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6th | 7th | 8th | 9th | 10th | 11th | 12th |
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Sequence | Series | Arithmeric Progression |
Arithmetic Mean | Geometric Progression | Geometric Mean |
Relationship Between AM and GM | Some Special Series |
Chapter 9 Sequences and Series (Concepts)
Embark on a structured exploration of ordered patterns and their summations in this chapter dedicated to Sequences and Series. While sets deal with unordered collections, a sequence introduces the crucial concept of order, representing a list of numbers arranged according to a specific rule or pattern. We distinguish sequences from sets and classify them as finite (having a specific last term) or infinite (continuing indefinitely). When we add the terms of a sequence together, we obtain a series. This chapter delves into two fundamental types of sequences – Arithmetic and Geometric Progressions – and develops powerful tools for analyzing their terms and calculating the sums of their corresponding series.
We begin by revisiting and significantly extending our understanding of Arithmetic Progression (AP). Recall that an AP is a sequence where each term (after the first) is obtained by adding a fixed constant, the common difference ($d$), to the preceding term. Denoting the first term by $a$, the general form is $a, a+d, a+2d, \dots$. We reinforce the key formulas:
- The $\mathbf{n^{th}}$ term (general term), $a_n$, is given by $\mathbf{a_n = a + (n-1)d}$.
- The sum of the first $n$ terms, $S_n$, has two useful forms:
- $\mathbf{S_n = \frac{n}{2} [2a + (n-1)d]}$
- $\mathbf{S_n = \frac{n}{2} [a + l]}$, where $l$ is the last ($n^{th}$) term ($l=a_n$).
The major focus then shifts to a different type of progression governed by multiplication: the Geometric Progression (GP). A sequence is a GP if each term (after the first) is obtained by multiplying the preceding term by a fixed, non-zero constant called the common ratio ($r$). With the first term as $a$, the general form is $a, ar, ar^2, ar^3, \dots$. Key formulas for GPs are:
- The $\mathbf{n^{th}}$ term, $a_n$, is given by $\mathbf{a_n = ar^{n-1}}$.
- The sum of the first $n$ terms, $S_n$, (for $r \neq 1$) is: $$ \mathbf{S_n = \frac{a(r^n - 1)}{r - 1}} \quad \text{or equivalently} \quad \mathbf{S_n = \frac{a(1 - r^n)}{1 - r}} $$
- For an infinite GP, if the magnitude of the common ratio is less than 1 (i.e., $\mathbf{|r| < 1}$), the sum converges to a finite value given by: $$ \mathbf{S_\infty = \frac{a}{1 - r}} $$
An important relationship between the Arithmetic Mean and Geometric Mean of two positive numbers $a$ and $b$ is established: the AM is always greater than or equal to the GM, i.e., $\mathbf{\frac{a+b}{2} \ge \sqrt{ab}}$. Equality holds if and only if $a=b$.
Finally, the chapter expands beyond APs and GPs to explore methods for finding the sum of certain special series. We derive and utilize formulas for the sums of powers of the first $n$ natural numbers:
- Sum of first $n$ natural numbers: $\mathbf{\sum\limits\limits_{k=1}^{n} k = 1 + 2 + \dots + n = \frac{n(n+1)}{2}}$
- Sum of squares of first $n$ natural numbers: $\mathbf{\sum\limits\limits_{k=1}^{n} k^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}}$
- Sum of cubes of first $n$ natural numbers: $\mathbf{\sum\limits\limits_{k=1}^{n} k^3 = 1^3 + 2^3 + \dots + n^3 = \left[ \frac{n(n+1)}{2} \right]^2 = (\sum\limits\limits_{k=1}^{n} k)^2}$
Sequence
In mathematics, a sequence is an ordered list of objects, typically numbers. The order in which the objects appear is crucial, and the same object can appear multiple times at different positions in the sequence. Formally, a sequence can be thought of as a function whose domain is a set of consecutive integers, usually the set of natural numbers.
Definition of a Sequence
A sequence is a function whose domain is the set of natural numbers ($\mathbb{N} = \{1, 2, 3, \dots\}$) or a finite subset of natural numbers of the form $\{1, 2, 3, \dots, k\}$. The range of the function, which consists of the output values, forms the elements of the sequence, known as its terms.
- An infinite sequence has the set of all natural numbers $\mathbb{N}$ as its domain.
- A finite sequence has a domain of the form $\{1, 2, 3, \dots, k\}$ for some positive integer $k$.
The terms of a sequence are denoted by a letter with a subscript indicating the position of the term. For example, $a_1$ is the first term, $a_2$ is the second term, and $a_n$ is the $n^{th}$ term. The entire sequence is often represented as $\{a_n\}$, $\langle a_n \rangle$, or by listing its terms: $a_1, a_2, a_3, \dots$.
Types and Examples of Sequences
Real Sequences
A sequence whose terms are all real numbers is called a real sequence. Most sequences encountered in introductory mathematics are real sequences.
Examples:
- Sequence of odd positive integers: $1, 3, 5, 7, \dots$
The general term is given by the formula $a_n = 2n-1$. This is an infinite sequence.
- Sequence of squares of the first 5 natural numbers: $1, 4, 9, 16, 25$.
The general term is $a_n = n^2$ for the domain $\{1, 2, 3, 4, 5\}$. This is a finite sequence.
