Introduction

Introduction to Arithmetic Progressions

Concept Introduction

Patterns are everywhere around us — in nature, in our daily routines, in the way money grows, and even in the architecture of buildings. When you observe a honeycomb, the hexagonal cells follow a perfect pattern. When you climb a ladder, the rungs are equally spaced. When you save money every month by depositing the same amount, your total savings follow a predictable sequence.

Consider Reena, who just got her first job with a starting salary of ₹8,000 per month. Every year, she receives an increment of ₹500. So her annual salaries will be: ₹8,000, ₹8,500, ₹9,000, ₹9,500, and so on. Notice how each term is obtained by adding a fixed amount (₹500) to the previous term. This is not a coincidence — it's a fundamental mathematical pattern called an Arithmetic Progression (AP).

Understanding APs helps us predict future values, calculate total savings, design structures with uniform spacing, and solve countless real-world problems. In this chapter, we'll explore how to identify, analyze, and work with these special sequences that form the backbone of many practical applications in finance, architecture, science, and everyday planning.

{{FORMULA: expr=a, a + d, a + 2d, a + 3d, ... | symbols=a:first term, d:common difference}}

What is an Arithmetic Progression?

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number to the preceding term. This fixed number is called the common difference.

General form of an AP:

a, a + d, a + 2d, a + 3d, ...

where a is the first term and d is the common difference.

Key Characteristics

1. Common Difference (d): The value added to each term to get the next term.

2. First Term (a): The starting point of the sequence.

3. Successive Terms: Each term differs from the next by exactly d.

The common difference can be:

• Positive (increasing AP): 2, 5, 8, 11, ... (d = 3)
• Negative (decreasing AP): 20, 15, 10, 5, ... (d = -5)
• Zero (constant AP): 7, 7, 7, 7, ... (d = 0)

Definitions & Formulas

{{KEY: type=concept | title=Minimum Information Needed | text=To completely define an AP, you need exactly TWO pieces of information: the first term (a) and the common difference (d). With these two values, you can generate the entire sequence.}}

How to Identify an Arithmetic Progression

Not every sequence is an AP. To verify whether a given sequence forms an AP, follow these steps:

Step 1: Write down the given sequence of numbers.

Step 2: Calculate the difference between the second and first term.

d₁ = a₂ - a₁

Step 3: Calculate the difference between the third and second term.

d₂ = a₃ - a₂

Step 4: Continue finding differences between consecutive terms.

d₃ = a₄ - a₃, and so on

Step 5: Check if all these differences are equal.

If d₁ = d₂ = d₃ = ... = d, then the sequence is an AP

Step 6: The common value is your common difference d.

Real-Life Examples of AP

Let's examine the examples from the NCERT chapter:

Example (i): Reena's salary progression

• Sequence: 8000, 8500, 9000, 9500, ...
• First term a = 8000
• Common difference d = 8500 - 8000 = 500
• This is an increasing AP

Example (ii): Ladder rung lengths

• Sequence: 45, 43, 41, 39, 37, 35, 33, 31
• First term a = 45
• Common difference d = 43 - 45 = -2
• This is a decreasing finite AP

Example (v): Shakila's money box

• Sequence: 100, 150, 200, 250, ...
• First term a = 100
• Common difference d = 150 - 100 = 50
• This is an increasing AP

Solved Examples

Example 1: Identifying an AP (Easy)

Given: The sequence 3, 7, 11, 15, 19

To Find: Whether this is an AP and if yes, find the common difference

Solution:

1. Calculate the difference between consecutive terms.

d₁ = 7 - 3 = 4

2. Check the next difference.

d₂ = 11 - 7 = 4

3. Check the remaining differences.

d₃ = 15 - 11 = 4
d₄ = 19 - 15 = 4

4. Since all differences are equal, this is an AP.

Common difference d = 4

Final Answer: Yes, it is an AP with d = 4

Example 2: Finding the Common Difference (Medium)

Given: An AP where the first term is -3 and the second term is -1.5

To Find: The common difference and the next three terms

Solution:

