This MCQ module is based on: Relations Among Number Sequences
TOPIC 2 OF 41
Relations Among Number Sequences
🎓 Class 6
Mathematics
CBSE
Theory
Ch 1 — Patterns in Mathematics
⏱ ~30 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Relations Among Number Sequences
Targeting Class 6 level in Number Patterns, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
1.4 Relations among Number Sequences
Number sequences are not isolated — they are often connected to each other in surprising and beautiful ways. Let us discover some of these hidden relationships?.
Adding Odd Numbers gives Square Numbers
What happens when we start adding up odd numbers? one by one?
Pattern — Sum of Odd Numbers
\(1 = 1 = 1^2\)
\(1 + 3 = 4 = 2^2\)
\(1 + 3 + 5 = 9 = 3^2\)
\(1 + 3 + 5 + 7 = 16 = 4^2\)
\(1 + 3 + 5 + 7 + 9 = 25 = 5^2\)
\(1 + 3 + 5 + 7 + 9 + 11 = 36 = 6^2\)
\(1 + 3 = 4 = 2^2\)
\(1 + 3 + 5 = 9 = 3^2\)
\(1 + 3 + 5 + 7 = 16 = 4^2\)
\(1 + 3 + 5 + 7 + 9 = 25 = 5^2\)
\(1 + 3 + 5 + 7 + 9 + 11 = 36 = 6^2\)
⚠️ Key Discovery
The sum of the first \(n\) odd numbers always equals \(n^2\). This is not a coincidence — it has a beautiful visual explanation!
To understand why this works, think of building a square grid step by step. Each time you add the next odd number of dots, they form an L-shaped border around the existing square, making it one size larger.
Why Odd Numbers Sum to Squares — Visual Proof
Each colour shows the L-shaped addition of the next odd number of dots
Odd Numbers → Square Numbers Simulator
Bloom: L4 AnalyseDrag the slider to see how adding odd numbers builds perfect squares:
4
Adding Counting Numbers Up and Down
Here is another surprising relationship. What happens when you add counting numbers up and then back down?
Pattern — Up-and-Down Sums
\(1 = 1 = 1^2\)
\(1 + 2 + 1 = 4 = 2^2\)
\(1 + 2 + 3 + 2 + 1 = 9 = 3^2\)
\(1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 4^2\)
\(1 + 2 + 1 = 4 = 2^2\)
\(1 + 2 + 3 + 2 + 1 = 9 = 3^2\)
\(1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 4^2\)
💡 Did You Know?
This pattern works because adding counting numbers up and down is equivalent to placing dots in a diamond (rotated square) shape. The number of dots in a diamond of side \(n\) always equals \(n^2\)!
Consecutive Triangular Numbers Sum to Squares
What happens when you add pairs of consecutive triangular numbers??
Pattern — Consecutive Triangular Numbers
\(1 + 3 = 4 = 2^2\)
\(3 + 6 = 9 = 3^2\)
\(6 + 10 = 16 = 4^2\)
\(10 + 15 = 25 = 5^2\)
\(3 + 6 = 9 = 3^2\)
\(6 + 10 = 16 = 4^2\)
\(10 + 15 = 25 = 5^2\)
The sum of two consecutive triangular numbers always gives a square number?! This happens because you can fit two consecutive triangles together to form a square — one triangle facing up and the other rotated to face down.
T₃ + T₂ = 6 + 3 = 9 = 3²
Multiplying Triangular Numbers by 6
What happens when you multiply each triangular number by 6 and add 1?
| Triangular Number (T) | T × 6 + 1 | Result |
|---|---|---|
| 1 | 1 × 6 + 1 | 7 |
| 3 | 3 × 6 + 1 | 19 |
| 6 | 6 × 6 + 1 | 37 |
| 10 | 10 × 6 + 1 | 61 |
| 15 | 15 × 6 + 1 | 91 |
💡 Did You Know?
The sequence 1, 7, 19, 37, 61, 91, ... are called hexagonal numbers (centred). They represent the number of dots in concentric hexagons!
1.5 Patterns in Other Sequences
Adding Hexagonal Numbers
What happens when you add up hexagonal numbers?: 1, 7, 19, 37, ...?
Cumulative Sum of Centred Hexagonal Numbers
\(1 = 1 = 1^3\)
\(1 + 7 = 8 = 2^3\)
\(1 + 7 + 19 = 27 = 3^3\)
\(1 + 7 + 19 + 37 = 64 = 4^3\)
\(1 + 7 = 8 = 2^3\)
\(1 + 7 + 19 = 27 = 3^3\)
\(1 + 7 + 19 + 37 = 64 = 4^3\)
📖 Discovery
Adding up centred hexagonal numbers gives perfect cubes! The sum of the first \(n\) centred hexagonal numbers equals \(n^3\).
Figure it Out (Pages 7–9)
Q1. Can you find a similar pictorial explanation for why adding counting numbers up and down (1, 1+2+1, 1+2+3+2+1, ...) gives square numbers?