- Alternating sequence: $1, -1, 1, -1, \dots$
The general term is $a_n = (-1)^{n-1}$. This is an infinite sequence.
- Sequence of reciprocals: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$
The general term is $a_n = \frac{1}{n}$. This is an infinite sequence.
Methods for Defining Sequences
There are several ways to define or specify a sequence:
1. Explicit Formula for the $n^{th}$ Term:
This method provides a rule or formula, $a_n = f(n)$, that can directly calculate any term of the sequence simply by plugging in its position number $n$.
- For $a_n = n^2 + 1$, the sequence is $a_1 = 1^2+1=2$, $a_2 = 2^2+1=5$, $a_3 = 3^2+1=10, \dots$. The sequence is $2, 5, 10, \dots$.
2. Recurrence Relation:
This method defines a term based on one or more preceding terms. An initial value (or values) must be provided to start the sequence.
- An Arithmetic Progression is defined by $a_n = a_{n-1} + d$, where $d$ is a constant. Given $a_1=5$ and $d=3$, the sequence is $5, 8, 11, 14, \dots$.
- A Geometric Progression is defined by $a_n = r \cdot a_{n-1}$, where $r$ is a constant. Given $a_1=2$ and $r=3$, the sequence is $2, 6, 18, 54, \dots$.
- The Fibonacci sequence is defined by $F_n = F_{n-1} + F_{n-2}$ with $F_1=1$ and $F_2=1$. The sequence is $1, 1, 2, 3, 5, 8, \dots$.
3. Description in Words:
Some sequences are best described by a property their terms must satisfy, for which a simple formula may not exist.
- The sequence of prime numbers: $2, 3, 5, 7, 11, 13, \dots$.
- The sequence of digits in the decimal expansion of $\pi$: $3, 1, 4, 1, 5, 9, \dots$.
Example 1. Write the first 4 terms of the sequence defined by $a_n = 2n - 3$.
Answer:
We are given the explicit formula for the $n^{th}$ term as $a_n = 2n - 3$.
To find the first four terms, we substitute $n = 1, 2, 3,$ and $4$ into the formula:
For $n=1$: $a_1 = 2(1) - 3 = 2 - 3 = -1$.
For $n=2$: $a_2 = 2(2) - 3 = 4 - 3 = 1$.
For $n=3$: $a_3 = 2(3) - 3 = 6 - 3 = 3$.
For $n=4$: $a_4 = 2(4) - 3 = 8 - 3 = 5$.
The first four terms of the sequence are -1, 1, 3, 5.
Example 2. Write the first 5 terms of the sequence defined by $a_1 = 3$ and $a_n = 3a_{n-1} + 2$ for all $n > 1$.
Answer:
We are given the first term $a_1 = 3$ and the recurrence relation $a_n = 3a_{n-1} + 2$.
We use the given first term to find the subsequent terms:
The first term is given: $a_1 = 3$.
For $n=2$:
$a_2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11$.
For $n=3$:
$a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35$.
For $n=4$:
$a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107$.
For $n=5$:
$a_5 = 3a_{5-1} + 2 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323$.
The first five terms of the sequence are 3, 11, 35, 107, 323.
Series
The concept of a series is a natural and powerful extension of the concept of a sequence. While a sequence is an ordered list of terms, a series represents the process of adding those terms together. Think of a sequence as a list of individual payments you need to make, whereas the series represents the total amount paid after a certain number of payments.
Definition of a Series
For any given sequence, whether finite or infinite, we can form a corresponding series.
Let $a_1, a_2, a_3, \dots, a_n, \dots$ be a sequence. The expression formed by adding all the terms of this sequence, $a_1 + a_2 + a_3 + \dots + a_n + \dots$, is called a series.
- A finite series is the sum of the terms of a finite sequence. If the sequence has $n$ terms, the series is $a_1 + a_2 + \dots + a_n$. The sum of a finite series is always a specific, calculable value.
- An infinite series is the sum of the terms of an infinite sequence, written as $a_1 + a_2 + a_3 + \dots$. The idea of "summing" an infinite number of terms is more complex. In higher mathematics, we study whether such a sum approaches a finite limit (it is convergent) or grows without bound (it is divergent).
The Language of Series: Sigma Notation
Writing out long sums can be tedious. To represent series in a compact and precise way, mathematicians use summation notation, which features the Greek capital letter Sigma ($\Sigma$).
The sum of the first $n$ terms of a sequence $\{a_k\}$, which is called the $n^{th}$ partial sum and is denoted by $S_n$, is written as:
$\mathbf{S_n = \sum\limits_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \dots + a_n}$
... (i)
Let's break down this powerful notation:
- $\Sigma$: This is the summation symbol. It is an instruction to sum the terms that follow.
- $k$: This is the index of summation. It's a placeholder variable that takes on integer values.
- $k=1$: This is the lower limit of the summation. It tells us the starting value for the index $k$.
- $n$: This is the upper limit of the summation. It tells us the final value for the index $k$. The summation includes all integer values from the lower limit to the upper limit.
- $a_k$: This is the general term or the formula for the terms being added. For each value of $k$ from 1 to $n$, we calculate $a_k$ and add it to our sum.