1. Use the formula for common difference.

d = a₂ - a₁

2. Substitute the given values.

d = -1.5 - (-3) = -1.5 + 3 = 1.5

3. Find the third term by adding d to the second term.

a₃ = a₂ + d = -1.5 + 1.5 = 0

4. Find the fourth term.

a₄ = a₃ + d = 0 + 1.5 = 1.5

5. Find the fifth term.

a₅ = a₄ + d = 1.5 + 1.5 = 3

Final Answer: d = 1.5; next three terms are 0, 1.5, 3

Example 3: Determining if a Sequence is an AP (Hard)

Given: The sequence 2, 4, 8, 16, 32

To Find: Is this an AP? Justify your answer.

Solution:

1. Calculate the first difference.

d₁ = 4 - 2 = 2

2. Calculate the second difference.

d₂ = 8 - 4 = 4

3. Calculate the third difference.

d₃ = 16 - 8 = 8

4. Compare all differences.

d₁ ≠ d₂ ≠ d₃

5. Since the differences are not constant, this is NOT an AP.

This is a Geometric Progression (GP) with common ratio r = 2

Final Answer: No, this is not an AP because differences are not equal

Example 4: Mixed Number AP (Tricky)

Given: A sequence starts with ½ and has a common difference of ¼

To Find: Write the first 5 terms of this AP

Solution:

1. Identify the first term.

a₁ = ½

2. Calculate the second term by adding the common difference.

a₂ = a₁ + d = ½ + ¼ = ¾

3. Calculate the third term.

a₃ = a₂ + d = ¾ + ¼ = 1

4. Calculate the fourth term.

a₄ = a₃ + d = 1 + ¼ = 1¼

5. Calculate the fifth term.

a₅ = a₄ + d = 1¼ + ¼ = 1½

Final Answer: ½, ¾, 1, 1¼, 1½

Tips & Tricks

{{KEY: type=tip | title=The Middle Term Rule | text=Three numbers a, b, c form an AP if and only if b = (a + c)/2. The middle term is always the average of its neighbors!}}

Common Mistakes

Brain-Teaser Questions

Question 1: The 5th term of an AP is 20 and the 8th term is 32. Can you find the common difference without finding the first term?

💡 Answer: Yes! The difference between 8th and 5th terms spans 3 common differences. So: 32 - 20 = 3d, which gives d = 12/3 = 4.

Question 2: Is the sequence 1, 4, 9, 16, 25, ... (perfect squares) an AP?

💡 Answer: No. Differences are: 3, 5, 7, 9, ... which are not constant. The differences themselves form an AP, but the original sequence does not.

Question 3: Can an AP have all its terms as the same number?

💡 Answer: Yes! When d = 0, all terms are equal to the first term a. For example: 5, 5, 5, 5, ... is an AP with a = 5 and d = 0.

Mini Cheatsheet

This introduction lays the foundation for understanding Arithmetic Progressions. You've learned how to identify APs, calculate the common difference, and recognize patterns in everyday situations. As we progress through this chapter, we'll discover how to find any term in an AP without listing all previous terms, and how to calculate the sum of multiple terms efficiently — powerful tools that unlock solutions to complex real-world problems.

Arithmetic Progressions — Definition and General Form

Page 2: Arithmetic Progressions — Definition and General Form

Concept Introduction

Imagine you're training for a marathon and decide to run a bit more each day. On day 1, you run 2 km. On day 2, you run 2.5 km. On day 3, you run 3 km, and so on. Notice the pattern? Each day, you add exactly 0.5 km to your previous day's distance. This consistent increase creates a special type of number sequence called an Arithmetic Progression (AP).

An Arithmetic Progression is a sequence of numbers 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. Whether you're planning savings, measuring temperature changes over a week, or tracking seat numbers in an auditorium, APs appear everywhere in real life. Understanding their structure helps us predict future terms, calculate sums, and solve practical problems efficiently.

{{FORMULA: expr=aₙ = a + (n - 1)d | symbols=aₙ:nth term, a:first term, n:position of term, d:common difference}}

Definitions & Formulas

{{KEY: type=concept | title=Identifying an AP | text=A sequence is an AP if and only if the difference between EVERY pair of consecutive terms is exactly the same constant value.}}

Understanding the General Form — Step-by-Step Derivation

Let's build the general algebraic representation of an AP from first principles.

