝑛+1𝑎𝑛anan+1 for 𝑛=1,2,3,4,5n=1,2,3,4,5:
- 𝑎2𝑎1=11=1a1a2=11=1
- 𝑎3𝑎2=21=2a2a3=12=2
- 𝑎4𝑎3=32=1.5a3a4=23=1.5
- 𝑎5𝑎4=53=1.67a4a5=35=1.67
- 𝑎6𝑎5=85=1.6a5a6=58=1.6
So, 𝑎𝑛+1𝑎𝑛anan+1 for 𝑛=1,2,3,4,5n=1,2,3,4,5 are 1,2,1.5,1.67,1.61,2,1.5,1.67,1.6.
5.Geometric Progression (G.P.)
Definition
A Geometric Progression (G.P.) is a sequence of numbers in which the ratio of any two consecutive terms is always constant. This constant ratio is called the common ratio (r).
General Form
A geometric sequence can be written as: 𝑎,𝑎𝑟,𝑎𝑟2,𝑎𝑟3,…a,ar,ar2,ar3,… where:
- 𝑎a is the first term
- 𝑟r is the common ratio
- 𝑎𝑟𝑛arn represents the nth term
General Term of a G.P.
The general term (nth term) of a geometric progression can be found using the formula: 𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1 where:
- 𝑎𝑛an is the nth term
- 𝑎a is the first term
- 𝑟r is the common ratio
- 𝑛n is the term number
Example
Consider a G.P. with the first term 𝑎=3a=3 and common ratio 𝑟=2r=2.
To find the 5th term (𝑎5a5):
𝑎5=3⋅25−1=3⋅24=3⋅16=48a5=3⋅25−1=3⋅24=3⋅16=48
So, the 5th term is 48.
Sum to n Terms of a G.P.
The sum of the first 𝑛n terms of a geometric progression can be found using the formula: 𝑆𝑛=𝑎(1−𝑟𝑛)1−𝑟if 𝑟≠1Sn=a1−r(1−rn)ifr=1
If the common ratio 𝑟r is between -1 and 1 (∣𝑟∣<1∣r∣<1), the series can be summed to infinity: 𝑆=𝑎1−𝑟if ∣𝑟∣<1S=1−raif∣r∣<1
Example
Consider the same G.P. with the first term 𝑎=3a=3 and common ratio 𝑟=2r=2.
To find the sum of the first 5 terms (𝑆5S5):
𝑆5=3(1−25)1−2=3(1−32)−1=3−31−1=3×31=93S5=31−2(1−25)=3−1(1−32)=3−1−31=3×31=93
So, the sum of the first 5 terms is 93
6.Geometric Mean (G.M.)
Definition
The geometric mean (G.M.) of two numbers 𝑎a and 𝑏b is the square root of their product. More generally, the geometric mean of 𝑛n positive numbers 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an is the nth root of their product.
For two numbers 𝑎a and 𝑏b:
G.M.=𝑎𝑏G.M.=ab
For 𝑛n numbers 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an:
G.M.=𝑎1⋅𝑎2⋅…⋅𝑎𝑛𝑛G.M.=na1⋅a2⋅…⋅an
Properties of Geometric Mean
1. Positivity: The geometric mean of positive numbers is always positive.
2. Symmetry: The geometric mean is symmetric, meaning the order of the numbers does not matter.
3. Proportionality: The geometric mean of 𝑘k times each of the numbers 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an is 𝑘k times the geometric mean of 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an.