Examples of Series
- The sum of the first 100 positive integers: $1+2+3+\dots+100$. The general term is $a_k=k$. This can be written as $\sum\limits_{k=1}^{100} k$.
- The sum of the first 5 terms of the sequence of squares: $1+4+9+16+25$. The general term is $a_k = k^2$. This is written as $\sum\limits_{k=1}^{5} k^2$.
- The infinite geometric series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$. The general term is $a_k = (\frac{1}{2})^{k-1}$. This is written as $\sum\limits_{k=1}^{\infty} (\frac{1}{2})^{k-1}$.
Example 1. A series is defined by $\sum\limits_{k=1}^{\infty} \frac{1}{k^2}$. What are the first three terms of this series?
Answer:
The notation $\sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ represents the infinite sum $a_1 + a_2 + a_3 + \dots$, where the general term is $a_k = \frac{1}{k^2}$.
The question asks for the first three terms of the series, which are the first three numbers being added together ($a_1, a_2,$ and $a_3$).
We find these by substituting $k=1, 2,$ and $3$ into the formula for the general term:
For $k=1$: The first term is $a_1 = \frac{1}{1^2} = 1$.
For $k=2$: The second term is $a_2 = \frac{1}{2^2} = \frac{1}{4}$.
For $k=3$: The third term is $a_3 = \frac{1}{3^2} = \frac{1}{9}$.
Thus, the series begins as $1 + \frac{1}{4} + \frac{1}{9} + \dots$.
The first three terms are 1, $\frac{1}{4}$, $\frac{1}{9}$.
Example 2. Write the series $2 + 4 + 6 + \dots + 20$ in sigma notation.
Answer:
To express a series in sigma notation, $\sum\limits_{k=1}^{n} a_k$, we must identify two key components: the formula for the general term ($a_k$) and the number of terms being added ($n$).
Step 1: Find the general term ($a_k$).
The terms of the series are $2, 4, 6, \dots$. We can recognize a pattern: each term is 2 multiplied by its position number.
- 1st term: $2 = 2 \times 1$
- 2nd term: $4 = 2 \times 2$
- 3rd term: $6 = 2 \times 3$
This suggests that the $k^{th}$ term, $a_k$, is given by the formula $a_k = 2k$.
Step 2: Find the upper limit of the summation ($n$).
The last term in the series is 20. We use our general term formula to find which position this term occupies.
Set $a_k = 20 \implies 2k = 20$
$k = 10$
Since the last term is the 10th term, there are 10 terms in total, so our upper limit is $n=10$.
Step 3: Write the series in sigma notation.
With the general term $a_k = 2k$, a starting index of $k=1$, and an ending index of $n=10$, the series is written as:
$\sum\limits_{k=1}^{10} 2k$
Arithmetic Progression
An Arithmetic Progression (AP), also known as an arithmetic sequence, is one of the simplest and most important types of sequences in mathematics. It is defined by the property that the difference between any two consecutive terms is always the same. Think of it as a sequence where you take "constant steps" from one term to the next.
Definition and Common Difference
A sequence $a_1, a_2, a_3, \dots$ is called an Arithmetic Progression if there exists a constant number $d$, called the common difference, such that each term is obtained by adding $d$ to the previous term.
Mathematically, this can be expressed as:
$a_{n+1} = a_n + d \quad \text{or} \quad a_{n+1} - a_n = d$
The common difference $d$ can be positive, negative, or zero.
- If $d > 0$, the AP is increasing (e.g., $2, 5, 8, 11, \dots$ where $d=3$).
- If $d < 0$, the AP is decreasing (e.g., $10, 8, 6, 4, \dots$ where $d=-2$).
- If $d = 0$, the AP is constant (e.g., $7, 7, 7, 7, \dots$ where $d=0$).
The General Term ($n^{th}$ Term) of an AP
We often need a way to find any term in an AP without having to list all the terms before it. This is done using the formula for the general term, or the $n^{th}$ term.
Let the first term be $a$ and the common difference be $d$. Let's observe the pattern:
First term: $a_1 = a = a + (1-1)d$
Second term: $a_2 = a_1 + d = a + d = a + (2-1)d$
Third term: $a_3 = a_2 + d = (a + d) + d = a + 2d = a + (3-1)d$
Fourth term: $a_4 = a_3 + d = (a + 2d) + d = a + 3d = a + (4-1)d$
The pattern is clear: the coefficient of $d$ is always one less than the term number. Therefore, the formula for the $n^{th}$ term of an AP is:
$a_n = a + (n-1)d$
... (i)
Here, $a_n$ is the term we want to find, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.
Example 1. Find the 15th term of the AP: 3, 7, 11, 15, ...
Answer:
First, identify the components of the AP.
The first term is $a = 3$.
The common difference is $d = 7 - 3 = 4$.
We need to find the 15th term, so $n=15$.
Using the formula for the $n^{th}$ term, $a_n = a + (n-1)d$:
$a_{15} = 3 + (15 - 1) \times 4$
$= 3 + (14 \times 4)$
$= 3 + 56 = 59$
The 15th term of the AP is 59.
Sum of the First $n$ Terms of an AP
Let $S_n$ denote the sum of the first $n$ terms of an AP with first term $a$ and common difference $d$.