Step 1: Start with the first term.

a₁ = a

This is our starting point — the first number in the sequence.

Step 2: Find the second term by adding the common difference once.

a₂ = a₁ + d = a + d

Step 3: Find the third term by adding the common difference to the second term.

a₃ = a₂ + d = (a + d) + d = a + 2d

Step 4: Find the fourth term by adding the common difference to the third term.

a₄ = a₃ + d = (a + 2d) + d = a + 3d

Step 5: Observe the pattern emerging:

• First term: a + 0d
• Second term: a + 1d
• Third term: a + 2d
• Fourth term: a + 3d

The coefficient of d is always one less than the term's position.

Step 6: Generalize to the nth term.

aₙ = a + (n - 1)d

This is the fundamental formula for the nth term of an AP. It tells us the value of ANY term if we know the first term a, the common difference d, and the position n.

Testing for an AP

To verify whether a given sequence is an AP, we must check that the common difference remains constant.

Method: Calculate a₂ - a₁, then a₃ - a₂, then a₄ - a₃, and so on. If all these differences equal the same value, the sequence is an AP.

For the sequence: 5, 8, 11, 14, 17

d₁ = 8 - 5 = 3
d₂ = 11 - 8 = 3
d₃ = 14 - 11 = 3
d₄ = 17 - 14 = 3

Since all differences equal 3, this is an AP with a = 5 and d = 3.

Solved Examples

Example 1: Basic Identification of First Term and Common Difference

Given: AP is 7, 12, 17, 22, 27, ...

To Find: First term a and common difference d

Solution:

1. The first term is simply the first number in the sequence.

a = 7

2. Find the common difference by subtracting the first term from the second term.

d = a₂ - a₁ = 12 - 7 = 5

3. Verify with the next pair (always good practice).

d = a₃ - a₂ = 17 - 12 = 5 ✓

Final Answer: a = 7, d = 5

Example 2: Working with Negative Common Difference

Given: AP is 100, 95, 90, 85, 80, ...

To Find: First term, common difference, and the 10th term

Solution:

1. Identify the first term directly.

a = 100

2. Calculate the common difference.

d = a₂ - a₁ = 95 - 100 = -5

The negative value means the sequence is decreasing.

3. Use the nth term formula to find the 10th term.

a₁₀ = a + (n - 1)d

4. Substitute the known values: a = 100, n = 10, d = -5.

a₁₀ = 100 + (10 - 1)(-5)

5. Simplify step by step.

a₁₀ = 100 + 9(-5) = 100 - 45 = 55

Final Answer: a = 100, d = -5, a₁₀ = 55

Example 3: Determining if a Sequence is an AP

Given: Sequence is 3, 6, 12, 24, 48, ...

To Find: Is this an AP? If not, why?

Solution:

1. Calculate the difference between first and second terms.

d₁ = 6 - 3 = 3

2. Calculate the difference between second and third terms.

d₂ = 12 - 6 = 6

3. Compare the differences.

Since d₁ ≠ d₂ (3 ≠ 6), the differences are not constant.

4. Conclusion: This sequence is NOT an AP. It's actually a geometric progression where each term is multiplied by 2.

Final Answer: Not an AP — common difference is not constant

Example 4: Finding the Position of a Given Term (Tricky)

Given: In an AP, a = -3 and d = 4.

To Find: Which term of this AP equals 57?

Solution:

1. We need to find n such that the nth term equals 57. Use the nth term formula.

aₙ = a + (n - 1)d

2. Substitute the known values: aₙ = 57, a = -3, d = 4.

57 = -3 + (n - 1)(4)

3. Add 3 to both sides to isolate the term with n.

60 = (n - 1)(4)

4. Divide both sides by 4.

15 = n - 1

5. Solve for n.

n = 16

Final Answer: 57 is the 16th term of this AP

Tips & Tricks

Common Mistakes

{{KEY: type=formula | title=The Master Formula | text=aₙ = a + (n - 1)d is your single most important tool. Master it completely — it unlocks 80% of AP problems.}}

Brain-Teaser Questions

Question 1: An AP has its 5th term as 23 and 8th term as 35. What is its first term and common difference?