Calculation Examples
Example 1: Two Numbers
Consider two numbers 44 and 1616:
G.M.=4×16=64=8G.M.=4×16=64=8
Example 2: Multiple Numbers
Consider three numbers 2,8,2,8, and 3232:
G.M.=2×8×323=5123=8G.M.=32×8×32=3512=8
Applications of Geometric Mean
1. Finance: The geometric mean is used to calculate average rates of return over multiple periods. It accounts for the compounding effect.
2. Growth Rates: It is used to find the average growth rates of populations, investments, etc., over time.
3. Statistics: The geometric mean is used in various statistical measures, especially when dealing with ratios and percentages.
4. Geometry: In geometric constructions, the geometric mean can be used to find the length of the altitude in right-angled triangles.
Detailed Explanation
Geometric Mean of Two Numbers
Let's consider two positive numbers 𝑎a and 𝑏b. The geometric mean of these numbers is given by:
G.M.=𝑎𝑏G.M.=ab
Step-by-Step Calculation:
1. Multiply the Numbers: Multiply the two numbers together:
  Product=𝑎×𝑏Product=a×b
2. Take the Square Root: Take the square root of the product:
  G.M.=𝑎×𝑏G.M.=a×b
Example: For 𝑎=9a=9 and 𝑏=4b=4:
G.M.=9×4=36=6G.M.=9×4=36=6
Geometric Mean of Multiple Numbers
Let's consider 𝑛n positive numbers 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an. The geometric mean of these numbers is given by:
G.M.=𝑎1⋅𝑎2⋅…⋅𝑎𝑛𝑛G.M.=na1⋅a2⋅…⋅an
Step-by-Step Calculation:
1. Multiply All Numbers: Multiply all the numbers together:
  Product=𝑎1×𝑎2×…×𝑎𝑛Product=a1×a2×…×an
2. Take the nth Root: Take the nth root of the product:
  G.M.=𝑎1×𝑎2×…×𝑎𝑛𝑛G.M.=na1×a2×…×an
Example: For 𝑎1=2,𝑎2=4,a1=2,a2=4, and 𝑎3=8a3=8:
G.M.=2×4×83=643=4G.M.=32×4×8=364=4
Real-Life Applications of Geometric Mean
1. Finance:
  - Investment Returns: When calculating the average return on investment over multiple periods, the geometric mean is preferred over the arithmetic mean because it accounts for compounding.
  - Example: If an investment grows by 10% in the first year and 20% in the second year, the geometric mean provides a more accurate measure of the average growth rate.
2. Population Growth:
  - Example: If a population grows by 3% one year and 5% the next year, the geometric mean gives a better average growth rate, accounting for compounding effects.
3. Science and Engineering:
  - Example: In biology, the geometric mean is used to calculate the average growth rate of bacteria or other microorganisms over time.
