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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Terms Related to Arithmetic Progression | General Term of an AP | Sum of First n Terms of an AP |
Chapter 5 Arithmetic Progressions (Concepts)
Welcome to the study of sequences and series! This chapter introduces a fundamental and widely applicable type of sequence known as an Arithmetic Progression, commonly abbreviated as AP. While we encounter various lists of numbers (sequences) in mathematics, an AP is characterized by a remarkably simple and consistent pattern of growth or decay. Understanding APs provides a foundation for analyzing patterns, calculating sums efficiently, and modeling numerous real-world situations involving regular increments or decrements.
An Arithmetic Progression (AP) is formally defined as a sequence of numbers where each term, starting from the second term, is obtained by adding a fixed constant number to the term immediately preceding it. This fixed number is the defining characteristic of the AP and is called the common difference, universally denoted by the letter $d$. The very first number in the sequence is called the first term, typically denoted by $a$. Given the first term $a$ and the common difference $d$, the general form of an arithmetic progression unfolds as follows: $$ a, \quad a+d, \quad a+2d, \quad a+3d, \quad a+4d, \quad \dots $$ The common difference $d$ can be positive (indicating an increasing AP), negative (a decreasing AP), or zero (a constant sequence).
A key objective is to develop formulas that allow us to easily find any specific term in the sequence without listing all preceding terms, and also to efficiently calculate the sum of a certain number of terms. Two crucial formulas are derived and utilized extensively:
- The $n^{th}$ Term (General Term) of an AP ($a_n$): This formula allows us to directly calculate the value of the term at any position 'n' in the sequence. It is given by: $$ \mathbf{a_n = a + (n - 1)d} $$ Here, $a_n$ is the $n^{th}$ term, $a$ is the first term, $n$ is the term number (position in the sequence), and $d$ is the common difference.
- The Sum of the First $n$ Terms of an AP ($S_n$): This formula calculates the sum of all terms from the first term up to the $n^{th}$ term. Two equivalent forms of the formula exist, useful in different scenarios:
- When the first term ($a$), common difference ($d$), and number of terms ($n$) are known: $$ \mathbf{S_n = \frac{n}{2} [2a + (n - 1)d]} $$
- When the first term ($a$), the $n^{th}$ term ($a_n$, also often denoted as the last term 'l' if summing the entire finite sequence), and the number of terms ($n$) are known: $$ \mathbf{S_n = \frac{n}{2} [a + a_n]} \quad \text{or} \quad \mathbf{S_n = \frac{n}{2} [a + l]} $$
A primary skill developed in this chapter is the ability to identify whether a given sequence of numbers constitutes an AP. This is done by checking if the difference between consecutive terms is constant throughout the sequence. If $a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots$, then the sequence is an AP with that constant difference as $d$. Once an AP is identified, or its parameters ($a, d, n, a_n, S_n$) are partially given, we practice finding the unknown quantities. This often involves setting up and solving equations (usually linear, but sometimes quadratic when dealing with $S_n$ formulas involving $n^2$) derived directly from the $a_n$ and $S_n$ formulas.
The chapter strongly emphasizes the application of these concepts to solve practical problems. Arithmetic progressions naturally model situations involving consistent, additive change. Examples include:
- Calculating the salary in a future year given a starting salary and a fixed annual increment.
- Determining the total amount saved over a period with a regular saving pattern.
- Calculating the total amount repaid under certain installment plans.
- Finding the number of objects arranged in rows where each row has a fixed number more (or less) than the previous row (e.g., logs stacked, seats in an auditorium).
Basic Terms Related to Arithmetic Progression
In mathematics, we often work with lists of numbers that follow a specific, predictable pattern. These ordered lists are called sequences. One of the most common and important types of sequences is the Arithmetic Progression, which is built on the simple, consistent pattern of adding or subtracting the same number over and over again.
Sequence and Terms
A sequence is an arrangement of numbers in a definite order, according to some rule. Each number in the sequence is called a term. Think of it as a train where each car is a term, and the cars are arranged in a specific order.