Derivation of the Formula:
We can write the sum in two ways:
$S_n = a + (a+d) + (a+2d) + \dots + (a+(n-1)d)$
... (a)
And in reverse order (the last term is $a_n = a+(n-1)d$):
$S_n = (a+(n-1)d) + (a+(n-2)d) + \dots + a$
... (b)
Adding the two equations term by term:
$2S_n = [a + (a+(n-1)d)] + [(a+d) + (a+(n-2)d)] + \dots$
Each pair of terms sums to the same value: $2a + (n-1)d$. Since there are $n$ such pairs, we have:
$2S_n = n \times [2a + (n-1)d]$
Dividing by 2 gives the primary formula for the sum:
$S_n = \frac{n}{2}[2a + (n-1)d]$
... (ii)
We can also write this formula in another useful way. Since $a_n = a + (n-1)d$, we can substitute this into the sum formula:
$S_n = \frac{n}{2}[a + a + (n-1)d] = \frac{n}{2}[a + a_n]$
$S_n = \frac{n}{2}(a + a_n)$
... (iii)
This second form is very useful when you know the first term, the last term, and the number of terms.
Example 2. Find the sum of the first 20 terms of the AP: 1, 5, 9, 13, ...
Answer:
First, identify the components of the AP.
The first term is $a = 1$.
The common difference is $d = 5 - 1 = 4$.
The number of terms to sum is $n=20$.
Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$S_{20} = \frac{20}{2}[2(1) + (20 - 1) \times 4]$
$= 10[2 + (19 \times 4)]
$= 10[2 + 76] = 10[78] = 780$
The sum of the first 20 terms is 780.
Properties of Arithmetic Progression
1. Effect of Adding/Subtracting a Constant
If a constant $k$ is added to (or subtracted from) every term of an AP, the resulting sequence is also an AP with the same common difference.
Example: Consider the AP $3, 7, 11, 15, \dots$ with $a=3$ and $d=4$.
Let's add a constant $k=5$ to each term:
New sequence: $(3+5), (7+5), (11+5), (15+5), \dots = 8, 12, 16, 20, \dots$
This is an AP with first term $a' = 8$ and common difference $d' = 12 - 8 = 4$. The common difference remains unchanged.
2. Effect of Multiplying/Dividing by a Non-Zero Constant
If every term of an AP is multiplied (or divided) by a non-zero constant $k$, the resulting sequence is also an AP. The new common difference will be $d \times k$ (or $d/k$).
Example: Consider the same AP $3, 7, 11, 15, \dots$ with $a=3$ and $d=4$.
Let's multiply each term by a constant $k=2$:
New sequence: $(3 \times 2), (7 \times 2), (11 \times 2), (15 \times 2), \dots = 6, 14, 22, 30, \dots$
This is an AP with first term $a' = 6$ and common difference $d' = 14 - 6 = 8$.
The new common difference is $d' = 8 = 4 \times 2 = d \times k$.
3. Arithmetic Mean
For any three consecutive terms $a, b, c$ in an AP, the middle term is the arithmetic mean of the other two. That is, $b = \frac{a+c}{2}$, which is equivalent to $2b = a+c$.
4. Sum of Equidistant Terms
In a finite AP, the sum of terms equidistant from the beginning and the end is always constant and equal to the sum of the first and last terms.
i.e., $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \dots$
Example: In the AP $2, 5, 8, 11, 14, 17$, we have $a_1=2, a_6=17$.
$a_1 + a_6 = 2 + 17 = 19$
$a_2 + a_5 = 5 + 14 = 19$
$a_3 + a_4 = 8 + 11 = 19$
Arithmetic Mean
The term Arithmetic Mean (AM) is commonly known as the average in statistics. In the context of sequences, it refers to a number that is placed between two other numbers to form an Arithmetic Progression (AP). This concept can be extended to inserting multiple numbers between two given numbers to create a longer AP.
Arithmetic Mean Between Two Numbers
The Arithmetic Mean (A) between two numbers, $a$ and $b$, is the single value that, when placed between them, creates a three-term Arithmetic Progression: $a, A, b$.
For this sequence to be an AP, the common difference between consecutive terms must be constant.
$A - a = d$
(Difference between 2nd and 1st term)
$b - A = d$
(Difference between 3rd and 2nd term)
Since the common difference is the same, we can set the expressions equal to each other:
$A - a = b - A$
Solving for A:
$2A = a + b$
$A = \frac{a+b}{2}$
... (i)
This shows that the single Arithmetic Mean between two numbers is simply their average. For example, the AM between 10 and 20 is $\frac{10+20}{2} = 15$, and the sequence 10, 15, 20 is an AP.
Inserting $n$ Arithmetic Means Between Two Numbers
A more general task is to insert a specified number of terms, say $n$ terms, between two numbers $a$ and $b$ so that the entire sequence forms an AP. These $n$ inserted terms are called the $n$ arithmetic means between $a$ and $b$.
Let the $n$ means be $A_1, A_2, \dots, A_n$. The resulting AP is:
$a, A_1, A_2, \dots, A_n, b$
To find the values of these means, we first need to determine the common difference ($d$) of this new, longer AP.
Derivation of the Common Difference
Let's analyze the structure of our new AP:
- The first term is $a$.
- The last term is $b$.
- The total number of terms is $n$ (the inserted means) plus the original two numbers, making a total of $n+2$ terms.
