Applied Mathematics for Class 11th & 12th (Concepts and Questions) | ||
---|---|---|
11th | Concepts | Questions |
12th | Concepts | Questions |
Content On This Page | ||
---|---|---|
Sequence and Series | Arithmetic Progression | Geometric Progression |
Applications of AP and GP |
Chapter 5 Sequences and Series (Concepts)
Welcome to this exploration of Sequences and Series, a vital area within Applied Mathematics focused on understanding ordered patterns in numbers and the cumulative effects of summing them. Recognizing and analyzing sequences, and calculating the sums of series, are fundamental skills applicable across numerous quantitative disciplines, including finance, economics, computer science, physics, and data analysis. This chapter will equip you with the tools to identify specific types of progressions, predict future terms, and calculate sums efficiently, moving beyond simple lists of numbers to structured mathematical analysis. We will primarily focus on two cornerstone types: Arithmetic Progressions (AP) and Geometric Progressions (GP), alongside methods for summing specific types of series.
A sequence is essentially an ordered list of numbers, often generated according to a specific rule or pattern. When we add the terms of a sequence together, we form a series. Understanding the underlying rule of a sequence allows us to predict its behavior and analyze its properties. We begin by revisiting Arithmetic Progressions (AP). An AP is a sequence where the difference between any two successive terms is constant; this constant difference is known as the common difference, denoted by $d$. If the first term is $a$, the sequence proceeds as $a, a+d, a+2d, \dots$. Key formulas govern APs:
- The $n^{th}$ term is given by $a_n = a + (n-1)d$.
- The sum of the first $n$ terms, $S_n$, can be calculated using $S_n = \frac{n}{2}[2a + (n-1)d]$ or, if the last term $l$ (which is $a_n$) is known, $S_n = \frac{n}{2}[a+l]$.
We will also explore the concept of inserting Arithmetic Means (AMs) between two given numbers. Applications of APs are common, such as modeling scenarios involving simple interest accumulation or linear growth patterns.
Next, we turn our attention to Geometric Progressions (GP). In a GP, each term after the first is obtained by multiplying the preceding term by a fixed, non-zero constant called the common ratio, denoted by $r$. A GP takes the form $a, ar, ar^2, \dots$. The essential formulas for GPs are:
- The $n^{th}$ term is given by $a_n = ar^{n-1}$.
- The sum of the first $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1}$ (or equivalently $\frac{a(1 - r^n)}{1 - r}$), provided $r \neq 1$.
A particularly important concept within GPs is the sum of an infinite geometric series. This sum converges to a finite value only when the absolute value of the common ratio is less than 1 (i.e., $|r| < 1$). In such cases, the sum is given by the simple formula $S_\infty = \frac{a}{1-r}$. We will also discuss inserting Geometric Means (GMs). GPs are crucial for modeling phenomena involving exponential growth or decay, such as compound interest calculations, population dynamics, or radioactive decay.
Beyond APs and GPs, we will touch upon the relationship between the Arithmetic Mean (AM) and Geometric Mean (GM) of positive numbers, stating the important inequality $AM \ge GM$. Furthermore, we will equip you with techniques for summing certain special series by utilizing standard formulas for the sum of the first $n$ natural numbers ($\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$), the sum of their squares ($\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$), and the sum of their cubes ($\sum\limits_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$). These formulas are invaluable for calculating the sum of series where the $n^{th}$ term can be expressed as a polynomial in $n$. This chapter aims to build robust skills in identifying numerical patterns, applying appropriate formulas, and calculating sums, laying a strong foundation for quantitative analysis in applied contexts.
Sequence and Series
In mathematics, we often deal with ordered collections of numbers. These ordered collections, especially when they follow a discernible pattern or rule, are central to the concepts of Sequences and Series. These ideas are fundamental and have extensive applications in various fields, including algebra, calculus, physics, finance, and computer science.
Sequences
A Sequence is an ordered list of numbers or other mathematical objects. The key characteristic of a sequence is that the order of the elements matters. Each number in the sequence is called a term or an element of the sequence. The terms are usually denoted by using subscripts to indicate their position in the list.
A sequence is typically represented as $a_1, a_2, a_3, ..., a_n, ...$
- $a_1$ is the first term.
- $a_2$ is the second term.
- $a_3$ is the third term.
- ...
- $a_n$ is the $n$-th term, also known as the general term or formula term of the sequence.
The terms of a sequence often follow a specific pattern or rule, which can sometimes be expressed as a formula for the $n$-th term ($a_n$) in terms of its position $n$. If such a formula exists, it allows us to find any term of the sequence by simply substituting the desired position number for $n$.
Types of Sequences based on Number of Terms
Finite Sequence: A sequence that has a limited or fixed number of terms. The list of terms has a definite beginning and a definite end.
Example: The sequence of odd numbers less than 10: 1, 3, 5, 7, 9.
Example: The list of ranks of students in a class in a test: 1st, 2nd, 3rd, ..., up to the last rank.
Infinite Sequence: A sequence that has an unlimited number of terms. The list of terms continues indefinitely.
Example: The sequence of all positive even numbers: 2, 4, 6, 8, 10, ... (The ellipsis '...' indicates that the sequence continues without end).
Example: The sequence of reciprocals of natural numbers: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$
Representing Sequences
Sequences can be represented in several ways, depending on how the terms are generated or described:
Listing the Terms (Roster Method): Writing out the first few terms of the sequence. This is effective when the pattern is clear from the initial terms, or for finite sequences where all terms can be listed. For infinite sequences with a clear pattern, an ellipsis (...) is used after the initial terms.
Example: 1, 4, 9, 16, ... (It is clear that the $n$-th term is $n^2$)
Example: 5, 10, 15, 20 (A finite sequence)
Using a Formula for the General Term ($a_n$): Providing an explicit expression for the $n$-th term $a_n$ as a function of $n$. This formula allows us to find any term of the sequence directly by substituting the position number $n$.
Example: Consider the sequence defined by $a_n = 2n$. The terms are generated by substituting $n=1, 2, 3, ...$
For $n=1, a_1 = 2(1) = 2$.
For $n=2, a_2 = 2(2) = 4$.
For $n=3, a_3 = 2(3) = 6$.