Answer: Arrange the dots in a diamond (rotated square) shape. For \(1+2+3+2+1 = 9\): place 1 dot in the first row, 2 dots in the second, 3 dots in the third (widest), then 2 and 1 again. This diamond has 3 dots on each diagonal — it is a 3×3 square rotated by 45°. So the total is always \(n^2\).
Q2. By imagining a large version of this picture, can you say what is the sum of the first 100 odd numbers?
Answer: The sum of the first 100 odd numbers = \(100^2 = \mathbf{10{,}000}\).
This follows from the pattern: the sum of the first \(n\) odd numbers is \(n^2\).
This follows from the pattern: the sum of the first \(n\) odd numbers is \(n^2\).
Q3. What do you get when you add up the first n counting numbers (1 + 2 + 3 + ... + n)? Which sequence do these sums form?
Answer: The sum of the first \(n\) counting numbers gives the triangular numbers:
\(1 = 1,\; 1+2 = 3,\; 1+2+3 = 6,\; 1+2+3+4 = 10, \ldots\)
Formula: \(1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\)
\(1 = 1,\; 1+2 = 3,\; 1+2+3 = 6,\; 1+2+3+4 = 10, \ldots\)
Formula: \(1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\)
Q4. What do you get when you add up pairs of consecutive counting numbers starting from 1? (i.e., 1+2, 2+3, 3+4, 4+5, ...) Which sequence?
Answer: 1+2=3, 2+3=5, 3+4=7, 4+5=9, ...
You get the odd numbers starting from 3: 3, 5, 7, 9, 11, ...
You get the odd numbers starting from 3: 3, 5, 7, 9, 11, ...
Q5. What happens when you add up pairs of consecutive triangular numbers? (1+3, 3+6, 6+10, 10+15, ...) Which sequence do you get? Why?
Answer: 1+3=4, 3+6=9, 6+10=16, 10+15=25, ...
You get square numbers: 4, 9, 16, 25, ...
Why: Two consecutive triangular numbers \(T_n\) and \(T_{n-1}\) fit together to form a square of side \(n\). \(T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2\).
You get square numbers: 4, 9, 16, 25, ...
Why: Two consecutive triangular numbers \(T_n\) and \(T_{n-1}\) fit together to form a square of side \(n\). \(T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2\).
Q6. What happens when you take squares of consecutive triangular numbers and find their differences? (3²−1², 6²−3², 10²−6², ...) Which sequence?
Answer: 9−1=8, 36−9=27, 100−36=64, 225−100=125, ...
You get the cube numbers: 8, 27, 64, 125, ...
In general: \(T_n^2 - T_{n-1}^2 = n^3\)
You get the cube numbers: 8, 27, 64, 125, ...
In general: \(T_n^2 - T_{n-1}^2 = n^3\)
Q7. What happens when you multiply the triangular numbers by 6 and add 1? Which sequence do you get? Can you explain it with a picture?
Answer: 1×6+1=7, 3×6+1=19, 6×6+1=37, 10×6+1=61, ...
You get the centred hexagonal numbers: 1, 7, 19, 37, 61, 91, ...
Pictorial explanation: A centred hexagon of ring \(n\) has 6 triangular arrangements around a centre dot. Each triangle contributes \(T_{n-1}\) dots, and with the centre: \(6 \times T_{n-1} + 1\).
You get the centred hexagonal numbers: 1, 7, 19, 37, 61, 91, ...
Pictorial explanation: A centred hexagon of ring \(n\) has 6 triangular arrangements around a centre dot. Each triangle contributes \(T_{n-1}\) dots, and with the centre: \(6 \times T_{n-1} + 1\).
Q8. What happens when you start to add up hexagonal numbers: 1, 1+7, 1+7+19, 1+7+19+37, ...? Which sequence do you get?
Answer: 1=1, 1+7=8, 1+7+19=27, 1+7+19+37=64, ...
You get the cube numbers: 1, 8, 27, 64, 125, ...
The sum of the first \(n\) centred hexagonal numbers equals \(n^3\).
You get the cube numbers: 1, 8, 27, 64, 125, ...
The sum of the first \(n\) centred hexagonal numbers equals \(n^3\).
🔍 Activity 1.2 — Discovering Hidden Connections
Bloom: L3 Apply
Materials needed: Grid paper (or plain notebook), coloured pencils
🤔 PREDICT FIRST: If the sum of the first 5 odd numbers is 25 (which is 5²), what do you think the sum of the first 20 odd numbers will be?
- On grid paper, colour a 1×1 square using colour 1
- Add an L-shaped border of 3 cells using colour 2 → you now have a 2×2 square
- Add an L-shaped border of 5 cells using colour 3 → you now have a 3×3 square
- Continue with borders of 7, 9, 11 cells using different colours until you reach a 6×6 square
- Write the sum equation next to each step: e.g., 1+3+5+7 = 16 = 4²
✅ Observation & Explanation
Each L-shaped border adds exactly the next odd number of cells (1, 3, 5, 7, 9, 11...). After adding \(n\) L-borders, you have an \(n \times n\) square. So the sum of the first \(n\) odd numbers = \(n^2\).Prediction answer: Sum of first 20 odd numbers = \(20^2 = 400\).