So, in this AP, $b$ is the $(n+2)^{th}$ term. We can use the general term formula for an AP, $a_m = \text{first term} + (m-1)d$, where $m=n+2$.
$b = a + ((n+2) - 1)d$
$b = a + (n+1)d$
Now, we solve this equation for $d$:
$b - a = (n+1)d$
$d = \frac{b-a}{n+1}$
... (ii)
Once we have calculated $d$, we can find each of the inserted means by starting with $a$ and successively adding $d$:
$A_1 = a + d$
$A_2 = a + 2d$
$A_3 = a + 3d$
... and so on, up to ...
$A_n = a + nd$
Sum of $n$ Arithmetic Means
There is an elegant formula for the sum of the $n$ arithmetic means inserted between $a$ and $b$. The sum of these means is simply $n$ times the single Arithmetic Mean of $a$ and $b$.
Sum of $n$ AMs = $A_1 + A_2 + \dots + A_n$.
$\sum\limits_{k=1}^{n} A_k = n \left(\frac{a+b}{2}\right)$
... (iii)
Example 1. Insert 5 arithmetic means between 2 and 14.
Answer:
We need to insert $n=5$ arithmetic means between $a=2$ and $b=14$.
Let the means be $A_1, A_2, A_3, A_4, A_5$. The resulting sequence will be $2, A_1, A_2, A_3, A_4, A_5, 14$. This is an AP with a total of $5+2=7$ terms.
Step 1: Find the common difference ($d$).
Using the formula $d = \frac{b-a}{n+1}$ with $a=2, b=14,$ and $n=5$:
$d = \frac{14 - 2}{5 + 1} = \frac{12}{6} = 2$
The common difference of the AP is 2.
Step 2: Find the arithmetic means.
We start with the first term, $a=2$, and repeatedly add the common difference, $d=2$.
$A_1 = a + d = 2 + 2 = 4$
$A_2 = a + 2d = 2 + 2(2) = 6$
$A_3 = a + 3d = 2 + 3(2) = 8$
$A_4 = a + 4d = 2 + 4(2) = 10$
$A_5 = a + 5d = 2 + 5(2) = 12$
The complete AP is $2, 4, 6, 8, 10, 12, 14$.
The five arithmetic means are 4, 6, 8, 10, 12.
Geometric Progression
A Geometric Progression (GP), also known as a geometric sequence, is another fundamental type of sequence. While an Arithmetic Progression is characterized by a constant difference between terms, a Geometric Progression is characterized by a constant ratio between consecutive terms. This means each term is found by multiplying the previous term by a fixed, non-zero number.
Definition and Common Ratio
A sequence of non-zero numbers $a_1, a_2, a_3, \dots$ is called a Geometric Progression if there exists a constant number $r$, called the common ratio, such that the ratio of any term to its preceding term is always $r$.
Mathematically, this can be expressed as:
$\frac{a_{n+1}}{a_n} = r \quad \text{or} \quad a_{n+1} = a_n \cdot r$
The common ratio $r$ can be any non-zero real number.
- If $|r| > 1$, the terms of the GP grow in magnitude (e.g., $2, 6, 18, \dots$ where $r=3$).
- If $|r| < 1$, the terms of the GP shrink in magnitude towards zero (e.g., $16, 8, 4, \dots$ where $r=1/2$).
- If $r$ is negative, the terms of the GP alternate in sign (e.g., $5, -10, 20, \dots$ where $r=-2$).
The General Term ($n^{th}$ Term) of a GP
To find any term in a GP without listing all the preceding ones, we use the formula for the general term. Let the first term be $a$ and the common ratio be $r$.
Let's observe the pattern formed by the terms:
First term: $a_1 = a = a \cdot r^{1-1}$
Second term: $a_2 = a_1 \cdot r = a \cdot r = a \cdot r^{2-1}$
Third term: $a_3 = a_2 \cdot r = (ar) \cdot r = ar^2 = a \cdot r^{3-1}$
Fourth term: $a_4 = a_3 \cdot r = (ar^2) \cdot r = ar^3 = a \cdot r^{4-1}$
The pattern shows that the exponent of $r$ for any term is always one less than its position number. Therefore, the formula for the $n^{th}$ term of a GP is:
$a_n = ar^{n-1}$
... (i)
Here, $a_n$ is the $n^{th}$ term, $a$ is the first term, $r$ is the common ratio, and $n$ is the position of the term.
Example 1. Find the 8th term of the GP: 5, 10, 20, ...
Answer:
First, we identify the key components of the given GP.
The first term is $a = 5$.
The common ratio is $r = \frac{10}{5} = 2$.
We need to find the 8th term, so the position is $n=8$.
Using the general term formula $a_n = ar^{n-1}$:
$a_8 = 5 \cdot (2)^{8-1} = 5 \cdot 2^7$
Now, we calculate the value:
$a_8 = 5 \cdot 128 = 640$
The 8th term of the GP is 640.
Sum of the First $n$ Terms of a GP
Let $S_n$ denote the sum of the first $n$ terms of a GP with first term $a$ and common ratio $r$.
$S_n = a + ar + ar^2 + \dots + ar^{n-1}$
Case 1: The Common Ratio is 1 ($r=1$)
If $r=1$, every term in the sequence is identical to the first term, $a$. The sum is simply $a$ added to itself $n$ times.