The sequence is 2, 4, 6, 8, ...Example: Consider the sequence defined by $a_n = (-1)^{n-1} \frac{1}{n}$.
For $n=1, a_1 = (-1)^{1-1} \frac{1}{1} = (-1)^0 \times 1 = 1 \times 1 = 1$.
For $n=2, a_2 = (-1)^{2-1} \frac{1}{2} = (-1)^1 \times \frac{1}{2} = - \frac{1}{2}$.
For $n=3, a_3 = (-1)^{3-1} \frac{1}{3} = (-1)^2 \times \frac{1}{3} = 1 \times \frac{1}{3} = \frac{1}{3}$.
The sequence is $1, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, ...$Using a Recurrence Relation (Recursive Definition): Defining a term in the sequence based on the value(s) of one or more preceding terms. This method requires specifying the value(s) of the initial term(s) to start the sequence.
Example: The Fibonacci sequence is defined by $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$, with initial conditions $a_1 = 1$ and $a_2 = 1$.
To find the terms:
$a_1 = 1$
$a_2 = 1$
$a_3 = a_2 + a_1 = 1 + 1 = 2$.
$a_4 = a_3 + a_2 = 2 + 1 = 3$.
$a_5 = a_4 + a_3 = 3 + 2 = 5$.
The sequence is 1, 1, 2, 3, 5, 8, 13, ...
Example 1. Write the first 5 terms of the sequence whose $n$-th term is given by $a_n = 4n - 3$.
Answer:
The general term of the sequence is $a_n = 4n - 3$.
To find the first 5 terms, we substitute $n=1, 2, 3, 4, 5$ into the formula $a_n = 4n - 3$.
For the 1st term ($n=1$):
$$a_1 = 4(1) - 3 = 4 - 3 = 1$$
For the 2nd term ($n=2$):
$$a_2 = 4(2) - 3 = 8 - 3 = 5$$
For the 3rd term ($n=3$):
$$a_3 = 4(3) - 3 = 12 - 3 = 9$$
For the 4th term ($n=4$):
$$a_4 = 4(4) - 3 = 16 - 3 = 13$$
For the 5th term ($n=5$):
$$a_5 = 4(5) - 3 = 20 - 3 = 17$$
... (i)
The first 5 terms of the sequence are 1, 5, 9, 13, and 17.
Example 2. Find the 6th term of the sequence defined by $a_1 = 3$ and $a_n = a_{n-1} + 5$ for $n > 1$. Write the first 6 terms.
Answer:
We are given the first term $a_1 = 3$ and a recurrence relation $a_n = a_{n-1} + 5$ for $n > 1$. This means each term after the first is obtained by adding 5 to the previous term.
We need to find the 6th term ($a_6$) and list the first 6 terms by calculating them sequentially.
1st term: $a_1 = 3$ (Given)
2nd term ($n=2$): Using the relation $a_2 = a_1 + 5$.
$$a_2 = 3 + 5 = 8$$
3rd term ($n=3$): Using $a_2 = 8$.
$$a_3 = a_2 + 5 = 8 + 5 = 13$$
4th term ($n=4$): Using $a_3 = 13$.
$$a_4 = a_3 + 5 = 13 + 5 = 18$$
5th term ($n=5$): Using $a_4 = 18$.
$$a_5 = a_4 + 5 = 18 + 5 = 23$$
6th term ($n=6$): Using $a_5 = 23$.
$$a_6 = a_5 + 5 = 23 + 5 = 28$$
... (i)
The 6th term of the sequence is 28.
The first 6 terms of the sequence are 3, 8, 13, 18, 23, and 28.
Note that this sequence follows a pattern where the difference between consecutive terms is constant (always 5). This type of sequence is called an Arithmetic Progression (AP), which will be discussed in the next section.
Series
A Series is simply the sum of the terms of a sequence. If we have a sequence of numbers $a_1, a_2, a_3, ..., a_n, ...$, the corresponding series is the expression obtained by adding these terms together: $a_1 + a_2 + a_3 + ... + a_n + ...$.
Series are conveniently represented using the Greek letter Sigma ($\sum$), which denotes summation. The sum of the first $n$ terms of a sequence $\{a_k\}$ is denoted by $S_n$.
$$S_n = a_1 + a_2 + a_3 + ... + a_n = \sum\limits_{k=1}^{n} a_k$$
... (38)
Here, $\sum\limits_{k=1}^{n} a_k$ means "the sum of $a_k$ for values of $k$ starting from 1 up to $n$". The index $k$ is a dummy variable and can be any letter (e.g., $\sum\limits_{i=1}^{n} a_i$).
Types of Series
Based on whether the corresponding sequence is finite or infinite:
Finite Series: A series formed by summing the terms of a finite sequence. A finite series always has a definite sum (a single numerical value).
Example: $1 + 3 + 5 + 7 + 9$. This is the sum of the finite sequence 1, 3, 5, 7, 9. The sum is $1+3+5+7+9 = 25$.
Infinite Series: A series formed by summing the terms of an infinite sequence. Represented as $a_1 + a_2 + a_3 + ...$, or using summation notation as $\sum\limits_{k=1}^{\infty} a_k$. The sum of an infinite series may converge to a finite value or diverge (not approach a finite value). The concept of convergence is a significant part of higher mathematics.
Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$ (The sum here approaches 2, so it converges)
Example: $1 + 2 + 3 + 4 + ...$ (The sum grows infinitely large, so it diverges)
In this chapter, we will primarily study specific types of sequences that follow simple patterns (Arithmetic Progression and Geometric Progression) and calculate the sums of their finite versions.
Example 3. Consider the sequence whose first 5 terms are 1, 5, 9, 13, 17 (from Example 1). Write the series for the first 5 terms and find its sum.
Answer:
The first 5 terms of the sequence are 1, 5, 9, 13, and 17.
The series for the first 5 terms, denoted by $S_5$, is the sum of these terms:
$$S_5 = 1 + 5 + 9 + 13 + 17$$
Calculate the sum by adding the numbers:
$$S_5 = 6 + 9 + 13 + 17 = 15 + 13 + 17 = 28 + 17 = 45$$
... (i)
The series for the first 5 terms is $1 + 5 + 9 + 13 + 17$, and its sum is 45.
