Arithmetic Progressions

In mathematics, a sequence is a systematically arranged list of numbers following a well-defined rule. Each number in a sequence is called a term, and its position determines its value.

Sequences form the foundation of Arithmetic Progressions (AP), which is a high-weightage topic in CBSE Class 10 Board Exams. Understanding sequences helps in solving real-life problems involving patterns, growth, and distribution.

Formal Definition

A sequence is a function whose domain is the set of natural numbers. It can be written as:

a₁, a₂, a₃, a₄, … , aₙ

where:

• a₁ = First term
• a₂ = Second term
• aₙ = nth term

Types of Sequences (Concept Booster)

• Increasing Sequence: Terms keep increasing (e.g., 2, 5, 8, 11, …)
• Decreasing Sequence: Terms keep decreasing (e.g., 20, 15, 10, 5, …)
• Constant Sequence: All terms are equal (e.g., 7, 7, 7, 7, …)
• Alternating Sequence: Terms change sign or pattern alternately

Illustrative Examples

• Example 1: 2, 5, 8, 11, 14, …
  Pattern: Add 3 each time → Increasing sequence
• Example 2: 20, 15, 10, 5, 0, …
  Pattern: Subtract 5 each time → Decreasing sequence
• Example 3: 3, 3, 3, 3, …
  Pattern: No change → Constant sequence

Visual Representation (Pattern Understanding)

Why Sequences Matter for Board Exams

• Forms the base of Arithmetic Progression formulas
• Used in finding nth term and sum of terms
• Frequently appears in case-study based questions (CBSE pattern)
• Helps in solving real-life problems like savings, patterns, and arrangements

Concept Bridge to Arithmetic Progression

When the difference between consecutive terms of a sequence is constant, the sequence becomes an Arithmetic Progression (AP). This constant difference is called the common difference (d).

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always constant. This constant value is known as the common difference.

In simple terms, each term in an AP is obtained by adding a fixed number to its previous term (except the first term).

Standard Form of an AP

a, a + d, a + 2d, a + 3d, … , a + (n − 1)d

• a = First term
• d = Common difference
• n = Number of terms
• aₙ = nth term of the AP

Common Difference (d)

The common difference is calculated as the difference between any term and its preceding term:

\[ d = a_2 - a_1 = a_3 - a_2 = a_n - a_{n-1} \]

Types of Arithmetic Progressions

• Increasing AP (d > 0): Terms increase
  Example: 2, 5, 8, 11, …
• Decreasing AP (d < 0): Terms decrease
  Example: 20, 15, 10, 5, …
• Constant AP (d = 0): All terms equal
  Example: 7, 7, 7, 7, …

Illustrative Examples (Exam-Oriented)

• Example 1: Find the common difference of 3, 7, 11, 15, …
  Solution: d = 7 − 3 = 4
• Example 2: Check whether 5, 9, 13, 18, … is an AP
  Solution: Differences are 4, 4, 5 → Not constant → Not an AP
• Example 3: Identify type of AP: 10, 7, 4, 1, …
  Solution: d = −3 → Decreasing AP

Visual Representation of an AP

Why AP is Important for Class 10 Boards

• Direct questions on nth term and sum formulas
• Forms base for real-life word problems
• Frequently appears in case-study questions
• Helps in pattern recognition and logical reasoning

Common Mistakes to Avoid

• Assuming sequence is AP without checking differences
• Using wrong formula for nth term
• Ignoring negative values of common difference
• Mixing up term position (n) with value

An Arithmetic Progression (AP) can be expressed in a general algebraic form, which helps in identifying any term of the sequence without writing all preceding terms.

\[ a,\; a + d,\; a + 2d,\; a + 3d,\; a + 4d,\; \cdots \]

Key Components Explained

• a = First term (starting value of the sequence)
• d = Common difference (fixed increment or decrement)
• n = Position of the term in the sequence
• aₙ = a + (n − 1)d = Formula to find the nth term

Nth Term Formula (Core Board Formula)

\[ a_n = a + (n - 1)d \]

Step-wise Term Formation

• 1st term: a
• 2nd term: a + d
• 3rd term: a + 2d
• 4th term: a + 3d
• nth term: a + (n − 1)d