4. Statistics:
  - Example: In statistical analyses, the geometric mean is used for datasets with exponential growth or data that are multiplicative in nature.
7.Relationship Between Arithmetic Mean (A.M.) and Geometric Mean (G.M.)
Definitions
1. Arithmetic Mean (A.M.): The arithmetic mean of two numbers 𝑎a and 𝑏b is given by:
  A.M.=𝑎+𝑏2A.M.=2a+b
2. Geometric Mean (G.M.): The geometric mean of two numbers 𝑎a and 𝑏b is given by:
  G.M.=𝑎𝑏G.M.=ab
Relationship Between A.M. and G.M.
For any two positive numbers 𝑎a and 𝑏b:
A.M.≥G.M.A.M.≥G.M.
with equality holding if and only if 𝑎=𝑏a=b.
This inequality is known as the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality).
Proof of AM-GM Inequality for Two Numbers
Consider two positive numbers 𝑎a and 𝑏b. According to the AM-GM Inequality, we need to show:
𝑎+𝑏2≥𝑎𝑏2a+b≥ab
1. Square Both Sides:
  (𝑎+𝑏2)2≥(𝑎𝑏)2(2a+b)2≥(ab)2
2. Simplify:
  (𝑎+𝑏)24≥𝑎𝑏4(a+b)2≥ab
3. Expand the Left Side:
  𝑎2+2𝑎𝑏+𝑏24≥𝑎𝑏4a2+2ab+b2≥ab
4. Multiply Both Sides by 4:
  𝑎2+2𝑎𝑏+𝑏2≥4𝑎𝑏a2+2ab+b2≥4ab
5. Simplify:
  𝑎2−2𝑎𝑏+𝑏2≥0a2−2ab+b2≥0
6. Factorize:
  (𝑎−𝑏)2≥0(a−b)2≥0
Since (𝑎−𝑏)2≥0(a−b)2≥0 for all real numbers 𝑎a and 𝑏b, the inequality holds true. The equality holds if and only if (𝑎−𝑏)2=0(a−b)2=0, which implies 𝑎=𝑏a=b.
Example
Consider two numbers 𝑎=4a=4 and 𝑏=16b=16:
- Arithmetic Mean:
  A.M.=4+162=202=10A.M.=24+16=220=10
- Geometric Mean:
  G.M.=4×16=64=8G.M.=4×16=64=8
Here, A.M.=10A.M.=10 and G.M.=8G.M.=8, so A.M.>G.M.A.M.>G.M..
Generalization to n Numbers
The AM-GM Inequality can be generalized to 𝑛n positive numbers 𝑎1,𝑎2,…,𝑎𝑛a1,a2,…,an:
𝑎1+𝑎2+…+𝑎𝑛𝑛≥𝑎1⋅𝑎2⋅…⋅𝑎𝑛𝑛na1+a2+…+an≥na1⋅a2⋅…⋅an
with equality if and only if 𝑎1=𝑎2=…=𝑎𝑛a1=a2=…=an.
Applications of AM-GM Inequality
1. Optimization Problems: Used in various optimization problems to find the minimum or maximum values.
2. Finance: Helps in understanding the average rates of return, especially in compound interest calculations.
3. Statistics: Useful in various statistical measures and analyses.
8.Exercise Questions
Question 1
Find the 20th and nth terms of the G.P. 52,54,58,…25,45,85,…
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
- First term 𝑎=52a=25
- Common ratio 𝑟=5452=54×25=12r=2545=45×52=21
To find the 20th term:
𝑎20=52⋅(12)20−1=52⋅(12)19=52⋅1524288=51048576a20=25⋅(21)20−1=25⋅(21)19=25⋅5242881=10485765
To find the nth term:
𝑎𝑛=52⋅(12)𝑛−1=52⋅12𝑛−1=52𝑛an=25⋅(21)n−1=25⋅2n−11=2n5
Question 2