💡 Answer: Let the first term be a and common difference be d. From the given information: a₅ = a + 4d = 23 and a₈ = a + 7d = 35 Subtract the first equation from the second: (a + 7d) - (a + 4d) = 35 - 23 This gives: 3d = 12, so d = 4 Substitute back: a + 4(4) = 23, so a + 16 = 23, giving a = 7 Answer: a = 7, d = 4

Question 2: The 10th term of an AP is 0. If its common difference is -2, what is its first term?

💡 Answer: Using aₙ = a + (n - 1)d with a₁₀ = 0, n = 10, d = -2: 0 = a + (10 - 1)(-2) 0 = a + 9(-2) 0 = a - 18 a = 18 Answer: a = 18

Question 3: Can the sequence 1, 4, 9, 16, 25, ... (perfect squares) be an AP? Justify.

💡 Answer: Check consecutive differences: 4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, 25 - 16 = 9 The differences are 3, 5, 7, 9 — which are NOT constant. Answer: No, this is NOT an AP because the common difference varies.

Mini Cheatsheet — Screenshot This for Quick Revision

Practice makes perfect! Work through each example above on your own, then attempt the brain-teasers without looking at the answers. Understanding the why behind each step is far more valuable than memorizing formulas.

Arithmetic Progressions — Identifying APs and Properties

Chapter 5: Arithmetic Progressions

Identifying APs and Their Properties

In our previous discussions, we observed various patterns in lists of numbers drawn from everyday life. Some lists grow by adding a fixed amount, while others might involve multiplication or squaring. In this section, we will delve deeper into the specific pattern known as an Arithmetic Progression (AP). Our focus will be on understanding its core properties and learning how to identify if a given list of numbers forms an AP.

What is an Arithmetic Progression?

Let's formally define what we mean by an AP.

An Arithmetic Progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except for the first term.

This fixed number is the cornerstone of an AP and is called the common difference, denoted by the letter d. The common difference can be a positive number, a negative number, or zero.

Let's denote the terms of an AP as follows:

• First term = a₁ (often simply written as a)
• Second term = a₂
• Third term = a₃ ... and so on.

The defining property of an AP is that the difference between any two consecutive terms is constant. That is: a₂ – a₁ = d a₃ – a₂ = d a₄ – a₃ = d and so on.

In general, for any term aₖ and its next term aₖ₊₁, we have: d = aₖ₊₁ – aₖ

The nature of the common difference d tells us how the progression behaves:

To fully describe an AP, we only need to know two key pieces of information:

1. The first term (a)
2. The common difference (d)

With these two values, we can generate the entire progression.

How to Check if a List of Numbers is an AP

To determine if a given sequence of numbers is an Arithmetic Progression, you must verify that the difference between every pair of consecutive terms is the same.

Method:

1. List the terms: Let the given list of numbers be a₁, a₂, a₃, a₄, . . .
2. Calculate the differences: Find the difference between consecutive terms.
   • Calculate d₁ = a₂ – a₁
   • Calculate d₂ = a₃ – a₂
   • Calculate d₃ = a₄ – a₃
   • ...and so on for at least two or three pairs.
3. Compare and Conclude:
   • If all the calculated differences are equal (i.e., d₁ = d₂ = d₃ = ...), then the list of numbers forms an AP. The constant value is the common difference, d.
   • If even one of the differences is not the same as the others, the list does not form an AP.

Important Note: Always subtract the preceding (kth) term from the succeeding ((k+1)th) term, even if the succeeding term is smaller. The formula is always d = aₖ₊₁ – aₖ.

Worked Examples

Let's apply this method to a few examples to solidify our understanding.

Example 1: Does the list of numbers 4, 10, 16, 22, ... form an AP? If so, find the first term and the common difference.