📋
Competency-Based Questions
Scenario: Arjun is building a staircase pattern with blocks. In step 1 he uses 1 block, step 2 he uses 3 blocks (total 1+3=4), step 3 he uses 5 blocks (total 1+3+5=9), and so on, always adding the next odd number of blocks. His friend Meera notices something about the total at each step.
Q1. After step 8, how many blocks will Arjun have used in total?
L3 Apply
Answer: (C) 64 — Sum of first 8 odd numbers = \(8^2 = 64\).
Q2. Meera says "The total blocks at any step always form a perfect square." Arjun says "That's because each odd number forms an L-shape around the previous square." Who is giving the better mathematical explanation and why?
L5 Evaluate
Model Answer: Arjun gives the better explanation because he explains why the pattern works, not just what the pattern is. Meera states the observation (totals are perfect squares), but Arjun provides the reasoning — each odd number of blocks forms an L-shaped border that extends a square by one row and one column, creating the next larger square.
Q3. Is it true that the sum of any 3 consecutive triangular numbers is always divisible by 3? Verify with at least 3 examples.
L4 Analyse
Answer:
1+3+6 = 10 (not divisible by 3 — ✗)
So the statement is false! However, the sum of 3 consecutive triangular numbers follows the pattern: \(T_n + T_{n+1} + T_{n+2}\). Checking: 1+3+6=10, 3+6+10=19, 6+10+15=31. None of these are divisible by 3. The claim is incorrect.
1+3+6 = 10 (not divisible by 3 — ✗)
So the statement is false! However, the sum of 3 consecutive triangular numbers follows the pattern: \(T_n + T_{n+1} + T_{n+2}\). Checking: 1+3+6=10, 3+6+10=19, 6+10+15=31. None of these are divisible by 3. The claim is incorrect.
HOT Q. Discover a new relationship between two different number sequences not discussed in the chapter. Write down the pattern, verify it for at least 4 terms, and try to explain why it works.
L6 Create
Hint: Try exploring what happens when you take differences between consecutive cube numbers (8−1, 27−8, 64−27, ...). Or try squaring the Virahanka numbers and looking for a pattern. Many beautiful relationships are waiting to be discovered!
⚖️ Assertion–Reason Questions
Options:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Assertion (A): The sum of the first 50 odd numbers is 2500.
Reason (R): The sum of the first \(n\) odd numbers is always equal to \(n^2\).
Answer: (A) — Both are true. \(50^2 = 2500\), and R correctly explains why: each odd number adds an L-shaped border to the previous square.
Assertion (A): Adding two consecutive triangular numbers always gives a square number.
Reason (R): Triangular numbers are always odd.
Answer: (C) — A is true (\(T_n + T_{n-1} = n^2\)). R is false — triangular numbers are not always odd (e.g., 6, 10, 28 are even).
Frequently Asked Questions
How is the sum of odd numbers related to square numbers?
The sum of the first n odd numbers always equals n squared. For instance, 1+3=4 which is 2 squared, 1+3+5=9 which is 3 squared, and 1+3+5+7=16 which is 4 squared. This pattern holds for all natural numbers and is a key relationship taught in NCERT Class 6 Maths Chapter 1.
What is the relation between consecutive triangular numbers?
When you add two consecutive triangular numbers, you always get a square number. For example, the 2nd triangular number (3) plus the 3rd triangular number (6) equals 9, which is 3 squared. Similarly, 6 plus 10 equals 16 which is 4 squared. This is covered in NCERT Class 6 Ganita Prakash.
Why do number sequence patterns matter in mathematics?
Number sequence patterns matter because they develop pattern recognition, logical reasoning and algebraic thinking. Understanding relations among sequences helps students grasp advanced concepts like series, progressions and mathematical proofs in later classes. NCERT introduces these foundations in Class 6 Chapter 1.
Can every square number be written as a sum of odd numbers?
Yes, every perfect square number can be expressed as the sum of consecutive odd numbers starting from 1. The number n squared equals the sum of the first n odd numbers. This identity works for every natural number n and is proven using visual dot arrangements in NCERT Class 6 Maths.
What patterns exist between different types of number sequences?
Several patterns connect number sequences in Class 6 Maths. Square numbers equal sums of consecutive odd numbers. Two consecutive triangular numbers sum to a square number. Powers of 2 double each time. These interconnections reveal the deep structure underlying seemingly different number families.
Frequently Asked Questions — Patterns in Mathematics
What is Relations Among Number Sequences in NCERT Class 6 Mathematics?
Relations Among Number Sequences is a key concept covered in NCERT Class 6 Mathematics, Chapter 1: Patterns in Mathematics. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Relations Among Number Sequences step by step?
To solve problems on Relations Among Number Sequences, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 6 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Patterns in Mathematics?
The essential formulas of Chapter 1 (Patterns in Mathematics) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Relations Among Number Sequences important for the Class 6 board exam?
Relations Among Number Sequences is part of the NCERT Class 6 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Relations Among Number Sequences?
Common mistakes in Relations Among Number Sequences include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Relations Among Number Sequences?
End-of-chapter NCERT exercises for Relations Among Number Sequences cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
AI Tutor
Mathematics Class 6 — Ganita Prakash
Ready