Key Distinction between Sequence and Series
It is important to clearly differentiate between a sequence and a series:
- A Sequence is an ordered list of terms (e.g., 2, 4, 6, 8). It focuses on the individual terms and their order.
- A Series is the sum of the terms of a sequence (e.g., $2 + 4 + 6 + 8$). It focuses on the combined value obtained by adding the terms.
Think of a sequence as a collection of numbers laid out in order, and a series as the result of adding those numbers together.
Arithmetic Progression
An Arithmetic Progression (AP) is a specific type of sequence in which the difference between any term and its preceding term is constant throughout the sequence. This constant difference is called the common difference and is typically denoted by the letter $d$.
Definition and General Form of an AP
A sequence $a_1, a_2, a_3, ..., a_n, ...$ is said to be an Arithmetic Progression if $a_{n+1} - a_n = d$ for all positive integers $n$, where $d$ is a constant.
If the first term of an AP is denoted by $a$ (or $a_1$), and the common difference is $d$, then the terms of the AP can be generated sequentially:
- First term ($a_1$) $= a$
- Second term ($a_2$) $= a_1 + d = a + d$
- Third term ($a_3$) $= a_2 + d = (a + d) + d = a + 2d$
- Fourth term ($a_4$) $= a_3 + d = (a + 2d) + d = a + 3d$
- ... and so on.
Thus, the general form of an AP is:
$$a, a+d, a+2d, a+3d, ..., a+(n-1)d, ...$$
... (39)
The common difference $d$ can be any real number: positive (increasing AP), negative (decreasing AP), or zero (constant AP).
Identifying an AP
To check if a given sequence is an AP, calculate the difference between consecutive terms. If $a_2-a_1 = a_3-a_2 = a_4-a_3 = ... = d$, then the sequence is an AP with common difference $d$.
Example: The sequence 5, 10, 15, 20, 25, ...
$10-5 = 5$
$15-10 = 5$
$20-15 = 5$
$25-20 = 5$
Since the difference is constantly 5, it is an AP with first term $a=5$ and common difference $d=5$.
Example: The sequence 20, 17, 14, 11, 8, ...
$17-20 = -3$
$14-17 = -3$
$11-14 = -3$
$8-11 = -3$
Since the difference is constantly -3, it is an AP with first term $a=20$ and common difference $d=-3$.
Example: The sequence $\frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, ...$
$1 - \frac{1}{2} = \frac{1}{2}$
$\frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2}$
$2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2}$
$\frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$
Since the difference is constantly $\frac{1}{2}$, it is an AP with first term $a=\frac{1}{2}$ and common difference $d=\frac{1}{2}$.
nth Term (General Term) of an AP
The formula for the $n$-th term of an AP, denoted by $a_n$, provides a direct way to calculate any term in the sequence without having to list all the preceding terms.
Derivation of the Formula for $a_n$:
Let the first term of the AP be $a_1 = a$, and the common difference be $d$.
By the definition of an AP, each subsequent term is obtained by adding the common difference $d$ to the previous term:
$$a_2 = a_1 + d = a + 1d$$
... (a)
$$a_3 = a_2 + d = (a + d) + d = a + 2d$$
... (b)
$$a_4 = a_3 + d = (a + 2d) + d = a + 3d$$
... (c)
We observe a pattern here: the coefficient of $d$ in each term is one less than the term number.
- For the 2nd term ($n=2$), the coefficient of $d$ is 1 ($=2-1$).
- For the 3rd term ($n=3$), the coefficient of $d$ is 2 ($=3-1$).
- For the 4th term ($n=4$), the coefficient of $d$ is 3 ($=4-1$).
Following this pattern, for the $n$-th term ($a_n$), the coefficient of $d$ will be $(n-1)$.
Thus, the formula for the $n$-th term is:
$$a_n = a + (n-1)d$$
... (40)
Where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term in the sequence ($n = 1, 2, 3, ...$).
Example 1. Find the 10th term of the AP: 2, 7, 12, ...
Answer:
The given AP is 2, 7, 12, ...
The first term is $a = a_1 = 2$.
To find the common difference $d$, subtract any term from its succeeding term:
$$d = a_2 - a_1 = 7 - 2 = 5$$
Check with the next pair: $a_3 - a_2 = 12 - 7 = 5$. The common difference is indeed 5.
We need to find the 10th term of this AP, so $n = 10$.
Using the formula for the $n$-th term of an AP: $a_n = a + (n-1)d$ (Equation 40).
Substitute $a=2$, $d=5$, and $n=10$:
$$a_{10} = 2 + (10 - 1) \times 5$$
Simplify the expression:
$$a_{10} = 2 + 9 \times 5$$
$$a_{10} = 2 + 45$$
$$a_{10} = 47$$
... (i)
The 10th term of the AP is 47.
Sum of the First n Terms of an AP
The sum of the first $n$ terms of an Arithmetic Progression is denoted by $S_n$.
Derivation of the Formula for $S_n$:
Let the AP be $a_1, a_2, ..., a_n$, with first term $a_1 = a$ and common difference $d$.
The terms are $a, a+d, a+2d, ..., a+(n-1)d$.
Let $l$ be the last term, i.e., $l = a_n = a + (n-1)d$.
The sum $S_n$ is the sum of these $n$ terms:
$$S_n = a + (a+d) + (a+2d) + ... + (a+(n-2)d) + (a+(n-1)d)$$
... (a)
We can also write the sum in terms of the last term $l$. The term before the last is $l-d$, the term before that is $l-2d$, and so on, until the first term $a$.
$$S_n = l + (l-d) + (l-2d) + ... + (l-(n-2)d) + (l-(n-1)d)$$
... (b)
Now, let's add equation (a) and equation (b) term by term vertically:
$$2S_n = (a+l) + ([a+d]+[l-d]) + ([a+2d]+[l-2d]) + ... + ([a+(n-1)d]+[l-(n-1)d])$$
Notice that in each pair of brackets, the common difference terms cancel out:
$a+d + l-d = a+l$
$a+2d + l-2d = a+l$
... and so on.