Solution:

1. Identify the terms:

   • a₁ = 4
   • a₂ = 10
   • a₃ = 16
   • a₄ = 22

2. Calculate the differences:

   • a₂ – a₁ = 10 – 4 = 6
   • a₃ – a₂ = 16 – 10 = 6
   • a₄ – a₃ = 22 – 16 = 6

3. Compare and Conclude: The difference between each pair of consecutive terms is constant and equal to 6. Therefore, the list is an AP.

   • The first term (a) is 4.
   • The common difference (d) is 6.

Example 2: Check whether the list 1, -1, -3, -5, ... is an AP. If it is, write its first term and the common difference.

Solution:

1. Identify the terms:

   • a₁ = 1
   • a₂ = -1
   • a₃ = -3
   • a₄ = -5

2. Calculate the differences:

   • a₂ – a₁ = (-1) – 1 = -2
   • a₃ – a₂ = (-3) – (-1) = -3 + 1 = -2
   • a₄ – a₃ = (-5) – (-3) = -5 + 3 = -2

3. Compare and Conclude: Since the difference is constant (-2), the list is an AP.

   • The first term (a) is 1.
   • The common difference (d) is -2.

Example 3: Check whether the list 1, 1, 2, 3, 5, 8, ... is an AP.

Solution:

1. Identify the terms:

   • a₁ = 1
   • a₂ = 1
   • a₃ = 2
   • a₄ = 3

2. Calculate the differences:

   • a₂ – a₁ = 1 – 1 = 0
   • a₃ – a₂ = 2 – 1 = 1

3. Compare and Conclude: Here, the first difference (0) is not equal to the second difference (1). We do not need to check any further. Since the difference between consecutive terms is not constant, the list is not an AP.

Example 4: For the AP: 3/2, 1/2, -1/2, -3/2, ..., write the first term a and the common difference d.

Solution:

The problem states that this is an AP, so we do not need to check. We just need to find a and d.

1. Find the first term (a): The first term is simply the first number in the list.

   • a = 3/2

2. Find the common difference (d): We can find d by subtracting any term from its succeeding term. Let's use the first two terms.

   • d = a₂ – a₁ = (1/2) – (3/2) = (1 – 3) / 2 = -2 / 2 = -1

   To be certain, we could also check the next pair:

   • a₃ – a₂ = (-1/2) – (1/2) = (-1 – 1) / 2 = -2 / 2 = -1

   The common difference is consistent.

   • a = 3/2 and d = -1.

Finite and Infinite APs

Arithmetic Progressions can be categorized based on the number of terms they contain.

• Finite AP: An AP that has a limited number of terms is called a finite AP. Such an AP always has a last term.

  • Example: The heights of students in a queue: 147, 148, 149, ..., 157. This AP starts at 147, ends at 157, and has a finite number of students.

• Infinite AP: An AP that has an unlimited number of terms is called an infinite AP. The list of numbers continues indefinitely, indicated by "...". Such an AP does not have a last term.

  • Example: The list of positive integers: 1, 2, 3, 4, ... This list goes on forever.

nth Term of an AP — Derivation and Formula

nth Term of an AP — Derivation and Formula

Concept Introduction

Imagine you join a company with a starting salary of ₹8,000 per month, and every year your salary increases by ₹500. What will your salary be in the 10th year? Or the 25th year? Calculating this step-by-step would be tedious. This is where the nth term formula of an Arithmetic Progression becomes powerful.

In real life, we encounter patterns like salary increments, distances covered in successive intervals, or marks scored in sequential tests. Instead of adding the common difference repeatedly, the nth term formula gives us a direct way to find any term in the sequence without computing all the previous terms.

The formula aₙ = a + (n – 1)d is one of the most important tools in AP — it unlocks the ability to jump to any position in the sequence instantly. Understanding how this formula is derived not only builds mathematical confidence but also sharpens logical reasoning.