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- 𝑎8=192a8=192
- 𝑟=2r=2
To find the first term 𝑎a:
192=𝑎⋅28−1=𝑎⋅27=𝑎⋅128 ⟹ 𝑎=192128=1.5192=a⋅28−1=a⋅27=a⋅128⟹a=128192=1.5
To find the 12th term:
𝑎12=𝑎⋅212−1=1.5⋅211=1.5⋅2048=3072a12=a⋅212−1=1.5⋅211=1.5⋅2048=3072
Question 3
The 5th, 8th, and 11th terms of a G.P. are 𝑝,𝑞,p,q, and 𝑠s, respectively. Show that 𝑞2=𝑝𝑠q2=ps.
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- 𝑎5=𝑝a5=p
- 𝑎8=𝑞a8=q
- 𝑎11=𝑠a11=s
Expressing in terms of 𝑎a and 𝑟r:
𝑝=𝑎⋅𝑟4p=a⋅r4𝑞=𝑎⋅𝑟7q=a⋅r7𝑠=𝑎⋅𝑟10s=a⋅r10
To show 𝑞2=𝑝𝑠q2=ps:
𝑞2=(𝑎⋅𝑟7)2=𝑎2⋅𝑟14q2=(a⋅r7)2=a2⋅r14𝑝𝑠=(𝑎⋅𝑟4)(𝑎⋅𝑟10)=𝑎2⋅𝑟14ps=(a⋅r4)(a⋅r10)=a2⋅r14
Therefore, 𝑞2=𝑝𝑠q2=ps.
Question 4
The 4th term of a G.P. is the square of its second term, and the first term is -3. Determine its 7th term.
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- 𝑎4=(𝑎2)2a4=(a2)2
- 𝑎=−3a=−3
Expressing in terms of 𝑎a and 𝑟r:
𝑎4=𝑎⋅𝑟3a4=a⋅r3𝑎2=𝑎⋅𝑟a2=a⋅r
Given 𝑎4=(𝑎2)2a4=(a2)2:
𝑎⋅𝑟3=(𝑎⋅𝑟)2 ⟹ 𝑎⋅𝑟3=𝑎2⋅𝑟2 ⟹ 𝑟3=𝑎⋅𝑟2 ⟹ 𝑟=𝑎=−3a⋅r3=(a⋅r)2⟹a⋅r3=a2⋅r2⟹r3=a⋅r2⟹r=a=−3
Since 𝑟=−3r=−3:
𝑎7=𝑎⋅𝑟6=−3⋅(−3)6=−3⋅729=−2187a7=a⋅r6=−3⋅(−3)6=−3⋅729=−2187
Question 5
Which term of the following sequences is 128128 and 729729?
(a) 2,22,4,…2,22,4,… is 128128?
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- First term 𝑎=2a=2
- Common ratio 𝑟=2r=2
To find 𝑛n such that 𝑎𝑛=128an=128:
2⋅(2)𝑛−1=128 ⟹ (2)𝑛−1=64 ⟹ (21/2)𝑛−1=26 ⟹ 2(𝑛−1)/2=26 ⟹ 𝑛−12=6 ⟹ 𝑛−1=12 ⟹ 𝑛=132⋅(2)n−1=128⟹(2)n−1=64⟹(21/2)n−1=26⟹2(n−1)/2=26⟹2n−1=6⟹n−1=12⟹n=13
So, the 13th term is 128128.
(b) 3,3,33,…3,3,33,… is 729729?
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- First term 𝑎=3a=3
- Common ratio 𝑟=3r=3
To find 𝑛n such that 𝑎𝑛=729an=729:
3⋅(3)𝑛−1=729 ⟹ (3)𝑛=729 ⟹ (31/2)𝑛=36 ⟹ 3𝑛/2=36 ⟹ 𝑛2=6 ⟹ 𝑛=123⋅(3)n−1=729⟹(3)n=729⟹(31/2)n=36⟹3n/2=36⟹2n=6⟹n=12
So, the 12th term is 729729.
9.Exercise Questions
Question 01
Show that the products of the corresponding terms of the sequences 𝑎,𝑎𝑟,𝑎𝑟2,…,𝑎𝑟𝑛−1a,ar,ar2,…,arn−1 and 𝐴,𝐴𝑅,𝐴𝑅2,…,𝐴𝑅𝑛−1A,AR,AR2,…,ARn−1 form a G.P., and find the common ratio.
Solution:
Consider two geometric progressions:
- Sequence 1: 𝑎,𝑎𝑟,𝑎𝑟2,…,𝑎𝑟𝑛−1a,ar,ar2,…,arn−1
- Sequence 2: 𝐴,𝐴𝑅,𝐴𝑅2,…,𝐴𝑅𝑛−1A,AR,AR2,…,ARn−1
We need to find the product of the corresponding terms of these sequences. The product sequence will be:
𝑎𝐴,𝑎𝑟⋅𝐴𝑅,𝑎𝑟2⋅𝐴𝑅2,…,𝑎𝑟𝑛−1⋅𝐴𝑅𝑛−1aA,ar⋅AR,ar2⋅AR2,…,arn−1⋅ARn−1
Let's simplify these terms:
- First term: 𝑎𝐴aA
- Second term: 𝑎𝑟⋅𝐴𝑅=𝑎𝐴⋅𝑟𝑅ar⋅AR=aA⋅rR