So, each pair of terms sums to $(a+l)$. There are $n$ such pairs (since there are $n$ terms in the sum).
$$2S_n = (a+l) + (a+l) + (a+l) + ... + (a+l) \text{ (n times)}$$$$
$$2S_n = n \times (a+l)$$$$
Divide both sides by 2 to get the formula for $S_n$:
$$S_n = \frac{n}{2}(a+l)$$
... (41)
This formula is useful if the first term ($a$), the last term ($l$), and the number of terms ($n$) are known.
We can also express $S_n$ in terms of $a$ and $d$. Since the last term $l$ is the $n$-th term $a_n$, we know that $l = a + (n-1)d$ (from Equation 40).
Substitute this expression for $l$ into formula (41):
$$S_n = \frac{n}{2}(a + [a + (n-1)d])$$
Simplify the expression inside the brackets:
$$S_n = \frac{n}{2}(2a + (n-1)d)$$
... (42)
This is the formula for the sum of the first $n$ terms of an AP in terms of the first term ($a$), the common difference ($d$), and the number of terms ($n$). This is the most frequently used formula for the sum of an AP.
Example 2. Find the sum of the first 15 terms of the AP: 5, 8, 11, 14, ...
Answer:
The given AP is 5, 8, 11, 14, ...
The first term is $a = 5$.
The common difference is $d = 8 - 5 = 3$. (Check: $11-8=3$, $14-11=3$. The common difference is indeed 3).
We need to find the sum of the first 15 terms, so the number of terms is $n = 15$.
Using the formula for the sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}(2a + (n-1)d)$ (Equation 42).
Substitute the values $a=5$, $d=3$, and $n=15$ into the formula:
$$S_{15} = \frac{15}{2} (2 \times 5 + (15 - 1) \times 3)$$$$
Simplify the expression inside the brackets:
$$S_{15} = \frac{15}{2} (10 + 14 \times 3)$$$$
$$S_{15} = \frac{15}{2} (10 + 42)$$$$
$$S_{15} = \frac{15}{2} (52)$$$$
Perform the multiplication and division:
$$S_{15} = 15 \times \left(\frac{52}{2}\right) = 15 \times 26$$
Calculate $15 \times 26$:
$$\begin{array}{cc}& & 2 & 6 \\ \times & & 1 & 5 \\ \hline & 1 & 3 & 0 \\ 2 & 6 & \times & \phantom{0} \\ \hline 3 & 9 & 0 \\ \hline \end{array}$$$$S_{15} = 390$$
... (i)
The sum of the first 15 terms of the given AP is 390.
Example 3. If the first term of an AP is 12 and the last term is 108, and there are 33 terms in total, find the sum of the AP.
Answer:
Given:
First term, $a = 12$.
Last term, $l = 108$.
Number of terms, $n = 33$.
We need to find the sum of these 33 terms ($S_{33}$). Since we know the first term, the last term, and the number of terms, we can use the formula $S_n = \frac{n}{2}(a+l)$ (Equation 41).
Substitute $n=33$, $a=12$, and $l=108$:
$$S_{33} = \frac{33}{2} (12 + 108)$$$$
Calculate the sum inside the brackets:
$$S_{33} = \frac{33}{2} (120)$$$$
Perform the multiplication and division:
$$S_{33} = 33 \times \left(\frac{120}{2}\right) = 33 \times 60$$
Calculate $33 \times 60$:
$$S_{33} = 1980$$
... (i)
The sum of the first 33 terms of the AP is 1980.
Properties of Arithmetic Progression
Arithmetic Progressions have several useful properties:
If a constant quantity is added to or subtracted from each term of an AP, the resulting sequence is also an AP with the same common difference.
Example: If 1, 3, 5, 7 is an AP (d=2), then adding 10 to each term gives 11, 13, 15, 17, which is also an AP (d=2).
If each term of an AP is multiplied or divided by a non-zero constant, the resulting sequence is also an AP. If the original AP had a common difference $d$, the new AP will have a common difference $kd$ (if multiplied by $k$) or $d/k$ (if divided by $k$).
Example: If 2, 4, 6, 8 is an AP (d=2), then multiplying by 3 gives 6, 12, 18, 24, which is an AP with common difference $3 \times 2 = 6$.
In a finite AP, the sum of terms equidistant from the beginning and the end is constant and is equal to the sum of the first and last terms. For an AP $a_1, a_2, ..., a_n$, the sum $a_k + a_{n-k+1}$ is always equal to $a_1 + a_n$ for $k=1, 2, ..., n$.
Example: In the AP 2, 5, 8, 11, 14, 17 ($n=6$). $a_1=2, a_6=17$. $a_1+a_6 = 19$.
$k=2$: $a_2=5, a_{6-2+1}=a_5=14$. $a_2+a_5 = 5+14=19$.
$k=3$: $a_3=8, a_{6-3+1}=a_4=11$. $a_3+a_4 = 8+11=19$.If three numbers $a, b, c$ are in AP, then the middle term $b$ is the arithmetic mean of the other two terms. $b = \frac{a+c}{2}$, which implies $2b = a+c$.
Arithmetic Mean (AM)
When we insert numbers between two terms of an AP such that the entire sequence remains in AP, the inserted numbers are called Arithmetic Means. If $A_1, A_2, ..., A_m$ are $m$ numbers inserted between $a$ and $b$ such that $a, A_1, A_2, ..., A_m, b$ form an AP, then these $A_i$ are $m$ arithmetic means between $a$ and $b$.
If we insert a single number $A$ between $a$ and $b$ such that $a, A, b$ are in AP, then $A = \frac{a+b}{2}$. This is the simple average we discussed earlier, also known as the single arithmetic mean between $a$ and $b$.
If $a$ and $b$ are two numbers, and $d$ is the common difference of the AP $a, A_1, ..., A_m, b$, then $b$ is the $(m+2)$-th term. $b = a_{(m+2)} = a + (m+2-1)d = a + (m+1)d$. So, $b-a = (m+1)d \implies d = \frac{b-a}{m+1}$. The $m$ arithmetic means are $A_1 = a+d, A_2 = a+2d, ..., A_m = a+md$.
Geometric Progression
A Geometric Progression (GP) is another important type of sequence, characterized by a constant ratio between consecutive terms. Unlike an AP where a fixed value is added, in a GP, a fixed value is multiplied to get the next term. This constant value is called the common ratio, denoted by the letter $r$.