{{FORMULA: expr=aₙ = a + (n – 1)d | symbols=aₙ:nth term of AP, a:first term, d:common difference, n:position of term}}

Definitions & Formulas

{{KEY: type=formula | title=Core Formula for nth Term | text=aₙ = a + (n – 1)d — This single formula replaces the need to calculate all intermediate terms. Memorize it thoroughly.}}

Derivation of the nth Term Formula

Let's derive the formula step-by-step using logical pattern recognition:

Step 1: Write the first few terms of an AP with first term a and common difference d.

a₁ = a

Step 2: The second term is obtained by adding the common difference once.

a₂ = a + d = a + (2 – 1)d

Step 3: The third term is obtained by adding the common difference twice to the first term.

a₃ = a + d + d = a + 2d = a + (3 – 1)d

Step 4: The fourth term follows the same pattern — add the common difference three times.

a₄ = a + 3d = a + (4 – 1)d

Step 5: Observe the pattern emerging — the coefficient of d is always one less than the term's position.

For the nth term, we add the common difference (n – 1) times to the first term.

aₙ = a + (n – 1)d

Step 6: This formula is valid for any positive integer n.

If the AP has m terms, then the last term can be denoted as l = aₘ = a + (m – 1)d.

Solved Examples

Example 1: Finding a Specific Term (Easy)

Given: AP is 2, 7, 12, 17, ...

To Find: The 10th term

Solution:

1. Identify the first term and common difference.

a = 2
d = 7 – 2 = 5
n = 10

2. Apply the nth term formula.

a₁₀ = a + (n – 1)d

3. Substitute the values.

a₁₀ = 2 + (10 – 1) × 5

4. Simplify step-by-step.

a₁₀ = 2 + 9 × 5
a₁₀ = 2 + 45
a₁₀ = 47

Final Answer: 47

Example 2: Finding Which Term Equals a Given Value (Medium)

Given: AP is 21, 18, 15, 12, ...

To Find: Which term of this AP is –81?

Solution:

1. Identify the first term and common difference.

a = 21
d = 18 – 21 = –3
aₙ = –81

2. Apply the nth term formula.

aₙ = a + (n – 1)d

3. Substitute known values.

–81 = 21 + (n – 1)(–3)

4. Rearrange to solve for n.

–81 = 21 – 3(n – 1)
–81 – 21 = –3(n – 1)
–102 = –3(n – 1)

5. Divide both sides by –3.

n – 1 = 34
n = 35

Final Answer: 35th term

Example 3: Determining an AP from Two Terms (Hard)

Given: The 3rd term of an AP is 5 and the 7th term is 9.

To Find: The AP itself.

Solution:

1. Write equations using the nth term formula for both given terms.

a₃ = a + 2d = 5 ... (Equation 1)
a₇ = a + 6d = 9 ... (Equation 2)

2. Subtract Equation 1 from Equation 2 to eliminate a.

(a + 6d) – (a + 2d) = 9 – 5
4d = 4
d = 1

3. Substitute d = 1 back into Equation 1.

a + 2(1) = 5
a + 2 = 5
a = 3

4. Write the AP using a = 3 and d = 1.

AP: 3, 4, 5, 6, 7, 8, ...

Final Answer: 3, 4, 5, 6, 7, ...

Example 4: Checking if a Number is a Term of an AP (Tricky)

Given: AP is 5, 11, 17, 23, ...

To Find: Is 301 a term of this AP?

Solution:

1. Identify the first term and common difference.

a = 5
d = 11 – 5 = 6

2. Assume 301 is the nth term and set up the equation.

301 = 5 + (n – 1) × 6

3. Rearrange to solve for n.

301 – 5 = (n – 1) × 6
296 = (n – 1) × 6
n – 1 = 296 / 6
n – 1 = 49.333...

4. Check if n is a positive integer.

Since n = 50.333... is not an integer, 301 cannot be a term of this AP.

Final Answer: No, 301 is NOT a term of this AP

Tips & Tricks

{{KEY: type=strategy | title=Quick Validation | text=Always verify that n comes out to be a POSITIVE INTEGER. If n is fractional or negative, the given number is not a term of that AP.}}

Common Mistakes

Brain-Teaser Questions

Q1: If the 5th term of an AP is 20 and the 10th term is 35, find the 1st term and common difference.

💡 Answer:

Set up equations: a + 4d = 20 and a + 9d = 35. Subtracting: 5d = 15, so d = 3. Substitute back: a + 12 = 20, so a = 8. First term = 8, d = 3.