For example, in the sequence of odd numbers: 1, 3, 5, 7, 9, ...
- The first term, denoted as $a_1$, is 1.
- The second term, denoted as $a_2$, is 3.
- The third term, denoted as $a_3$, is 5, and so on.
The "..." at the end indicates that the sequence is infinite, meaning it continues forever. A sequence with a specific end point is called a finite sequence (e.g., 1, 3, 5, 7, 9).
Arithmetic Progression (AP)
An Arithmetic Progression (AP) is a special type of sequence where each term after the first is obtained by adding a fixed number to the previous term. This fixed number is called the common difference.
Imagine walking up or down a staircase where every step is the exact same height. The position of each step is a term in the sequence, and the height of each step is the common difference.
Key Components of an AP
- First Term ($a$ or $a_1$): This is the starting point of the progression. In our staircase analogy, this is the level you start on.
- Common Difference ($d$): This is the fixed amount you add to get from one term to the next. It's the "step size" of the staircase. To find the common difference, simply subtract any term from the term that immediately follows it. This difference must be the same for all consecutive pairs.
$d = (\text{Any Term}) - (\text{Its Previous Term})$
For example: $d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots$
General Form of an AP
Any AP can be written in a general form using its first term 'a' and common difference 'd'. This shows the underlying structure of the progression:
$a, \quad a+d, \quad a+2d, \quad a+3d, \quad \dots$
- First Term ($a_1$) = $a$
- Second Term ($a_2$) = $a + 1d$
- Third Term ($a_3$) = $a + 2d$
- Fourth Term ($a_4$) = $a + 3d$
Notice that for any term, the number multiplying 'd' is always one less than the term number.
Types of AP based on the Common Difference ($d$)
- Increasing AP ($d > 0$): The terms continuously get larger. This is like climbing up the staircase.
Example: 3, 7, 11, 15, ... (Here $d = 4$) - Decreasing AP ($d < 0$): The terms continuously get smaller. This is like walking down the staircase.
Example: 20, 17, 14, 11, ... (Here $d = -3$) - Constant AP ($d = 0$): All the terms are the same. This is like walking on a flat surface.
Example: 5, 5, 5, 5, ... (Here $d = 0$)
Worked Examples
Example 1. Check if the sequence 1, -1, -3, -5, ... is an AP. If it is, find the common difference and the first term.
Answer:
Solution
The given sequence is 1, -1, -3, -5, ...
The terms are $a_1 = 1, a_2 = -1, a_3 = -3, a_4 = -5$.
To check if it is an AP, we must calculate the difference between consecutive terms and verify that this difference is constant.
Check 1: Difference between the second and first term.
$a_2 - a_1 = -1 - 1 = -2$
Check 2: Difference between the third and second term.
$a_3 - a_2 = -3 - (-1) = -3 + 1 = -2$
Check 3: Difference between the fourth and third term.
$a_4 - a_3 = -5 - (-3) = -5 + 3 = -2$
Since the difference is constant and equal to -2 in all checks, the sequence is indeed an Arithmetic Progression.
The first term is the starting number of the sequence, which is 1.
Answer: Yes, the sequence is an AP. The first term $a = 1$ and the common difference $d = -2$.
Example 2. Which of the following lists of numbers form an AP? If they form an AP, write the next two terms.
(i) 4, 10, 16, 22, ...
(ii) 1, -1, 2, -2, ...
Answer:
Part (i): 4, 10, 16, 22, ...
Let's check the differences between consecutive terms:
$10 - 4 = 6$
$16 - 10 = 6$
$22 - 16 = 6$
The difference is constant, so this is an AP with a common difference $d=6$.
To find the next two terms, we continue adding the common difference to the last known term.
Next term ($a_5$) = Last term + $d = 22 + 6 = 28$.
Next term ($a_6$) = Previous term + $d = 28 + 6 = 34$.