Visual Pattern of Term Growth

Illustrative Examples

• Example 1: Find the 10th term of AP: 2, 5, 8, …
  a = 2, d = 3
  a₁₀ = 2 + (10 − 1)×3 = 29
• Example 2: Find the nth term of AP: 7, 10, 13, …
  a = 7, d = 3
  aₙ = 7 + (n − 1)×3
• Example 3: Which term of AP 3, 7, 11, … is 99?
  99 = 3 + (n − 1)×4 → n = 25

Board Exam Importance

• Direct formula-based questions on nth term
• Finding position of a given term (reverse application)
• Used in word problems and case studies

Common Errors to Avoid

• Writing nth term as a + nd instead of a + (n − 1)d
• Using wrong value of n (position confusion)
• Ignoring negative common difference in decreasing AP
• Calculation mistakes in larger n values

A Finite Arithmetic Progression (Finite AP) is an AP that contains a fixed and limited number of terms. It always has a clearly defined last term.

General Representation

\[ a,\; a + d,\; a + 2d,\; a + 3d,\; \cdots,\; a + (n-2)d,\; a + (n-1)d \]

• a = First term
• d = Common difference
• n = Total number of terms
• l = Last term

Last Term Formula

\[ l = a + (n - 1)d \]

Visual Structure of a Finite AP

Illustrative Example (Corrected & Explained)

• Given AP: 4, 9, 14, 19, 24
• First term: a = 4
• Common difference: d = 9 − 4 = 5
• Number of terms: n = 5
• Last term: l = 24

Key Applications in Board Exams

• Finding number of terms (n) using last term
• Determining missing terms in sequences
• Solving real-life problems involving fixed intervals
• Base for sum of n terms (Sₙ) formulas

Quick Concept Checks

• If last term is known, use l = a + (n − 1)d
• If n is unknown, rearrange formula:
  n = (l − a)/d + 1
• Always verify constant difference before applying formulas

Common Mistakes to Avoid

• Writing incorrect last term expression (missing (n − 1)d)
• Using wrong difference in irregular sequences
• Arithmetic errors in large n calculations
• Not checking whether sequence is actually finite AP

An Infinite Arithmetic Progression (Infinite AP) is a sequence of numbers that continues indefinitely, with a constant difference between consecutive terms.

General Form

\[ a,\; a + d,\; a + 2d,\; a + 3d,\; \cdots \]

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

Mathematical Properties

• No fixed number of terms: The sequence does not terminate.
• Constant difference: For all n ≥ 2,
  \(a_n - a_{n-1} = d\)
• nth term formula applies:
  \(a_n = a + (n - 1)d\)
• No finite sum: Unlike finite AP, the sum of infinite AP is generally undefined (diverges).

Visual Understanding (Endless Growth)

Illustrative Examples

Finite AP vs Infinite AP (Concept Clarity)

• Finite AP: Has last term (l) and fixed n
• Infinite AP: No last term, n → ∞
• Finite AP: Sum can be calculated
• Infinite AP: Sum generally not defined

Board Exam Relevance

• Helps distinguish between finite and infinite sequences
• Useful in concept-based MCQs and case studies
• Strengthens understanding for sum of AP (finite only)

Common Mistakes to Avoid

• Trying to apply sum formulas on infinite AP
• Confusing infinite AP with geometric progression
• Assuming existence of last term in infinite sequences
• Ignoring sign of common difference

For the AP \(\frac{3}{2},\; \frac{1}{2},\; -\frac{1}{2},\; -\frac{3}{2},\; \cdots\), find the first term (a) and the common difference (d).