Q2: How many two-digit numbers are divisible by 7?

💡 Answer:

First two-digit multiple of 7 is 14, last is 98. This forms an AP: 14, 21, 28, ..., 98 with a = 14, d = 7. Using 98 = 14 + (n – 1) × 7, we get 84 = 7(n – 1), so n – 1 = 12, giving n = 13. There are 13 such numbers.

Q3: In an AP, the sum of the 3rd and 7th terms is 50, and the 5th term is 23. Find the 10th term.

💡 Answer:

Let a₃ + a₇ = 50. This gives (a + 2d) + (a + 6d) = 50, so 2a + 8d = 50. Also, a₅ = a + 4d = 23. Multiply second equation by 2: 2a + 8d = 46. But first gives 2a + 8d = 50 — contradiction suggests re-check. Actually, solve system: from a + 4d = 23 → a = 23 – 4d. Substitute in 2a + 8d = 50: 2(23 – 4d) + 8d = 50 → 46 = 50, error in problem setup. Corrected: if conditions are a₃ + a₇ = 46, then a = 7, d = 4, so a₁₀ = 7 + 9(4) = 43.

Mini Cheatsheet

Remember: The nth term formula is your direct access pass to any term in an AP. Master it, and you can solve 80% of AP problems in seconds without tedious step-by-step addition!

nth Term of an AP — Application with Examples

Chapter 5: Arithmetic Progressions (Page 5/5)

5.5 The nth Term of an Arithmetic Progression

In our previous lessons, we learned to identify an Arithmetic Progression (AP) and find its first term (a) and common difference (d). We can write out the terms of an AP easily if we know these two values. For example, if a = 5 and d = 3, the AP is 5, 8, 11, 14, and so on.

But what if we need to find the 25th term? Or the 100th term? Writing out all the terms would be tedious and time-consuming. We need a more efficient method—a formula to directly calculate any term in the sequence. This is where the formula for the nth term comes in.

Deriving the Formula for the nth Term

Let's look at the general form of an AP and observe the pattern:

• 1st term (a₁): a = a + 0d = a + (1 - 1)d
• 2nd term (a₂): a + d = a + (2 - 1)d
• 3rd term (a₃): a + 2d = a + (3 - 1)d
• 4th term (a₄): a + 3d = a + (4 - 1)d
• 5th term (a₅): a + 4d = a + (5 - 1)d

Do you see the pattern? The coefficient of the common difference d is always one less than the term number. Following this pattern, we can generalize for any term number, which we call n.

The nth term (aₙ) of an AP with first term a and common difference d is given by the formula:

aₙ = a + (n - 1)d

Here:

• aₙ is the term you want to find (the nth term).
• a is the first term of the AP.
• n is the position of the term in the sequence (e.g., 5th, 20th, 100th).
• d is the common difference.

This formula is also known as the general term of the AP because it allows us to find any term in the sequence.

Note: For a finite AP with 'm' terms, the last term is the mth term. This last term is often denoted by l. So, for a finite AP, the last term l is given by l = a + (m - 1)d.

Worked Examples: Applying the nth Term Formula

Let's solve some problems to see how this powerful formula works in different situations.

Worked Example 1: Finding a Specific Term

Question: Find the 10th term of the AP: 2, 7, 12, . . .

Solution: Our goal is to find the 10th term (a₁₀) of the given AP.

1. Identify the first term (a), common difference (d), and term number (n).

   • The first term, a = 2.
   • The common difference, d = a₂ - a₁ = 7 - 2 = 5.
   • We need to find the 10th term, so n = 10.

2. State the formula for the nth term.

   • aₙ = a + (n - 1)d

3. Substitute the values of a, d, and n into the formula.

   • a₁₀ = 2 + (10 - 1) × 5

4. Calculate the result.

   • a₁₀ = 2 + (9) × 5
   • a₁₀ = 2 + 45
   • a₁₀ = 47

Answer: The 10th term of the AP is 47.

Worked Example 2: Finding the Term Number

Question: Which term of the AP: 21, 18, 15, . . . is -81? Also, is any term 0? Give reason for your answer.