$S_n = na$
... (ii)
Case 2: The Common Ratio is not 1 ($r \neq 1$)
We can derive a formula for the sum with an algebraic trick. First, write the sum:
$S_n = a + ar + ar^2 + \dots + ar^{n-1}$
... (a)
Next, multiply the entire equation by the common ratio $r$:
$rS_n = ar + ar^2 + ar^3 + \dots + ar^n$
... (b)
Now, subtract equation (a) from equation (b). Most of the terms on the right side will cancel out.
$rS_n - S_n = (ar + ar^2 + \dots + ar^n) - (a + ar + \dots + ar^{n-1})$
$S_n(r-1) = ar^n - a$
Factoring out $a$ on the right side:
$S_n(r-1) = a(r^n - 1)$
Since $r \neq 1$, we can divide by $(r-1)$ to get the formula for the sum:
$S_n = \frac{a(r^n - 1)}{r - 1}$
... (iii)
An equivalent version, obtained by multiplying the numerator and denominator by -1, is often used when $|r|<1$ to keep the denominator positive:
$S_n = \frac{a(1 - r^n)}{1 - r}$
... (iv)
Example 2. Find the sum of the first 6 terms of the GP: 3, 6, 12, ...
Answer:
First, identify the components of the GP.
The first term is $a = 3$.
The common ratio is $r = \frac{6}{3} = 2$.
The number of terms to sum is $n=6$.
Since $r=2 \neq 1$, we use the sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$.
$S_6 = \frac{3(2^6 - 1)}{2 - 1}$
$= \frac{3(64 - 1)}{1} = 3(63) = 189$
The sum of the first 6 terms is 189.
Sum of an Infinite Geometric Series
It is possible to find the sum of an infinite GP, but only under a specific condition. Consider the sum formula $S_n = \frac{a(1 - r^n)}{1 - r}$. As we add more and more terms ($n \to \infty$), the value of the sum depends entirely on what happens to the $r^n$ part.
If the magnitude of the common ratio is less than 1 (i.e., $-1 < r < 1$ or $|r| < 1$), the term $r^n$ gets progressively closer to zero as $n$ becomes infinitely large. For example, $(\frac{1}{2})^{10} \approx 0.001$ and $(\frac{1}{2})^{20}$ is even smaller. We say that $\lim\limits_{n \to \infty} r^n = 0$ for $|r|<1$.
Under this condition, the sum formula for an infinite number of terms ($S_\infty$) becomes:
$S_\infty = \lim\limits_{n \to \infty} \frac{a(1 - r^n)}{1 - r} = \frac{a(1 - 0)}{1 - r}$
$S_\infty = \frac{a}{1 - r}$, for $|r| < 1$
... (v)
If $|r| \geq 1$, the term $r^n$ does not approach zero, and the sum does not approach a finite value. In this case, the infinite series is said to diverge, and it has no sum.
Example 3. Find the sum of the infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...
Answer:
First, we identify the components of the infinite GP.
The first term is $a = 1$.
The common ratio is $r = \frac{1/2}{1} = \frac{1}{2}$.
We check the condition for the sum to exist: $|r| = |\frac{1}{2}| = \frac{1}{2}$. Since this is less than 1, the sum of the infinite series exists.
Using the formula $S_\infty = \frac{a}{1 - r}$:
$S_\infty = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}}$
$= 2$
The sum of the infinite series is 2.
Geometric Mean
Similar to how the Arithmetic Mean relates to Arithmetic Progressions, the Geometric Mean (GM) relates to Geometric Progressions. It is a type of average that is particularly useful for sets of positive numbers whose values are multiplicative in nature, such as rates of growth.
Geometric Mean Between Two Positive Numbers
The Geometric Mean (G) between two positive numbers, $a$ and $b$, is the single positive value that, when placed between them, creates a three-term Geometric Progression: $a, G, b$.
For this sequence to be a GP, the ratio of consecutive terms must be constant (the common ratio, $r$).
$\frac{G}{a} = r$
(Ratio of 2nd to 1st term)
$\frac{b}{G} = r$
(Ratio of 3rd to 2nd term)
Since the common ratio is the same, we can set the expressions equal to each other:
$\frac{G}{a} = \frac{b}{G}$
To solve for G, we cross-multiply:
$G^2 = ab$
Taking the positive square root (as GMs between positive numbers are conventionally positive):
$G = \sqrt{ab}$
... (i)
The Geometric Mean of two positive numbers is the positive square root of their product. For example, the GM between 4 and 9 is $\sqrt{4 \times 9} = \sqrt{36} = 6$. The sequence 4, 6, 9 is a GP with $r=1.5$.
Inserting $n$ Geometric Means Between Two Positive Numbers
We can also insert a specified number of terms, say $n$ terms, between two positive numbers $a$ and $b$ to form a longer GP. These $n$ inserted terms are called the $n$ geometric means.
Let the $n$ means be $G_1, G_2, \dots, G_n$. The resulting GP is:
$a, G_1, G_2, \dots, G_n, b$
To find the values of these means, we first need to determine the common ratio ($r$) of this new GP.
Derivation of the Common Ratio
Let's analyze the structure of our new GP:
- The first term is $a$.