Step-by-Step Solution

• Step 1: Identify the first term
  \(a = \frac{3}{2}\)
• Step 2: Apply common difference formula
  \(d = a_2 - a_1\)

\[ \begin{aligned} d &= \frac{1}{2} - \frac{3}{2} \\ &= \frac{1 - 3}{2} \\ &= \frac{-2}{2} \\ &= -1 \end{aligned} \]

Quick Verification

Add \(d = -1\) repeatedly:

• \(\frac{3}{2} - 1 = \frac{1}{2}\)
• \(\frac{1}{2} - 1 = -\frac{1}{2}\)
• \(-\frac{1}{2} - 1 = -\frac{3}{2}\)

Common Mistakes to Avoid

• Subtracting in wrong order (using \(a_1 - a_2\))
• Errors in fraction subtraction
• Ignoring negative sign in common difference

Check whether the following list of numbers forms an Arithmetic Progression (AP). If yes, find the next two terms:

\[ 4,\; 10,\; 16,\; 22,\; \cdots \]

Step-by-Step Verification

\[ \begin{aligned} a_2 - a_1 &= 10 - 4 = 6 \\ a_3 - a_2 &= 16 - 10 = 6 \\ a_4 - a_3 &= 22 - 16 = 6 \end{aligned} \]

Since the difference between consecutive terms is constant, the given sequence forms an Arithmetic Progression.

Finding the Next Two Terms

To find the next terms, keep adding the common difference \(d = 6\) to the last term.

\[ \begin{aligned} \text{Next term} &= 22 + 6 = 28 \\ \text{Next term} &= 28 + 6 = 34 \end{aligned} \]

Quick Verification Pattern

• 4 → 10 (+6)
• 10 → 16 (+6)
• 16 → 22 (+6)
• 22 → 28 (+6)
• 28 → 34 (+6)

Common Mistakes to Avoid

• Checking only one difference and assuming AP
• Arithmetic errors in subtraction
• Forgetting to extend sequence using same common difference

In an Arithmetic Progression (AP), each term is obtained by adding a constant value called the common difference (d) to the previous term. The first term is denoted by a.

Derivation of the Nth Term Formula

Consider the AP:

\[ a,\; a+d,\; a+2d,\; a+3d,\; \cdots \]

• 1st term: \(a_1 = a\)
• 2nd term: \(a_2 = a + d\)
• 3rd term: \(a_3 = a + 2d\)
• 4th term: \(a_4 = a + 3d\)

We observe a pattern: each term increases by adding d repeatedly.

\[ a_n = a + (n - 1)d \]

Visual Pattern of Term Growth

Illustrative Examples

• Example 1: Find the 10th term of AP: 3, 7, 11, …
  a = 3, d = 4
  \(a_{10} = 3 + (10 - 1)\times 4 = 39\)
• Example 2: Find nth term of AP: 5, 8, 11, …
  a = 5, d = 3
  \(a_n = 5 + (n - 1)\times 3\)

Board Exam Importance

• Direct questions on finding nth term
• Used to find missing terms
• Helps in solving word problems and case studies

Common Mistakes to Avoid

• Writing formula as \(a + nd\) instead of \(a + (n - 1)d\)
• Incorrect identification of common difference
• Substitution errors in large values of n
• Ignoring negative values of d

Determine the Arithmetic Progression (AP) whose 3rd term is 5 and 7th term is 9.

Step 1: Form Equations Using Given Terms

For the 3rd term:

\[ a_3 = a + (3 - 1)d = a + 2d = 5 \quad \text{(1)} \]

For the 7th term:

\[ a_7 = a + (7 - 1)d = a + 6d = 9 \quad \text{(2)} \]

Step 2: Solve the Equations

Subtract equation (1) from equation (2):

\[ \begin{aligned} (a + 6d) - (a + 2d) &= 9 - 5 \\ a + 6d - a - 2d &= 4 \\ 4d &= 4 \\ d &= 1 \end{aligned} \]

Step 3: Find First Term (a)

Substitute \(d = 1\) into equation (1):

\[ \begin{aligned} a + 2(1) &= 5 \\ a + 2 &= 5 \\ a &= 3 \end{aligned} \]

Required AP

\[ 3,\; 4,\; 5,\; 6,\; 7,\; 8,\; \cdots \]

Quick Verification

• 3rd term: \(3 + 2 = 5\) ✓
• 7th term: \(3 + 6 = 9\) ✓

Common Mistakes to Avoid

• Writing incorrect term expression (like using wrong n)
• Arithmetic mistakes during subtraction
• Forgetting to substitute d back to find a

How many two-digit numbers are divisible by 3?