- Third term: 𝑎𝑟2⋅𝐴𝑅2=𝑎𝐴⋅(𝑟𝑅)2ar2⋅AR2=aA⋅(rR)2
- nth term: 𝑎𝑟𝑛−1⋅𝐴𝑅𝑛−1=𝑎𝐴⋅(𝑟𝑅)𝑛−1arn−1⋅ARn−1=aA⋅(rR)n−1
This sequence is:
𝑎𝐴,𝑎𝐴(𝑟𝑅),𝑎𝐴(𝑟𝑅)2,…,𝑎𝐴(𝑟𝑅)𝑛−1aA,aA(rR),aA(rR)2,…,aA(rR)n−1
This is clearly a geometric progression with the first term 𝑎𝐴aA and the common ratio 𝑟𝑅rR.
Question 02
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution:
Let the four numbers in the geometric progression be 𝑎,𝑎𝑟,𝑎𝑟2,𝑎𝑟3a,ar,ar2,ar3.
Given:
- Third term is greater than the first term by 9:
  𝑎𝑟2=𝑎+9ar2=a+9
- Second term is greater than the fourth term by 18:
  𝑎𝑟=𝑎𝑟3+18ar=ar3+18
From the first condition:
𝑎𝑟2=𝑎+9 ⟹ 𝑎𝑟2−𝑎=9 ⟹ 𝑎(𝑟2−1)=9 ⟹ 𝑎=9𝑟2−1(i)ar2=a+9⟹ar2−a=9⟹a(r2−1)=9⟹a=r2−19(i)
From the second condition:
𝑎𝑟=𝑎𝑟3+18 ⟹ 𝑎𝑟(1−𝑟2)=18 ⟹ 𝑎=18𝑟(1−𝑟2)(ii)ar=ar3+18⟹ar(1−r2)=18⟹a=r(1−r2)18(ii)
Equating (i) and (ii):
9𝑟2−1=18𝑟(1−𝑟2)r2−19=r(1−r2)18
Simplify:
9𝑟(1−𝑟2)=18(𝑟2−1)9r(1−r2)=18(r2−1)9𝑟−9𝑟3=18𝑟2−189r−9r3=18r2−189𝑟3+18𝑟2−9𝑟−18=09r3+18r2−9r−18=0𝑟3+2𝑟2−𝑟−2=0r3+2r2−r−2=0
Factoring:
(𝑟−1)(𝑟2+3𝑟+2)=0(r−1)(r2+3r+2)=0
Solving:
𝑟−1=0 ⟹ 𝑟=1(Not valid, as the numbers are distinct)r−1=0⟹r=1(Not valid, as the numbers are distinct)𝑟2+3𝑟+2=0 ⟹ 𝑟=−1or𝑟=−2r2+3r+2=0⟹r=−1orr=−2
If 𝑟=−2r=−2:
𝑎=9(−2)2−1=94−1=3a=(−2)2−19=4−19=3
So, the four numbers are 3,−6,12,−243,−6,12,−24.
Question 03
If the 𝑝pth, 𝑞qth, and 𝑟rth terms of a G.P. are 𝑎,𝑏,a,b, and 𝑐c, respectively. Prove that 𝑎𝑞−𝑟⋅𝑏𝑟−𝑝⋅𝑐𝑝−𝑞=1aq−r⋅br−p⋅cp−q=1.
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- 𝑎𝑝=𝑎=𝑎⋅𝑟𝑝−1ap=a=a⋅rp−1
- 𝑎𝑞=𝑏=𝑎⋅𝑟𝑞−1aq=b=a⋅rq−1
- 𝑎𝑟=𝑐=𝑎⋅𝑟𝑟−1ar=c=a⋅rr−1
We need to prove:
𝑎𝑞−𝑟⋅𝑏𝑟−𝑝⋅𝑐𝑝−𝑞=1aq−r⋅br−p⋅cp−q=1
Substitute the values:
(𝑎⋅𝑟𝑝−1)𝑞−𝑟⋅(𝑎⋅𝑟𝑞−1)𝑟−𝑝⋅(𝑎⋅𝑟𝑟−1)𝑝−𝑞(a⋅rp−1)q−r⋅(a⋅rq−1)r−p⋅(a⋅rr−1)p−q
Simplify:
𝑎𝑞−𝑟⋅𝑟(𝑝−1)(𝑞−𝑟)⋅𝑎𝑟−𝑝⋅𝑟(𝑞−1)(𝑟−𝑝)⋅𝑎𝑝−𝑞⋅𝑟(𝑟−1)(𝑝−𝑞)aq−r⋅r(p−1)(q−r)⋅ar−p⋅r(q−1)(r−p)⋅ap−q⋅r(r−1)(p−q)
Combine the exponents:
𝑎𝑞−𝑟+𝑟−𝑝+𝑝−𝑞⋅𝑟(𝑝−1)(𝑞−𝑟)+(𝑞−1)(𝑟−𝑝)+(𝑟−1)(𝑝−𝑞)aq−r+r−p+p−q⋅r(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)
Simplify the exponents of 𝑎a:
𝑎0⋅𝑟(𝑝−1)(𝑞−𝑟)+(𝑞−1)(𝑟−𝑝)+(𝑟−1)(𝑝−𝑞)a0⋅r(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)
Simplify the exponents of 𝑟r:
(𝑝−1)(𝑞−𝑟)+(𝑞−1)(𝑟−𝑝)+(𝑟−1)(𝑝−𝑞)=0(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)=0
Thus:
𝑎0⋅𝑟0=1a0⋅r0=1
Hence proved:
𝑎𝑞−𝑟⋅𝑏𝑟−𝑝⋅𝑐𝑝−𝑞=1aq−r⋅br−p⋅cp−q=1
Question 04
If the first and the nth term of a G.P. are 𝑎a and 𝑏b, respectively, and if 𝑃P is the product of 𝑛n terms, prove that 𝑃2=(𝑎𝑏)𝑛P2=(ab)n.
Solution:
The general term of a G.P. is given by:
𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1
Given:
- First term 𝑎1=𝑎a1=a
- nth term 𝑎𝑛=𝑏an=b
Since 𝑎𝑛=𝑎⋅𝑟𝑛−1an=a⋅rn−1:
𝑏=𝑎⋅𝑟𝑛−1 ⟹ 𝑟𝑛−1=𝑏𝑎 ⟹ 𝑟=(𝑏𝑎)1𝑛−1b=a⋅rn−1⟹rn−1=ab⟹r=(ab)n−11
The product of the first 𝑛n terms:
𝑃=𝑎⋅𝑎𝑟⋅𝑎𝑟2⋅…⋅𝑎𝑟𝑛−1P=a⋅ar⋅ar2⋅…⋅arn−1𝑃=𝑎𝑛⋅𝑟0+1+2+…+(𝑛−1)P=an⋅r0+1+2+…+(n−1)
The sum of the exponents of 𝑟r is:
0+1+2+…+(𝑛−1)=(𝑛−1)𝑛20+1+2+…+(n−1)=2(n−1)n
Thus:
𝑃=𝑎𝑛⋅𝑟(𝑛−1)𝑛2P=an⋅r2(n−1)n
Since 𝑟=(𝑏𝑎)1𝑛−1r=(ab)n−11:
𝑃=𝑎𝑛⋅((𝑏𝑎)1𝑛−1)(𝑛−1)𝑛2=𝑎𝑛⋅(𝑏𝑎)𝑛2P=an⋅((ab)n−11)2(n−1)n=an⋅(ab)2n
Simplify:
𝑃=𝑎𝑛⋅𝑏𝑛2𝑎𝑛2=𝑎𝑛−𝑛2⋅𝑏𝑛2=𝑎𝑛2⋅𝑏𝑛2P=an⋅a2nb2n=an−2n⋅b2n=a2n⋅b2n𝑃=(𝑎⋅𝑏)𝑛2P=(a⋅b)2n
Square both sides:
𝑃2=(𝑎𝑏)𝑛P2=(ab)n
Hence proved:
𝑃2=(𝑎𝑏)𝑛P2=(ab)n
10.Exercise Questions
Question 01
If 𝐴A and 𝐺G be A.M. and G.M., respectively, between two positive numbers, prove that the numbers are 𝐴±(𝐴+𝐺)(𝐴−𝐺)A±(A+G)(A−G).
Solution:
Let the two positive numbers be 𝑎a and 𝑏b.
The arithmetic mean (A.M.) 𝐴A between 𝑎a and 𝑏b is:
𝐴=𝑎+𝑏2A=2a+b
The geometric mean (G.M.) 𝐺G between 𝑎a and 𝑏b is:
𝐺=𝑎𝑏G=ab
We need to prove that:
𝑎,𝑏=𝐴±(𝐴+𝐺)(𝐴−𝐺)a,b=A±(A+G)(A−G)
Step-by-Step Proof:
1. Express 𝑎a and 𝑏b in terms of 𝐴A and 𝐺G:
  From the definition of A.M.:
  𝐴=𝑎+𝑏2 ⟹ 𝑎+𝑏=2𝐴A=2a+b⟹a+b=2A
  From the definition of G.M.:
  𝐺=𝑎𝑏 ⟹ 𝐺2=𝑎𝑏G=ab⟹G2=ab
2. Form a quadratic equation using 𝑎a and 𝑏b:
  Consider the quadratic equation with roots 𝑎a and 𝑏b:
  𝑥2−(𝑎+𝑏)𝑥+𝑎𝑏=0x2−(a+b)x+ab=0
  Substitute 𝑎+𝑏=2𝐴a+b=2A and 𝑎𝑏=𝐺2ab=G2:
  𝑥2−2𝐴𝑥+𝐺2=0x2−2Ax+G2=0
3. Solve the quadratic equation:
  The roots of the quadratic equation 𝑥2−2𝐴𝑥+𝐺2=0x2−2Ax+G2=0 are given by:
  𝑥=2𝐴±(2𝐴)2−4𝐺22=2𝐴±4𝐴2−4𝐺22=2𝐴±2𝐴2−𝐺22=𝐴±𝐴2−𝐺2x=22A±(2A)2−4G2=22A±4A2−4G2=22A±2A2−G2=A±A2−G2
4. Rewrite 𝐴2−𝐺2A2−G2:
  Note that:
  𝐴2−𝐺2=(𝐴+𝐺)(𝐴−𝐺)A2−G2=(A+G)(A−G)
5. Final form of the numbers:
  Thus, the two numbers 𝑎a and 𝑏b can be written as:
  𝑎,𝑏=𝐴±(𝐴+𝐺)(𝐴−𝐺)a,b=A±(A+G)(A−G)
Hence proved.
Question 02
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour, and nth hour?
Solution:
The bacteria count doubles every hour, which means this follows a geometric progression (G.P.) with the first term 𝑎=30a=30 and common ratio 𝑟=2r=2.
The general term for the number of bacteria after 𝑛n hours is given by:
𝑎𝑛=𝑎⋅𝑟𝑛an=a⋅rn
For the 2nd hour:
𝑎2=30⋅22=30⋅4=120a2=30⋅22=30⋅4=120
For the 4th hour:
𝑎4=30⋅24=30⋅16=480a4=30⋅24=30⋅16=480
For the nth hour:
𝑎𝑛=30⋅2𝑛an=30⋅2n
So, the number of bacteria at the end of the 2nd hour, 4th hour, and nth hour are 120, 480, and 30⋅2𝑛30⋅2n respectively.