Definition and General Form of a GP
A sequence of non-zero numbers $a_1, a_2, a_3, ..., a_n, ...$ is said to be a Geometric Progression if the ratio of any term to its preceding term is constant for all positive integers $n$.
Mathematically, this means $\frac{a_{n+1}}{a_n} = r$ for all $n \geq 1$, where $r$ is a constant and $r \neq 0$.
If the first term of a GP is denoted by $a$ (or $a_1$), and the common ratio is $r$, then the terms of the GP can be generated sequentially by multiplying the previous term by $r$:
- First term ($a_1$) $= a$
- Second term ($a_2$) $= a_1 \times r = a \times r = ar$
- Third term ($a_3$) $= a_2 \times r = (ar) \times r = ar^2$
- Fourth term ($a_4$) $= a_3 \times r = (ar^2) \times r = ar^3$
- ... and so on.
Thus, the general form of a GP is:
$$a, ar, ar^2, ar^3, ..., ar^{n-1}, ...$$
... (43)
It is important that all terms in a GP are non-zero. If the first term $a=0$, all terms would be 0, which trivializes the concept of a ratio. If the common ratio $r=0$, all terms after the first would be 0, which also loses the characteristic pattern unless specified as a special case. Therefore, we typically assume $a \neq 0$ and $r \neq 0$ for a GP.
The common ratio $r$ can be positive or negative, leading to different types of GPs (e.g., terms alternating in sign if $r$ is negative).
Identifying a GP
To check if a given sequence of non-zero numbers is a GP, calculate the ratio of each term to its preceding term. If this ratio is constant throughout the sequence, it is a GP with that constant ratio $r$.
Example: The sequence 2, 6, 18, 54, 162, ...
$\frac{6}{2} = 3$
$\frac{18}{6} = 3$
$\frac{54}{18} = 3$
$\frac{162}{54} = 3$
Since the ratio is constantly 3, it is a GP with first term $a=2$ and common ratio $r=3$.
Example: The sequence 16, -8, 4, -2, 1, ...
$\frac{-8}{16} = -\frac{1}{2}$
$\frac{4}{-8} = -\frac{1}{2}$
$\frac{-2}{4} = -\frac{1}{2}$
$\frac{1}{-2} = -\frac{1}{2}$
Since the ratio is constantly $-\frac{1}{2}$, it is a GP with first term $a=16$ and common ratio $r=-\frac{1}{2}$. The terms alternate in sign.
nth Term (General Term) of a GP
The formula for the $n$-th term of a GP, denoted by $a_n$, provides a direct method to calculate any term in the sequence using its position number $n$, the first term $a$, and the common ratio $r$.
Derivation of the Formula for $a_n$:
Let the first term of the GP be $a_1 = a$, and the common ratio be $r$.
By the definition of a GP, each subsequent term is obtained by multiplying the previous term by the common ratio $r$:
$$a_2 = a_1 \times r = a \times r = ar^{1}$$
... (a)
$$a_3 = a_2 \times r = (ar) \times r = ar^2$$
... (b)
$$a_4 = a_3 \times r = (ar^2) \times r = ar^3$$
... (c)
We observe a pattern in the exponent of the common ratio $r$: the exponent is always one less than the term number.
- For the 2nd term ($n=2$), the exponent of $r$ is 1 ($=2-1$).
- For the 3rd term ($n=3$), the exponent of $r$ is 2 ($=3-1$).
- For the 4th term ($n=4$), the exponent of $r$ is 3 ($=4-1$).
Following this pattern, for the $n$-th term ($a_n$), the exponent of $r$ will be $(n-1)$.
Thus, the formula for the $n$-th term is:
$$a_n = ar^{n-1}$$
... (44)
Where $a$ is the first term, $r$ is the common ratio, and $n$ is the position of the term in the sequence ($n = 1, 2, 3, ...$).
For $n=1$, $a_1 = ar^{1-1} = ar^0 = a \times 1 = a$ (assuming $r \neq 0$), which matches the first term.
Example 1. Find the 7th term of the GP: 3, 6, 12, ...
Answer:
The given GP is 3, 6, 12, ...
The first term is $a = a_1 = 3$.
To find the common ratio $r$, divide any term by its preceding term:
$$r = \frac{a_2}{a_1} = \frac{6}{3} = 2$$
Check with the next pair: $\frac{a_3}{a_2} = \frac{12}{6} = 2$. The common ratio is indeed 2.
We need to find the 7th term of this GP, so $n = 7$.
Using the formula for the $n$-th term of a GP: $a_n = ar^{n-1}$ (Equation 44).
Substitute $a=3$, $r=2$, and $n=7$:
$$a_7 = 3 \times 2^{7-1}$$
Simplify the exponent and calculate the power:
$$a_7 = 3 \times 2^6$$
$$a_7 = 3 \times 64$$
[Since $2^6 = 64$]
Perform the multiplication:
$$a_7 = 192$$
... (i)
The 7th term of the GP is 192.
Sum of the First n Terms of a GP
The sum of the first $n$ terms of a Geometric Progression is denoted by $S_n$.
Derivation of the Formula for $S_n$:
Let the GP be $a_1, a_2, ..., a_n$, with first term $a_1 = a$ and common ratio $r$.
The terms are $a, ar, ar^2, ..., ar^{n-1}$.
The sum $S_n$ is the sum of these $n$ terms:
$$S_n = a + ar + ar^2 + ... + ar^{n-2} + ar^{n-1}$$
... (a)
Multiply equation (a) by the common ratio $r$. Shift the terms in the sum to align like terms:
$$r S_n = \quad \quad ar + ar^2 + ar^3 + ... + ar^{n-1} + ar^n$$
... (b)
Now, subtract equation (b) from equation (a). Notice that many terms will cancel out:
$$S_n - r S_n = (a + ar + ar^2 + ... + ar^{n-1}) - (ar + ar^2 + ar^3 + ... + ar^n)$$$$
$$S_n - r S_n = a + (\cancel{ar - ar}) + (\cancel{ar^2 - ar^2}) + ... + (\cancel{ar^{n-1} - ar^{n-1}}) - ar^n$$
This leaves only the first term of (a) and the last term of (b):
$$S_n(1 - r) = a - ar^n$$
Factor out $a$ on the right side:
$$S_n(1 - r) = a(1 - r^n)$$
Case 1: If $r \neq 1$, we can divide both sides by $(1-r)$ to get the formula for $S_n$:
$$S_n = \frac{a(1 - r^n)}{1 - r} \quad (\text{for } r \neq 1)$$
... (45a)
This formula can also be written as $\frac{a(r^n - 1)}{r - 1}$ by multiplying the numerator and denominator by -1. This alternative form is often preferred when $|r| > 1$ because it keeps the denominator positive, but both formulas are algebraically equivalent for $r \neq 1$.