Step 1: Identify the AP

The smallest two-digit number divisible by 3 is 12, and the largest is 99.

\[ 12,\; 15,\; 18,\; \cdots,\; 99 \]

• First term: \(a = 12\)
• Common difference: \(d = 3\)
• Last term: \(l = 99\)

Step 2: Use nth Term Formula

\[ a_n = a + (n - 1)d \]

\[ \begin{aligned} 99 &= 12 + (n - 1)\times 3 \\ 99 - 12 &= 3(n - 1) \\ 87 &= 3(n - 1) \\ n - 1 &= 29 \\ n &= 30 \end{aligned} \]

Quick Concept Check

Alternatively, using:

\[ n = \frac{l - a}{d} + 1 = \frac{99 - 12}{3} + 1 = 29 + 1 = 30 \]

Common Mistakes to Avoid

• Starting from 10 instead of first divisible number (12)
• Missing the last term (99)
• Forgetting to add +1 in formula
• Calculation errors in subtraction

Find the 11th term from the last term (towards the first term) of the AP: \(10,\; 7,\; 4,\; \cdots,\; -62\)

Step 1: Identify AP Parameters

• First term: \(a = 10\)
• Common difference: \(d = 7 - 10 = -3\)
• Last term: \(l = -62\)

Step 2: Find Total Number of Terms (n)

\[ \begin{aligned} l &= a + (n - 1)d \\ -62 &= 10 + (n - 1)(-3) \\ -62 - 10 &= -3(n - 1) \\ -72 &= -3(n - 1) \\ n - 1 &= 24 \\ n &= 25 \end{aligned} \]

Step 3: Convert “From Last” to “From Beginning”

Required term = 11th from last

\[ \text{Position from start} = n - 11 + 1 = 25 - 11 + 1 = 15 \]

Step 4: Find the Required Term

\[ \begin{aligned} a_{15} &= a + (15 - 1)d \\ &= 10 + 14(-3) \\ &= 10 - 42 \\ &= -32 \end{aligned} \]

Quick Shortcut (High Scoring)

\[ \text{kth term from end} = l - (k - 1)d \]

\[ = -62 - (11 - 1)(-3) = -62 + 30 = -32 \]

Common Mistakes to Avoid

• Forgetting formula \(n - k + 1\)
• Sign errors when \(d\) is negative
• Using wrong value of n
• Arithmetic mistakes in multiplication

Definition

If an Arithmetic Progression (AP) has first term \(a\) and common difference \(d\), then the sum of its first \(n\) terms is denoted by \(S_n\). It represents the total of all terms from the first term up to the nth term.

Derivation (Pairing Method)

Consider the AP:

\[ a,\; a+d,\; a+2d,\; \cdots,\; a+(n-1)d \]

Write the sum in forward and reverse order:

\[ \begin{aligned} S_n &= a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] \\ S_n &= [a+(n-1)d] + [a+(n-2)d] + \cdots + a \end{aligned} \]

Add corresponding terms:

\[ 2S_n = [2a + (n-1)d] + [2a + (n-1)d] + \cdots \text{ (n terms)} \]

\[ 2S_n = n\bigl[2a + (n-1)d\bigr] \]

\[ S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr] \]

Visual Understanding (Pairing Concept)

Alternate Formula (Using Last Term)

If the last term is \(l = a + (n-1)d\), then:

\[ S_n = \frac{n}{2}(a + l) \]

Illustrative Example

• Find the sum of first 10 terms of AP: 2, 5, 8, …
  a = 2, d = 3
  \(S_{10} = \frac{10}{2}[2×2 + (10-1)×3] = 5[4 + 27] = 155\)

Board Exam Importance

• Direct questions on sum of n terms
• Used in word problems (money, rows, patterns)
• Appears in case-study and HOTS questions

Common Mistakes to Avoid

• Forgetting factor \(\frac{n}{2}\)
• Using wrong value of n
• Arithmetic errors in expansion
• Using sum formula for infinite AP

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Step 1: Identify Given Data

• \(S_{14} = 1050\)
• \(a = 10\)
• \(n = 14\)
• Common difference \(d = ?\)