- The last term is $b$.
- The total number of terms is $n$ (inserted means) + 2 (original numbers) = $n+2$.
Using the general term formula for a GP, $a_m = \text{first term} \cdot r^{m-1}$, with $b$ as the $(n+2)^{th}$ term:
$b = a \cdot r^{(n+2) - 1}$
$b = a \cdot r^{n+1}$
Now, we solve for the common ratio $r$:
$\frac{b}{a} = r^{n+1}$
$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$
... (ii)
With this common ratio, we can find all the inserted means:
$G_1 = ar, \quad G_2 = ar^2, \quad \dots, \quad G_n = ar^n$
Product of $n$ Geometric Means
A useful property relates the product of the inserted GMs to the single GM. The product of the $n$ geometric means inserted between $a$ and $b$ is equal to the $n^{th}$ power of the single Geometric Mean of $a$ and $b$.
Product = $G_1 \cdot G_2 \cdot \dots \cdot G_n$
$\prod\limits_{k=1}^{n} G_k = (\sqrt{ab})^n$
... (iii)
Example 1. Insert 3 geometric means between 1 and 256.
Answer:
We need to insert $n=3$ geometric means between $a=1$ and $b=256$.
The resulting sequence will be $1, G_1, G_2, G_3, 256$. This is a GP with $3+2=5$ terms.
Step 1: Find the common ratio ($r$).
Using the formula $r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ with $a=1, b=256,$ and $n=3$:
$r = \left(\frac{256}{1}\right)^{\frac{1}{3+1}} = (256)^{\frac{1}{4}}$
We need to find the fourth root of 256. Since $256 = 4^4$, the positive real root is:
$r = 4$
The common ratio is 4.
Step 2: Find the geometric means.
We start with the first term, $a=1$, and repeatedly multiply by the common ratio, $r=4$.
$G_1 = a \cdot r = 1 \cdot 4 = 4$
$G_2 = a \cdot r^2 = 1 \cdot 4^2 = 16$
$G_3 = a \cdot r^3 = 1 \cdot 4^3 = 64$
The complete GP is $1, 4, 16, 64, 256$.
The three geometric means are 4, 16, 64.
Relationship Between AM and GM
For any two non-negative real numbers, there exists a fundamental and important relationship between their Arithmetic Mean (AM) and their Geometric Mean (GM). This relationship, known as the AM-GM Inequality, states that the Arithmetic Mean of the numbers is always greater than or equal to their Geometric Mean.
The AM-GM Inequality for Two Numbers
Let $a$ and $b$ be any two non-negative real numbers.
- Their Arithmetic Mean (AM) is defined as: $\text{AM} = \frac{a+b}{2}$
- Their Geometric Mean (GM) is defined as: $\text{GM} = \sqrt{ab}$
The AM-GM Inequality states:
$\frac{a+b}{2} \geq \sqrt{ab}$
... (i)
Furthermore, the equality holds (i.e., AM = GM) if and only if the two numbers are equal ($a=b$). If the numbers are different ($a \neq b$), the inequality is strict (AM > GM).
Proof of the AM-GM Inequality
The proof is straightforward and relies on the basic principle that the square of any real number is non-negative.
Given: Two non-negative real numbers, $a$ and $b$.
To Prove: $\frac{a+b}{2} \geq \sqrt{ab}$.
Proof:
Since $a$ and $b$ are non-negative, their square roots, $\sqrt{a}$ and $\sqrt{b}$, are real numbers. Consider the square of their difference:
$(\sqrt{a} - \sqrt{b})^2 \geq 0$
(The square of a real number is always non-negative)
Expand the left side of the inequality:
$(\sqrt{a})^2 - 2(\sqrt{a})(\sqrt{b}) + (\sqrt{b})^2 \geq 0$
$a - 2\sqrt{ab} + b \geq 0$
Now, rearrange the terms to isolate the AM and GM components. Add $2\sqrt{ab}$ to both sides:
$a + b \geq 2\sqrt{ab}$
Finally, divide both sides by 2 (which is positive, so the inequality sign does not change):
$\frac{a+b}{2} \geq \sqrt{ab}$
This completes the proof.
Condition for Equality
The equality $\frac{a+b}{2} = \sqrt{ab}$ holds only when the initial inequality is an equality, which means $(\sqrt{a} - \sqrt{b})^2 = 0$. This is true if and only if $\sqrt{a} - \sqrt{b} = 0$, which implies $\sqrt{a} = \sqrt{b}$. Squaring both sides gives $a=b$. Thus, AM = GM if and only if the numbers are equal.
Generalized AM-GM Inequality
The AM-GM inequality can be generalized to a set of $n$ non-negative real numbers: $x_1, x_2, \dots, x_n$.
- The AM is $\frac{x_1 + x_2 + \dots + x_n}{n}$.
- The GM is $\sqrt[n]{x_1 x_2 \dots x_n}$.
The generalized inequality states:
$\frac{x_1 + x_2 + \dots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}$
... (ii)
Again, equality holds if and only if all the numbers are equal ($x_1 = x_2 = \dots = x_n$). This powerful inequality is a cornerstone of many proofs and optimization problems in mathematics.
Example 1. Verify the AM-GM inequality for the numbers 4 and 9.