$$S_n = \frac{a(r^n - 1)}{r - 1} \quad (\text{for } r \neq 1)$$
... (45b)
Case 2: If $r = 1$, the GP is $a, a \times 1, a \times 1^2, ..., a \times 1^{n-1}$, which simplifies to $a, a, a, ..., a$. In this case, the sum of the first $n$ terms is simply the sum of $n$ terms each equal to $a$.
$$S_n = a + a + a + ... + a \text{ (n times)}$$$$
$$S_n = na \quad (\text{for } r = 1)$$
... (45c)
So, there are two formulas for the sum of the first $n$ terms of a GP, depending on whether the common ratio $r$ is equal to 1 or not.
Example 2. Find the sum of the first 5 terms of the GP: 4, 12, 36, ...
Answer:
The given GP is 4, 12, 36, ...
The first term is $a = 4$.
The common ratio is $r = \frac{a_2}{a_1} = \frac{12}{4} = 3$. (Check: $\frac{a_3}{a_2} = \frac{36}{12} = 3$. The common ratio is indeed 3).
We need to find the sum of the first 5 terms, so $n = 5$.
Since the common ratio $r=3$, which is not equal to 1, we use one of the formulas for $r \neq 1$. Let's use $S_n = \frac{a(r^n - 1)}{r - 1}$ (Equation 45b) since $r > 1$.
Substitute $a=4$, $r=3$, and $n=5$ into the formula:
$$S_5 = \frac{4(3^5 - 1)}{3 - 1}$$
Calculate the exponent and the denominator:
$$S_5 = \frac{4(243 - 1)}{2}$$
[Since $3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 = 81 \times 3 = 243$]
Simplify the numerator:
$$S_5 = \frac{4(242)}{2}$$
Simplify further:
$$S_5 = \cancel{4}^{2} \times \frac{242}{\cancel{2}_{1}} = 2 \times 242$$
Perform the multiplication:
$$S_5 = 484$$
... (i)
The sum of the first 5 terms of the given GP is 484.
Example 3. Find the sum of the first 4 terms of the GP: 10, -5, 2.5, ...
Answer:
The given GP is 10, -5, 2.5, ...
The first term is $a = 10$.
The common ratio is $r = \frac{a_2}{a_1} = \frac{-5}{10} = -\frac{1}{2}$. (Check: $\frac{a_3}{a_2} = \frac{2.5}{-5} = \frac{5/2}{-5} = \frac{5}{-10} = -\frac{1}{2}$. The common ratio is indeed $-\frac{1}{2}$).
We need to find the sum of the first 4 terms, so $n = 4$.
Since the common ratio $r = -\frac{1}{2}$, which is not equal to 1, we use one of the formulas for $r \neq 1$. Let's use $S_n = \frac{a(1 - r^n)}{1 - r}$ (Equation 45a) since $|r| = |-\frac{1}{2}| = \frac{1}{2} < 1$.
Substitute $a=10$, $r=-\frac{1}{2}$, and $n=4$ into the formula:
$$S_4 = \frac{10(1 - (-\frac{1}{2})^4)}{1 - (-\frac{1}{2})}$$
Calculate the exponent and the denominator:
$$S_4 = \frac{10(1 - \frac{1}{16})}{1 + \frac{1}{2}}$$$$
[Since $(-\frac{1}{2})^4 = (-\frac{1}{2}) \times (-\frac{1}{2}) \times (-\frac{1}{2}) \times (-\frac{1}{2}) = \frac{1}{16}$]
Simplify the numerator and the denominator:
$$S_4 = \frac{10(\frac{16-1}{16})}{(\frac{2+1}{2})} = \frac{10(\frac{15}{16})}{(\frac{3}{2})}$$$$
Rewrite the division as multiplication by the reciprocal:
$$S_4 = 10 \times \frac{15}{16} \times \frac{2}{3}$$$$
Simplify by cancelling common factors:
$$S_4 = \cancel{10}^{5} \times \frac{\cancel{15}^{5}}{\cancel{16}_{8}} \times \frac{\cancel{2}^{1}}{\cancel{3}_{1}} = 5 \times \frac{5}{8} \times 1$$
Perform the multiplication:
$$S_4 = \frac{25}{8}$$
... (i)
The sum of the first 4 terms of the GP is $\frac{25}{8}$ or 3.125.
Sum of Infinite Terms of a GP ($S_\infty$)
For an infinite Geometric Progression ($a + ar + ar^2 + ar^3 + ...$), the sum of all infinite terms, denoted by $S_\infty$, exists and is a finite value only if the absolute value of the common ratio $r$ is strictly less than 1 (i.e., $|r| < 1$ or, equivalently, $-1 < r < 1$).
If $|r| < 1$, as the number of terms $n$ approaches infinity ($n \to \infty$), the term $r^n$ approaches 0 ($r^n \to 0$). Looking at the formula for $S_n = \frac{a(1 - r^n)}{1 - r}$, as $n$ becomes very large, $r^n$ becomes very small and approaches 0.
$$\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} \frac{a(1 - r^n)}{1 - r}$$
... (a)
Since $\lim\limits_{n \to \infty} r^n = 0$ when $|r| < 1$:
$$S_\infty = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r} \quad (\text{for } |r| < 1)$$
... (46)
This formula gives the sum of an infinite GP when it converges (i.e., when $|r| < 1$).
If $|r| \geq 1$ (and $r \neq 1$), the terms of the GP do not approach zero, and the sum of the infinite series grows infinitely large or oscillates. In this case, the infinite series diverges and does not have a finite sum. (If $r=1$, the sum is infinite unless $a=0$).