Sequences and Series — Class 11 Maths Notes

Topics covered in Sequences and Series

1.Introduction of Sequences and Series

Sequences

Definition

Types of Sequences

Series

Types of Series

Detailed Real-Life Example

Real-Life Applications and Careers

Detailed Step-by-Step Explanation of a Geometric Sequence

2.Sequences

Definition of a Sequence

Notation

Types of Sequences

Properties of Sequences

Real-Life Examples of Sequences

Practice Problems

Real-Life Applications and Careers

3.Series

Types of Series

Arithmetic Series

Definition

Sum of the First 𝑛n Terms

Example

Geometric Series

Definition

Sum of the First 𝑛n Terms

Sum of an Infinite Geometric Series

Example

Harmonic Series

Definition

Properties

Fibonacci Series

Definition

Example

Real-Life Examples of Series

Real-Life Applications and Careers

4.Exercise Questions

Exercises 1 to 6: Write the first five terms

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercises 7 to 10: Find the indicated terms

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Exercises 11 to 14: Write the first five terms and obtain the corresponding series

Exercise 11

Exercise 12

Exercise 13

Exercise 14

5.Geometric Progression (G.P.)

Definition

General Form

General Term of a G.P.

Example

Sum to n Terms of a G.P.

Example

6.Geometric Mean (G.M.)

Definition

Properties of Geometric Mean

Calculation Examples

Example 1: Two Numbers

Example 2: Multiple Numbers

Applications of Geometric Mean

Detailed Explanation

Geometric Mean of Two Numbers

Geometric Mean of Multiple Numbers

Real-Life Applications of Geometric Mean

7.Relationship Between Arithmetic Mean (A.M.) and Geometric Mean (G.M.)

Definitions

Relationship Between A.M. and G.M.

Proof of AM-GM Inequality for Two Numbers

Example

Generalization to n Numbers