Step 2: Use Sum Formula

\[ S_n = \frac{n}{2}\bigl[2a + (n - 1)d\bigr] \]

\[ \begin{aligned} 1050 &= \frac{14}{2}[2(10) + (14 - 1)d] \\ &= 7[20 + 13d] \\ 1050 &= 140 + 91d \\ 91d &= 1050 - 140 \\ 91d &= 910 \\ d &= 10 \end{aligned} \]

Step 3: Find the 20th Term

\[ a_n = a + (n - 1)d \]

\[ \begin{aligned} a_{20} &= 10 + (20 - 1)\times 10 \\ &= 10 + 190 \\ &= 200 \end{aligned} \]

Quick Strategy Insight

• Use \(S_n\) to find missing \(d\)
• Then apply \(a_n\) formula to find required term

Common Mistakes to Avoid

• Substituting wrong value of n in sum formula
• Arithmetic mistakes while solving for d
• Forgetting to use (n − 1) in nth term formula
• Skipping intermediate steps (loss of marks)

How many terms of the AP \(24,\; 21,\; 18,\; \cdots\) must be taken so that their sum is 78?

Step 1: Identify Given Values

• First term: \(a = 24\)
• Common difference: \(d = 21 - 24 = -3\)
• Sum: \(S_n = 78\)
• Number of terms: \(n = ?\)

Step 2: Apply Sum Formula

\[ S_n = \frac{n}{2}\bigl[2a + (n - 1)d\bigr] \]

\[ \begin{aligned} 78 &= \frac{n}{2}[2(24) + (n - 1)(-3)] \\ 156 &= n[48 - 3(n - 1)] \\ 156 &= n(48 - 3n + 3) \\ 156 &= n(51 - 3n) \\ 156 &= 51n - 3n^2 \\ 3n^2 - 51n + 156 &= 0 \end{aligned} \]

Step 3: Solve the Quadratic Equation

\[ \begin{aligned} 3n^2 - 51n + 156 &= 0 \\ n^2 - 17n + 52 &= 0 \\ (n - 13)(n - 4) &= 0 \\ n &= 13 \quad \text{or} \quad n = 4 \end{aligned} \]

Why Two Answers?

• First few terms are positive
• Later terms become negative
• Positive and negative terms cancel each other

Quick Verification

• Sum of first 4 terms = 24 + 21 + 18 + 15 = 78 ✓
• Sum of first 13 terms also = 78 (due to cancellation effect) ✓

Common Mistakes to Avoid

• Ignoring one root of quadratic equation
• Sign errors when \(d\) is negative
• Incorrect expansion of expressions
• Not verifying final answers

• Definition: An Arithmetic Progression (AP) is a sequence in which each term is obtained by adding a constant number d (common difference) to the previous term.
  General form: \[ a,\; a + d,\; a + 2d,\; a + 3d,\; \cdots \]
• Condition for AP: A sequence is an AP if: \[ a_{k+1} - a_k = d \quad \text{(constant for all } k\text{)} \]
• nth Term Formula: \[ a_n = a + (n - 1)d \]
• Sum of First n Terms: \[ S_n = \frac{n}{2}\bigl[2a + (n - 1)d\bigr] \]
• Sum Using Last Term: \[ S_n = \frac{n}{2}(a + l) \] where \(l = a + (n - 1)d\)
• Number of Terms (when last term is known): \[ n = \frac{l - a}{d} + 1 \]
• kth Term from the End: \[ \text{kth from end} = l - (k - 1)d \]
• Types of AP:
  • Increasing AP: \(d > 0\)
  • Decreasing AP: \(d < 0\)
  • Constant AP: \(d = 0\)
• Finite vs Infinite AP:
  • Finite AP: Has last term and fixed number of terms
  • Infinite AP: No last term, continues indefinitely

Monthly savings form an AP:

\[ 50,\; 70,\; 90,\; 110,\; \cdots \]

• First term: \(a = 50\)
• Common difference: \(d = 20\)

Questions

• Q1. How much will the student save in the 12th month?
  Solution: \(a_{12} = 50 + 11×20 = 270\)
• Q2. Total savings after 12 months?
  Solution: \(S_{12} = \frac{12}{2}[2×50 + 11×20] = 6(100 + 220) = 1920\)
• Q3. In which month will savings reach ₹450?
  Solution: \(450 = 50 + (n-1)20 → n = 21\)