Answer for (i): Yes, this is an AP. The next two terms are 28 and 34.
Part (ii): 1, -1, 2, -2, ...
Let's check the differences:
$a_2 - a_1 = -1 - 1 = -2$
$a_3 - a_2 = 2 - (-1) = 3$
As soon as we find that two differences are not the same ($-2 \neq 3$), we can stop. There is no common difference.
Answer for (ii): No, this is not an AP because the difference between consecutive terms is not constant.
Example 3. Find the first four terms of an AP whose first term is 10 and common difference is 5.
Answer:
Solution
We are given the starting point, $a = 10$, and the step size, $d = 5$.
We build the sequence term by term:
First term ($a_1$): $a = 10$
Second term ($a_2$): $a_1 + d = 10 + 5 = 15$
Third term ($a_3$): $a_2 + d = 15 + 5 = 20$
Fourth term ($a_4$): $a_3 + d = 20 + 5 = 25$
Answer: The first four terms of the AP are 10, 15, 20, 25.
General Term (n-th Term) of an AP
In an Arithmetic Progression (AP), the terms follow a predictable pattern. While you can find the next term by adding the common difference, this becomes inefficient if you want to find a term far down the sequence, like the 100th term. The formula for the general term, also known as the $n$-th term, gives you a direct and powerful way to calculate any term in an AP simply by knowing its position.
Derivation of the General Term Formula
Let's build the formula by observing the pattern. Let the first term of an AP be $a$ and the common difference be $d$.
- The 1st term is: $a_1 = a$
- The 2nd term is: $a_2 = a_1 + d = a + d$
- The 3rd term is: $a_3 = a_2 + d = (a + d) + d = a + 2d$
- The 4th term is: $a_4 = a_3 + d = (a + 2d) + d = a + 3d$
Let's look at the relationship between the term number and the number multiplying $d$:
- For the 1st term ($a_1$), we add $d$ zero times: $a_1 = a + 0d \ $$ = a + (1-1)d$.
- For the 2nd term ($a_2$), we add $d$ one time: $a_2 = a + 1d \ $$ = a + (2-1)d$.
- For the 3rd term ($a_3$), we add $d$ two times: $a_3 = a + 2d \ $$ = a + (3-1)d$.
- For the 4th term ($a_4$), we add $d$ three times: $a_4 = a + 3d \ $$ = a + (4-1)d$.
The pattern is clear: to get to the $n$-th term, we start with the first term $a$ and add the common difference $d$ a total of $(n-1)$ times.
This gives us the formula for the $n$-th term of an AP:
$a_n = a + (n-1)d$
In this formula:
- $a_n$ is the value of the term at the $n$-th position.
- $a$ is the first term of the AP.
- $n$ is the position of the term in the sequence (e.g., 5th, 20th, 100th). It must be a positive integer.
- $d$ is the common difference.
If an AP is finite and its last term is denoted by $l$, then $l$ is simply the $n$-th term, so we can also write $l = a + (n-1)d$, where $n$ is the total number of terms in the AP.
Worked Examples
Example 1. Find the 10th term of the AP: 2, 7, 12, ...
Answer:
Solution
First, identify the components of the AP from the given sequence 2, 7, 12, ...
First term ($a$): $a = 2$.
Common difference ($d$): $d = 7 - 2 = 5$.
We need to find the 10th term, so the term number ($n$) is 10.
Now, use the formula for the n-th term: $a_n = a + (n-1)d$.
Substitute the values:
$a_{10} = 2 + (10 - 1) \times 5$
$a_{10} = 2 + (9) \times 5$
$a_{10} = 2 + 45$
$a_{10} = 47$
Answer: The 10th term of the AP is 47.
Example 2. Which term of the AP 21, 18, 15, ... is -81? Also, is any term 0? If yes, find its position.
Answer:
Part 1: Which term is -81?
Identify the components of the AP: 21, 18, 15, ...
First term ($a$): $a = 21$.