Example 4. Find the sum of the infinite GP: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$
Answer:
The given infinite GP is $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$
The first term is $a = 1$.
The common ratio is $r = \frac{1/2}{1} = \frac{1}{2}$. (Check: $\frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2}$).
We check the condition for convergence: $|r| = |\frac{1}{2}| = \frac{1}{2}$. Since $\frac{1}{2} < 1$, the sum to infinity exists and is finite.
Using the formula for the sum of an infinite GP: $S_\infty = \frac{a}{1 - r}$ (Equation 46).
Substitute $a=1$ and $r=\frac{1}{2}$:
$$S_\infty = \frac{1}{1 - \frac{1}{2}}$$$$
Simplify the denominator:
$$S_\infty = \frac{1}{\frac{2-1}{2}} = \frac{1}{\frac{1}{2}}$$$$
Calculate the value:
$$S_\infty = 1 \times \frac{2}{1} = 2$$
... (i)
The sum of the infinite GP is 2.
Properties of Geometric Progression
Geometric Progressions have several useful properties:
If each term of a GP is multiplied or divided by a non-zero constant, the resulting sequence is also a GP with the same common ratio.
Example: If 2, 6, 18, ... is a GP (a=2, r=3), then multiplying by 5 gives 10, 30, 90, ..., which is a GP (a=10, r=3).
The reciprocals of the terms of a GP also form a GP. If the original GP has common ratio $r$, the sequence of reciprocals will have a common ratio $1/r$.
Example: If 2, 4, 8, 16 is a GP (r=2), then the reciprocals are $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}$, which form a GP with common ratio $\frac{1/4}{1/2} = \frac{1}{2} = 1/2$.
If three non-zero numbers $a, b, c$ are in GP, then the middle term $b$ is the geometric mean of the other two terms. $b^2 = ac$.
Example: 3, 6, 12 are in GP. $6^2 = 36$, $3 \times 12 = 36$. So $6^2 = 3 \times 12$.
If $a_1, a_2, ..., a_n$ is a GP with positive terms, then $\log a_1, \log a_2, ..., \log a_n$ is an AP.
Explanation: Let $a_k = ar^{k-1}$. Then $\log a_k = \log(ar^{k-1}) = \log a + \log(r^{k-1}) = \log a + (k-1)\log r$. This term is in the form $A + (k-1)D$, where $A = \log a$ and $D = \log r$. This is the general term of an AP with first term $\log a$ and common difference $\log r$.
Geometric Mean (GM)
When we insert numbers between two terms of a GP such that the entire sequence remains in GP, the inserted numbers are called Geometric Means. If $G_1, G_2, ..., G_m$ are $m$ numbers inserted between $a$ and $b$ such that $a, G_1, G_2, ..., G_m, b$ form a GP, then these $G_i$ are $m$ geometric means between $a$ and $b$.
If we insert a single number $G$ between $a$ and $b$ such that $a, G, b$ are in GP, then $\frac{G}{a} = \frac{b}{G}$, which implies $G^2 = ab$. For positive $a$ and $b$, $G = \sqrt{ab}$. This is the geometric mean of $a$ and $b$.
If $a$ and $b$ are two numbers, and $r$ is the common ratio of the GP $a, G_1, ..., G_m, b$, then $b$ is the $(m+2)$-th term. $b = a_{(m+2)} = ar^{m+2-1} = ar^{m+1}$. So, $\frac{b}{a} = r^{m+1} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{m+1}}$. The $m$ geometric means are $G_1 = ar, G_2 = ar^2, ..., G_m = ar^m$.
Applications of AP and GP
Arithmetic Progressions (AP) and Geometric Progressions (GP) are not merely theoretical constructs studied in mathematics classrooms. They are powerful tools for modeling and analyzing various phenomena in the real world. Understanding how quantities change additively (AP) or multiplicatively (GP) allows us to solve problems in finance, biology, physics, economics, and many other fields.
Applications of Arithmetic Progression (AP)
Arithmetic Progression is applicable in situations where a quantity increases or decreases by a fixed amount in each successive step, time period, or iteration.
Simple Interest: Simple interest is calculated only on the initial principal amount. This means the amount of interest earned remains constant for each interest period (e.g., each year). If this constant interest amount is added to the principal at the end of each period, the total amount (Principal + accumulated Simple Interest) at the end of successive periods forms an AP.
Let Principal amount be $P$.
Let Rate of simple interest be $R$ per annum (as a percentage).
Simple Interest for 1 year $= \frac{P \times R \times 1}{100} = \frac{PR}{100}$.
Amount at the beginning (Year 0) $= P$.
Amount after 1 year $= P + \frac{PR}{100}$.
Amount after 2 years $= P + \frac{PR}{100} + \frac{PR}{100} = P + 2 \left(\frac{PR}{100}\right)$.
Amount after 3 years $= P + 3 \left(\frac{PR}{100}\right)$.
Amount after $n$ years $= P + n \left(\frac{PR}{100}\right)$.
The sequence of amounts at the end of Year 0, Year 1, Year 2, ... is $P, P + \frac{PR}{100}, P + 2\frac{PR}{100}, P + 3\frac{PR}{100}, ..., P + n\frac{PR}{100}$.
This sequence is an AP with the first term (Year 0) as $P$ and the common difference $d = \frac{PR}{100}$. The amount after $n$ years is the $(n+1)$-th term of this AP (starting from the 0th term $a_0=P$) or it's the $n$-th term of the sequence of *amounts accumulated* starting from Year 1, $a_n = (P + \frac{PR}{100}) + (n-1)\frac{PR}{100} = P + \frac{PR}{100} + n\frac{PR}{100} - \frac{PR}{100} = P + n\frac{PR}{100}$.
Linear Growth or Decay: Many natural or artificial processes involve a quantity changing by a fixed amount per unit of time or per step. Examples include:
- The height of a plant increasing by a constant amount each week.
- The distance covered by an object moving with constant acceleration (displacement in equal time intervals).
- Depreciation of an asset by a fixed amount each year (straight-line depreciation).
- The balance in a bank account where a fixed amount is deposited or withdrawn regularly (ignoring interest).
Equal Instalments: In some financial arrangements like loan repayments or recurring deposits with fixed principal reduction/addition, the series of payments or balances might follow an AP or relate to the sum of an AP.