Solution: This time, we are given the value of a term (aₙ) and we need to find its position (n).

Part 1: Which term is -81?

1. Identify a, d, and aₙ.

   • First term, a = 21.
   • Common difference, d = 18 - 21 = -3.
   • The term we are looking for is aₙ = -81.
   • We need to find n.

2. Use the nth term formula.

   • aₙ = a + (n - 1)d

3. Substitute the known values.

   • -81 = 21 + (n - 1) × (-3)

4. Solve the equation for n.

   • -81 - 21 = (n - 1) × (-3)
   • -102 = (n - 1) × (-3)
   • -102 / -3 = n - 1
   • 34 = n - 1
   • n = 34 + 1
   • n = 35

Answer (Part 1): The 35th term of the given AP is -81.

Part 2: Is any term 0?

1. Set aₙ = 0 and solve for n.

   • We will use the same values for a and d: a = 21, d = -3.
   • Let's assume aₙ = 0.
   • 0 = 21 + (n - 1) × (-3)

2. Solve the equation.

   • -21 = (n - 1) × (-3)
   • -21 / -3 = n - 1
   • 7 = n - 1
   • n = 7 + 1
   • n = 8

Answer (Part 2): Yes, 0 is a term in this AP. Since n = 8 is a positive integer, it means the 8th term of the AP is 0. If 'n' had been a fraction or a negative number, then 0 would not have been a term in the sequence.

Worked Example 3: Finding the AP when Two Terms are Given

Question: Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution: To define an AP, we need to find its first term (a) and common difference (d). We are given two pieces of information that we can use to form two linear equations.

1. Translate the given information into equations using the nth term formula.

   • We are given a₃ = 5. Using the formula aₙ = a + (n - 1)d:
     • a + (3 - 1)d = 5
     • a + 2d = 5 --- (Equation 1)
   • We are given a₇ = 9.
     • a + (7 - 1)d = 9
     • a + 6d = 9 --- (Equation 2)

2. Solve the pair of linear equations. We can use the elimination method.

   • Subtract Equation 1 from Equation 2:
     • (a + 6d) - (a + 2d) = 9 - 5
     • 4d = 4
     • d = 1

3. Substitute the value of d back into one of the equations to find a.

   • Using Equation 1:
     • a + 2(1) = 5
     • a + 2 = 5
     • a = 3

4. Write the AP.

   • Now that we have a = 3 and d = 1, we can write the terms of the AP.
   • AP: a, a+d, a+2d, ...
   • AP: 3, 3+1, 3+2(1), ...
   • AP: 3, 4, 5, 6, . . .

Answer: The required AP is 3, 4, 5, 6, . . .

Worked Example 4: Real-Life Application

Question: Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

Solution: This real-life situation can be modelled as an Arithmetic Progression.

1. Formulate the AP from the given information.

   • Savings in the 1st week (a₁) = ₹5
   • Savings in the 2nd week (a₂) = ₹5 + ₹1.75 = ₹6.75
   • Savings in the 3rd week (a₃) = ₹6.75 + ₹1.75 = ₹8.50
   • The list of savings is: 5, 6.75, 8.50, . . . This is an AP.

2. Identify a, d, and aₙ.

   • First term, a = 5.
   • The weekly increase is the common difference, d = 1.75.
   • We are given that the savings in the nth week are ₹20.75, so aₙ = 20.75.
   • We need to find n.

3. Apply the nth term formula.

   • aₙ = a + (n - 1)d
   • 20.75 = 5 + (n - 1) × 1.75

4. Solve for n.

   • 20.75 - 5 = (n - 1) × 1.75
   • 15.75 = (n - 1) × 1.75
   • n - 1 = 15.75 / 1.75
   • n - 1 = 9
   • n = 10

Answer: In the 10th week, Ramkali's weekly savings will become ₹20.75.

Let's Check Your Understanding

1. Find the 20th term of the AP: -5, -1, 3, 7, . . .
2. The first term of an AP is 3, the common difference is 4, and the last term is 123. How many terms are in this AP?
3. If the 17th term of an AP exceeds its 10th term by 7, find the common difference.