Common difference ($d$): $d = 18 - 21 = -3$.
We are given the value of a term, $a_n = -81$, and we need to find its position, $n$.
Using the formula $a_n = a + (n-1)d$:
$-81 = 21 + (n-1)(-3)$
Subtract 21 from both sides:
$-81 - 21 = (n-1)(-3)$
$-102 = (n-1)(-3)$
Divide by -3:
$\frac{-102}{-3} = n - 1$
$34 = n - 1$
$n = 35$
Since $n=35$ is a positive integer, -81 is indeed a term in this AP.
Answer to Part 1: The 35th term of the AP is -81.
Part 2: Is any term 0?
We check if there is a positive integer $n$ for which $a_n = 0$.
Using the same formula with $a=21$ and $d=-3$:
$0 = 21 + (n-1)(-3)$
$-21 = (n-1)(-3)$
$\frac{-21}{-3} = n - 1$
$7 = n - 1$
$n = 8$
Since $n=8$ is a positive integer, 0 is a term in this AP.
Answer to Part 2: Yes, the 8th term of the AP is 0.
Example 3. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Answer:
Solution
We are given two terms of an AP, but not the first term ($a$) or the common difference ($d$). We must first find these values.
We can set up a system of two linear equations using the formula $a_n = a + (n-1)d$.
For the 3rd term ($a_3 = 4$):
$a + (3-1)d = 4 \implies a + 2d = 4$
... (1)
For the 9th term ($a_9 = -8$):
$a + (9-1)d = -8 \implies a + 8d = -8$
... (2)
Now we solve this system. Subtract Equation (1) from Equation (2) to eliminate $a$:
$(a + 8d) - (a + 2d) = -8 - 4$
$6d = -12 \implies d = -2$.
Substitute $d = -2$ back into Equation (1) to find $a$:
$a + 2(-2) = 4 \implies a - 4 = 4 \implies a = 8$.
Now we know the first term is $a=8$ and the common difference is $d=-2$. We want to find which term ($n$) is zero ($a_n = 0$).
$a_n = a + (n-1)d$
$0 = 8 + (n-1)(-2)$
$-8 = (n-1)(-2)$
$\frac{-8}{-2} = n - 1$
$4 = n - 1 \implies n = 5$.
Answer: The 5th term of this AP is zero.
Sum of First n Terms of an AP
Often in mathematics and real-world applications, we need to find the total of a certain number of terms in an Arithmetic Progression. For example, calculating the total savings over a year if you save a little more each month, or finding the total distance an object travels if its speed increases steadily. Instead of adding each term one by one, we can use a direct formula to find the sum of the first $n$ terms, denoted as $S_n$.
Derivation of the Sum Formula
The method used to derive this formula is a clever trick famously attributed to the young mathematician Carl Friedrich Gauss. Let's consider an AP with first term $a$, common difference $d$, and $n$ terms. The last term, or the $n$-th term, is $l = a + (n-1)d$.
Step 1: Write the sum in forward order.
The sum, $S_n$, is the sum of all terms from the first ($a$) to the last ($l$).
$S_n = a + (a+d) + (a+2d) + \dots + (l-d) + l$
... (1)
Step 2: Write the sum in reverse order.
Writing the same sum but starting from the last term and going backwards to the first:
$S_n = l + (l-d) + (l-2d) + \dots + (a+d) + a$
... (2)
Step 3: Add the two equations together.
We add Equation (1) and Equation (2) by pairing up the terms vertically:
$$S_n + S_n = [a + l] + [(a+d) + (l-d)] + [(a+2d) + (l-2d)] + \dots + [l + a]$$
$2S_n = (a+l) + (a+l) + (a+l) + \dots + (a+l)$
Notice that in each pair, the '$d$' terms cancel out, leaving just $(a+l)$. Since there are $n$ terms in the sequence, we are adding $(a+l)$ to itself $n$ times.
Step 4: Simplify the result.