Example 1. A person saves ₹500 in the first month, ₹600 in the second month, ₹700 in the third month, and so on. Find the total amount saved in 12 months.
Answer:
The amount saved each month forms a sequence:
Monthly Savings: 500, 600, 700, ...
Let's check if this is an AP by finding the difference between consecutive terms:
$600 - 500 = 100$
$700 - 600 = 100$
The difference is constant, which is ₹100. So, the monthly savings form an Arithmetic Progression.
The first term of this AP is $a = 500$.
The common difference is $d = 100$.
We need to find the total amount saved in 12 months. This is the sum of the first 12 terms of this AP. So, $n = 12$.
We use the formula for the sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}(2a + (n-1)d)$ (Equation 42).
Substitute $n=12$, $a=500$, and $d=100$ into the formula:
$$S_{12} = \frac{12}{2}(2 \times 500 + (12 - 1) \times 100)$$$$
Simplify the expression:
$$S_{12} = 6(1000 + 11 \times 100)$$$$
$$S_{12} = 6(1000 + 1100)$$$$
$$S_{12} = 6(2100)$$$$
Perform the multiplication:
$$S_{12} = 12600$$
... (i)
The total amount saved in 12 months is ₹12,600.
Applications of Geometric Progression (GP)
Geometric Progression is applicable in situations where a quantity increases or decreases by a constant percentage or factor in each successive step, time period, or iteration.
Compound Interest: Compound interest is calculated on the principal amount as well as the accumulated interest from previous periods. This leads to the amount growing by a constant multiplication factor each period. The amounts at the end of successive periods form a GP.
Let Principal amount be $P$.
Let Rate of compound interest be $R$ per annum (as a percentage), compounded annually.
The growth factor per year is $(1 + \frac{R}{100})$. Let $r = 1 + \frac{R}{100}$.
Amount at the beginning (Year 0) $= P$. (We can consider this as $a_0$).
Amount after 1 year $= P \times r = Pr$. (This is $a_1$ if $P$ is $a_0$, or $a_1$ if we start sequence from Year 1 amount).
Amount after 2 years $= (Pr) \times r = Pr^2$.
Amount after 3 years $= (Pr^2) \times r = Pr^3$.
Amount after $n$ years $= Pr^n = P \left(1 + \frac{R}{100}\right)^n$.
The sequence of amounts at the end of Year 0, Year 1, Year 2, ... is $P, P(1+\frac{R}{100}), P(1+\frac{R}{100})^2, ..., P(1+\frac{R}{100})^n$. This is a GP with first term $P$ and common ratio $(1+\frac{R}{100})$. The amount after $n$ years is the $(n+1)$-th term ($a_n$) of this sequence (if we start counting terms from $n=0$ for the initial principal) or the $n$-th term of the sequence of amounts starting from Year 1 as the first term ($a_1$).
Population Growth or Decay: If a population of people, animals, or bacteria increases or decreases by a fixed percentage each year or generation, the population size over time can be modeled using a GP.
Depreciation: If the value of an asset (like machinery, vehicles) decreases by a fixed percentage each year (reducing balance method of depreciation), its value over successive years forms a GP.
Radioactive Decay: The amount of a radioactive substance decreases by a fixed percentage over equal periods (half-life concept is related to GP with $r = 1/2$).
Example 2. A sum of ₹20,000 is invested at 10% per annum compound interest, compounded annually. Find the amount after 3 years.
Answer:
Given:
Principal amount, $P = \textsf{₹} 20,000$.
Annual compound interest rate, $R = 10\%$.
Number of years, $n = 3$.
The amounts at the end of each year form a GP. The common ratio is the growth factor per year, $r = (1 + \frac{R}{100})$.
$$r = 1 + \frac{10}{100} = 1 + 0.10 = 1.10$$
The amount after $n$ years ($A_n$) with initial principal $P$ is given by the compound interest formula, which is the $(n+1)$-th term of a GP starting with $P$ or the $n$-th term of a GP starting with the amount after 1 year. It is commonly expressed as:
$$\text{Amount after } n \text{ years} = P \left(1 + \frac{R}{100}\right)^n = P r^n$$
We need to find the amount after 3 years, so $n=3$. Substitute $P = 20000$ and $r = 1.10$:
$$A_3 = 20000 (1.10)^3$$
Calculate $(1.10)^3 = 1.1 \times 1.1 \times 1.1 = 1.21 \times 1.1 = 1.331$.
$$A_3 = 20000 \times 1.331$$
[Since $(1.10)^3 = 1.331$]
Perform the multiplication:
$$A_3 = 26620$$
... (i)
The amount after 3 years will be ₹26,620.
Example 3. A population of bacteria doubles every hour. If there are initially 100 bacteria, how many bacteria will there be after 5 hours?
Answer:
Given:
Initial number of bacteria $= 100$.
The population doubles every hour, which means it is multiplied by a factor of 2 each hour. This indicates a Geometric Progression.
Let the initial population be the first term (or 0th term). Let $a=100$.
The common ratio is $r=2$ (since it doubles). The time is measured in hours, and the growth happens every hour.
Population after 1 hour $= 100 \times 2 = 200$.
Population after 2 hours $= 200 \times 2 = 400$.
Population after 3 hours $= 400 \times 2 = 800$.
This sequence of population counts at the end of each hour (starting from hour 0) is 100, 200, 400, 800, ...
We can model this using the GP formula $a_n = ar^{n-1}$, where $a$ is the initial population and $n$ is the number of time periods (hours in this case) PLUS ONE (if we consider the initial population as $a_1$) or $a_n = ar^n$ if we consider $n$ as the number of periods elapsed from time 0 (starting with $a_0 = a$). Using the latter, $a_n$ is the population after $n$ hours.
Initial population (at time $n=0$) $= a_0 = 100$.
Population after $n$ hours $= a_n = a_0 \times r^n$.
We need to find the population after 5 hours, so $n=5$. Substitute $a_0=100$ and $r=2$:
$$a_5 = 100 \times 2^5$$
Calculate $2^5 = 32$.
$$a_5 = 100 \times 32$$
Perform the multiplication:
$$a_5 = 3200$$
... (i)
After 5 hours, there will be 3200 bacteria.