Answer:
We are given the numbers $a=4$ and $b=9$.
Step 1: Calculate the Arithmetic Mean (AM).
AM = $\frac{a+b}{2} = \frac{4+9}{2} = \frac{13}{2} = 6.5$
Step 2: Calculate the Geometric Mean (GM).
GM = $\sqrt{ab} = \sqrt{4 \times 9} = \sqrt{36} = 6$
Step 3: Compare the AM and GM.
We need to check if AM $\geq$ GM.
$6.5 \geq 6$
The statement is true. Since the numbers 4 and 9 are not equal, the inequality is strict (6.5 > 6), which is consistent with the theory.
The AM-GM inequality is verified for the given numbers.
Some Special Series
Beyond the standard arithmetic and geometric series, there are several other important series that occur frequently in mathematics. These special series, particularly the sums of powers of the first $n$ natural numbers, serve as fundamental building blocks for solving more complex summation problems.
Sum of the First $n$ Natural Numbers
This series is the sum of the first $n$ positive integers: $1 + 2 + 3 + \dots + n$. This is a simple Arithmetic Progression (AP) with the first term $a=1$, common difference $d=1$, and $n$ terms.
The series is denoted as:
$\sum\limits_{k=1}^{n} k = 1 + 2 + 3 + \dots + n$
We can derive the formula using the sum of an AP, $S_n = \frac{n}{2}(\text{first term} + \text{last term})$. Here, the first term is 1 and the last term is $n$.
$\sum\limits_{k=1}^{n} k = \frac{n}{2}(1 + n)$
This gives us the well-known formula:
$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$
... (i)
Sum of the Squares of the First $n$ Natural Numbers
This series is the sum of the squares of the first $n$ positive integers: $1^2 + 2^2 + 3^2 + \dots + n^2$.
The series is denoted as:
$\sum\limits_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \dots + n^2$
The formula for this sum is:
$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
... (ii)
Derivation of the Formula
The proof uses a clever technique involving a "telescoping sum". We start with the identity for the difference of cubes:
$k^3 - (k-1)^3 = k^3 - (k^3 - 3k^2 + 3k - 1) = 3k^2 - 3k + 1$
Now, we sum this identity for all integer values of $k$ from 1 to $n$:
$\sum\limits_{k=1}^{n} [k^3 - (k-1)^3] = \sum\limits_{k=1}^{n} (3k^2 - 3k + 1)$
The left side is a telescoping sum, where intermediate terms cancel out:
$(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \dots + (n^3 - (n-1)^3) = n^3 - 0^3 = n^3$.
The right side can be split using properties of summation:
$3\sum\limits_{k=1}^{n} k^2 - 3\sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1$
Equating the two sides and substituting the known formulas for $\sum k$ and $\sum 1$ (which is $n$):
$n^3 = 3\left(\sum\limits_{k=1}^{n} k^2\right) - 3\left(\frac{n(n+1)}{2}\right) + n$
Now, we algebraically solve for $\sum k^2$:
$3\left(\sum\limits_{k=1}^{n} k^2\right) = n^3 + \frac{3n(n+1)}{2} - n$
$= \frac{2n^3 + 3n^2 + 3n - 2n}{2} = \frac{2n^3 + 3n^2 + n}{2}$
$= \frac{n(2n^2 + 3n + 1)}{2} = \frac{n(n+1)(2n+1)}{2}$
Finally, dividing by 3 gives the desired formula:
$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of the Cubes of the First $n$ Natural Numbers
This series is the sum of the cubes of the first $n$ positive integers: $1^3 + 2^3 + 3^3 + \dots + n^3$.
The series is denoted as:
$\sum\limits_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \dots + n^3$
The formula for this sum has a beautiful connection to the sum of the first $n$ natural numbers:
$\sum\limits_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \left(\sum\limits_{k=1}^{n} k\right)^2$
... (iii)
In words, the sum of the first $n$ cubes is equal to the square of the sum of the first $n$ natural numbers.
Example 1. Find the sum of the series $1 + 2 + 3 + \dots + 50$.
Answer:
We need to find the sum of the first 50 natural numbers, so $n=50$.
Using the formula $\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$:
Sum = $\frac{50(50+1)}{2} = \frac{50 \times 51}{2}$
$= 25 \times 51 = 1275$
The sum is 1275.
Example 2. Find the sum of the series $1^2 + 2^2 + \dots + 10^2$.
Answer:
We need to find the sum of the squares of the first 10 natural numbers, so $n=10$.
Using the formula $\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:
Sum = $\frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \times 11 \times 21}{6}$
We can simplify the fraction before multiplying:
$= \frac{\cancel{10}^5 \times 11 \times \cancel{21}^7}{\cancel{6}_{~3~1}} = 5 \times 11 \times 7$
$= 55 \times 7 = 385$
The sum is 385.
Example 3. Find the sum of the series $1^3 + 2^3 + \dots + 8^3$.
Answer:
We need to find the sum of the cubes of the first 8 natural numbers, so $n=8$.
Using the formula $\sum\limits_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$:
Sum = $\left(\frac{8(8+1)}{2}\right)^2 = \left(\frac{8 \times 9}{2}\right)^2$
$= \left(\frac{72}{2}\right)^2 = (36)^2$
$= 1296$
The sum is 1296.