$2S_n = n \times (a+l)$
Step 5: Solve for $S_n$.
Dividing by 2 gives us our first formula for the sum:
$S_n = \frac{n}{2}(a+l)$
This formula is useful when you know the first term, the last term, and the number of terms.
To get a more general formula, we can substitute the expression for the last term, $l = a + (n-1)d$, into the formula above:
$S_n = \frac{n}{2}(a + [a + (n-1)d])$
$S_n = \frac{n}{2}(2a + (n-1)d)$
This gives us the second, and most commonly used, formula for the sum:
$S_n = \frac{n}{2}[2a + (n-1)d]$
This formula is useful when you know the first term, the common difference, and the number of terms.
Worked Examples
Example 1. Find the sum of the first 20 terms of the AP: 1, 3, 5, 7, ...
Answer:
Solution
First, identify the known values from the AP.
First term ($a$): $a=1$.
Common difference ($d$): $d = 3 - 1 = 2$.
Number of terms ($n$): $n=20$.
Since we know $a, d,$ and $n$, we use the second formula: $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substitute the values:
$S_{20} = \frac{20}{2}[2(1) + (20-1)(2)]$
$S_{20} = 10[2 + (19)(2)]$
$S_{20} = 10[2 + 38]$
$S_{20} = 10[40] = 400$
Answer: The sum of the first 20 terms is 400.
Example 2. Find the sum of the AP: 7 + 10.5 + 14 + ... + 84.
Answer:
Solution
In this problem, we are given the first and last terms, but not the number of terms ($n$). We must find $n$ first.
First term ($a$): $a=7$.
Common difference ($d$): $d = 10.5 - 7 = 3.5$.
Last term ($l$ or $a_n$): $l = 84$.
Use the n-th term formula, $a_n = a + (n-1)d$, to find $n$.
$84 = 7 + (n-1)(3.5)$
$84 - 7 = (n-1)(3.5)$
$77 = (n-1)(3.5)$
$n-1 = \frac{77}{3.5} = \frac{770}{35} = 22$
$n = 22 + 1 = 23$. So, there are 23 terms in this AP.
Now we can find the sum using the formula $S_n = \frac{n}{2}(a+l)$, as we know $n, a,$ and $l$.
$S_{23} = \frac{23}{2}(7 + 84)$
$S_{23} = \frac{23}{2}(91) = \frac{2093}{2} = 1046.5$
Answer: The sum of the given AP is 1046.5.
Example 3. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Answer:
Solution
We are given the sum and need to find the number of terms ($n$).
First term ($a$): $a=9$.
Common difference ($d$): $d = 17 - 9 = 8$.
Sum ($S_n$): $S_n = 636$.
We use the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ and solve for $n$.
$636 = \frac{n}{2}[2(9) + (n-1)(8)]$
$636 = \frac{n}{2}[18 + 8n - 8]$
$636 = \frac{n}{2}[10 + 8n]$
Multiply both sides by 2:
$1272 = n(10 + 8n)$
$1272 = 10n + 8n^2$
Rearrange into a standard quadratic equation:
$8n^2 + 10n - 1272 = 0$
Divide by 2 to simplify:
$4n^2 + 5n - 636 = 0$
We solve this quadratic equation for $n$ using the quadratic formula, where A=4, B=5, C=-636.
$D = B^2 - 4AC = 5^2 - 4(4)(-636) = 25 + 10176 = 10201$.
$\sqrt{D} = \sqrt{10201} = 101$.
$n = \frac{-B \pm \sqrt{D}}{2A} = \frac{-5 \pm 101}{2(4)} = \frac{-5 \pm 101}{8}$
The two possible values for $n$ are:
$n_1 = \frac{-5 + 101}{8} = \frac{96}{8} = 12$
$n_2 = \frac{-5 - 101}{8} = \frac{-106}{8}$
Since the number of terms ($n$) cannot be negative or a fraction, we reject $n_2$.
Answer: 12 terms must be taken to give a sum of 636.