What is the formula for the $n$-th term of an AP from the beginning and from the end?

From the beginning: a_n = a + (n - 1)d, where a is the first term and d is common difference. From the end (last term l): a_n(\text{from end}) = l - (n - 1)d.

What are the two formulas for the sum of the first $n$ terms of an AP?

1. S_n = \frac{n}{2}[2a + (n - 1)d] (used when first term a, difference d, and number of terms n are known). 2. S_n = \frac{n}{2}[a + l] (used when first term a and last term l = a_n are known).

How do you find the $n$-th term $a_n$ if only the sum formula $S_n$ is given?

Use the fundamental recurrence relation: a_n = S_n - S_{n-1} for n \ge 2, with a_1 = S_1. For example, if S_n = 3n^2 + 5n, then a_n = S_n - S_{n-1} = 6n + 2.

What convenient symmetric choices should be made when 3, 4, or 5 terms in AP are unknown?

- 3 terms in AP: (a - d), a, (a + d) (sum = 3a). - 4 terms in AP: (a - 3d), (a - d), (a + d), (a + 3d) with common difference 2d (sum = 4a). - 5 terms in AP: (a - 2d), (a - d), a, (a + d), (a + 2d) (sum = 5a).

Can $n$ (the number of terms in an AP) ever be negative or a fraction?

No. n represents the count or position of terms and must be a strictly positive integer (n \in \mathbb{N}). If solving a quadratic equation for n yields a negative or fractional value, it must be rejected.

Why do two different values of $n$ sometimes yield the exact same sum $S_n$?

This occurs when the AP has positive terms initially with a negative common difference d < 0. After crossing zero, the subsequent negative terms cancel out the positive terms added earlier, resulting in the same sum at two different positions (e.g., S_5 = S_{12}).

All Revision Notes

MathematicsCh-5 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 5

Arithmetic Progressions

General term of an AP ( $a_n = a + (n-1)d$ ), sum of first $n$ terms ( $S_n$ ), $n$ -th term from end, arithmetic mean, and real-world installment/ladder word problems.

Quick Key Takeaways:

Definition of AP: A sequence in which each term is obtained by adding a fixed number

d

(common difference) to the preceding term:

a, a+d, a+2d, \dots

General

n

-th Term Formula:

a_n = a + (n - 1)d

n

-th Term from the End:

a_n' = l - (n - 1)d, \quad \text{where } l \text{ is the last term}

Sum of First

n

Terms (

S_n

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

Relation Between

S_n

and

a_n

a_n = S_n - S_{n-1}

Three Numbers in AP: Taken as

(a - d), a, (a + d)

with common difference

d

AP Formula ReferenceQuick check for nth term and sum of first n terms formulas

Open Tool

Jump to Concept:1. 1. AP Sequence Architecture, General Term & Arithmetic Means 2. 2. Sum of $n$ Terms ($S_n$) Derivation & Essential Identities 3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Loan Repayment Installments Scheme 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. AP Sequence Architecture, General Term & Arithmetic Means

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Arithmetic Progressions.

• General Term & Common Difference Formulas

•

Common Difference:

d = a_k - a_{k-1}

for all

k \ge 2

. Note that

d

can be positive, negative, or zero.

•

$n$ -th Term:

a_n = a + (n - 1)d

•

$n$ -th Term from End: If an AP has

m

terms, the

n

-th term from the end is the

(m - n + 1)

-th term from the beginning, given by

l - (n - 1)d

•

Symmetric Variable Selections:

•

3 terms in AP:

(a - d), a, (a + d)

\implies

Sum

= 3a

•

4 terms in AP:

(a - 3d), (a - d), (a + d), (a + 3d)

\implies

Common diff

= 2d

, Sum

= 4a

📊 Arithmetic Progressions: General Term Ladder & Series SumVisual Model

Visual schematic mapping the arithmetic ladder progression, constant difference steps, and the series summation area formula.

2. Sum of $n$ Terms ($S_n$) Derivation & Essential Identities

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Arithmetic Progressions.

• Derivation of the Sum of First $n$ Terms ( $S_n$ )

•

Write the sum in forward order:

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

•

Write the sum in reverse order:

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

•

Adding (1) and (2) term by term:

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

2S_n = n[2a + (n - 1)d] \implies \mathbf{S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}[a + l]}

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: 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 (Given Data):

n = 14, S_{14} = 1050, a = 10

•

Step 2 (Formula for $S_n$ ):

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

1050 = \frac{14}{2}[2(10) + (14 - 1)d]

1050 = 7[20 + 13d]

150 = 20 + 13d \implies 13d = 130 \implies \mathbf{d = 10}

•

Step 3 (Finding 20th Term $a_{20}$ ):

a_{20} = a + (20 - 1)d = 10 + 19(10) = 10 + 190 = \mathbf{200}

•

Final Boxed Answer:

\mathbf{a_{20} = 200}

• 5-Mark Heavyweight Board Problem / Rider: The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

•

Step 1 (Variable Setup): Let the terms be

a_3 = a + 2d

and

a_7 = a + 6d

•

Step 2 (Sum Condition):

(a + 2d) + (a + 6d) = 6 \implies 2a + 8d = 6 \implies a + 4d = 3 \implies a = 3 - 4d \quad \text{--- (1)}

•

Step 3 (Product Condition):

(a + 2d)(a + 6d) = 8

Substitute

a = 3 - 4d

(3 - 4d + 2d)(3 - 4d + 6d) = 8 \implies (3 - 2d)(3 + 2d) = 8

9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d^2 = \frac{1}{4} \implies \mathbf{d = \pm \frac{1}{2}}

•

Step 4 (Case 1: $d = \frac{1}{2}$ ):

a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1

S_{16} = \frac{16}{2}[2(1) + (16 - 1)(\frac{1}{2})] = 8[2 + \frac{15}{2}] = 8[\frac{19}{2}] = 4 \times 19 = \mathbf{76}

•

Step 5 (Case 2: $d = -\frac{1}{2}$ ):

a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5

S_{16} = \frac{16}{2}[2(5) + (16 - 1)(-\frac{1}{2})] = 8[10 - \frac{15}{2}] = 8[\frac{5}{2}] = 4 \times 5 = \mathbf{20}

•

Final Boxed Answer:

\mathbf{S_{16} = 76 \quad \text{or} \quad S_{16} = 20}

4. CBSE Case-Study Modeling: Loan Repayment Installments Scheme

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Loan Repayment Installments Scheme

A person agrees to repay a total loan of Rs 1,18,000 by paying every month. In the first month, he pays Rs 1,000 and then increases the installment by Rs 100 every month.

•

Q1: Identify the sequence formed and write its first term and common difference.

\rightarrow

The monthly installments form an Arithmetic Progression with first term

a = \mathbf{1000}

and common difference

d = \mathbf{100}

•

Q2: What amount will be paid in the 30th installment?

\rightarrow

a_{30} = a + (30 - 1)d = 1000 + 29(100) = 1000 + 2900 = \mathbf{\text{Rs } 3,900}

•

Q3: What total amount will he have paid after 30 installments?

\rightarrow

S_{30} = \frac{30}{2}[2(1000) + 29(100)] = 15[2000 + 2900] = 15[4900] = \mathbf{\text{Rs } 73,500}

•

Q4: What balance loan amount still remains to be paid after 30 months?

\rightarrow

Remaining

= 1,18,000 - 73,500 = \mathbf{\text{Rs } 44,500}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: Stating the standard formula (

a_n = a + (n-1)d

S_n = n/2[2a + (n-1)d]

•

1 Mark: Correct algebraic substitution of

a, d, n

•

2 Marks: Evaluating multi-case solutions (e.g.

d = \pm 1/2

) completely.

•

1 Mark: Boxed final answer with units (Rupees, terms).

• Common Calculation Traps & Verification Checklist

•

Trap 1: Forgetting both

\pm

signs when taking square roots of

d^2

•

Trap 2: Confusing

a_n

(

n

-th term) with

S_n

(sum of

n

terms).

•

Trap 3: Using

a_n = S_n - S_{n-1}

incorrectly for

n = 1

Authentic Board Question (3 Marks)Topic: Arithmetic Progressions Algebraic Formulations & Exact Methods

Solve the algebraic equation, find unknown coefficients, and apply standard theorem methods for Arithmetic Progressions.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Standard Form Equation & Algebraic Setup: Convert given conditions into standard algebraic form (

ax^2+bx+c=0

a_n = a+(n-1)d

, or system of equations).

1 Mark

Step 2: Step-by-Step Factorisation / Reduction Method: Apply formal algebraic method (splitting the middle term, quadratic formula, elimination, or AP summation).

1 Mark

Step 3: Final Solution & Domain Verification: State the final values of

x, n, d

, verifying against real-life boundary conditions (e.g.

n > 0

, positive speeds).

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Arithmetic Progressions in CBSE Mathematics:

1. Algebraic Setup: Express the given problem into standard algebraic form.
2. Solving Technique: Carry out step-by-step algebraic manipulation showing all factorization or formula substitution lines.
3. Boxed Result: State and box the final root values, discarding any non-viable negative or fractional answers where context dictates.

Examiner Mark Deduction Traps:

•Always check boundary conditions (e.g., number of terms

n

in an AP must be a positive natural number).

•Show all factorization steps clearly—do not jump directly from equation to roots.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

An AP is a sequence of numbers where the difference between any two consecutive terms is constant. Common difference

d = a_{n} - a_{n-1} = a_2 - a_1

d

can be positive (increasing AP), negative (decreasing AP), or zero (constant AP).

Arithmetic Progressions

1. AP Sequence Architecture, General Term & Arithmetic Means

2. Sum of $n$ Terms ($S_n$) Derivation & Essential Identities

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Loan Repayment Installments Scheme

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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