NCERT Solutions for Class 6 Maths Chapter 1: Patterns in Mathematics — Complete Guide

A pattern is a sequence of numbers, shapes, or objects that follows a definite rule. The rule tells you how to get from one term to the next. Once you know the rule, you can extend the pattern as far as you like.

For example, in the sequence $5, 10, 15, 20, 25, \ldots$, the rule is **add $5$** to get the next term. This is the sequence of multiples of $5$.

Patterns can be:
- Repeating patterns: The same group repeats over and over, like $A, B, C, A, B, C, A, B, C, \ldots$
- Growing patterns: Each term is larger (or smaller) than the previous one, like $1, 4, 9, 16, 25, \ldots$
- Number patterns: Sequences of numbers following a rule
- Shape patterns: Sequences of geometric figures following a rule

A sequence is an ordered list of numbers following a rule. Each number in the sequence is called a term.

We use subscript notation to refer to specific terms:
- $a_1$ = first term
- $a_2$ = second term
- $a_n$ = the $n$th term (general term)

For example, in the sequence $3, 7, 11, 15, 19, \ldots$:

The general term (or $n$th term formula) lets you find any term directly. For this sequence, $a_n = 4n - 1$. Check: $a_1 = 4(1) - 1 = 3$ and $a_5 = 4(5) - 1 = 19$.

An arithmetic sequence is one where the difference between consecutive terms is always the same. This constant difference is called the common difference ($d$).

Examples:
- $2, 5, 8, 11, 14, \ldots$ has common difference $d = 3$
- $20, 17, 14, 11, 8, \ldots$ has common difference $d = -3$ (decreasing)
- $7, 7, 7, 7, \ldots$ has common difference $d = 0$ (constant)

The general term of an arithmetic sequence is:

For $2, 5, 8, 11, \ldots$: $a_n = 2 + (n-1) \times 3 = 3n - 1$.

So the 100th term is $a_{100} = 3(100) - 1 = 299$.

A geometric sequence is one where each term is obtained by multiplying the previous term by a fixed number called the common ratio ($r$).

Examples:
- $2, 6, 18, 54, 162, \ldots$ has common ratio $r = 3$
- $1000, 100, 10, 1, \ldots$ has common ratio $r = \frac{1}{10}$
- $3, -6, 12, -24, \ldots$ has common ratio $r = -2$

The general term of a geometric sequence is:

For $2, 6, 18, 54, \ldots$: $a_n = 2 \times 3^{n-1}$.

So the 6th term is $a_6 = 2 \times 3^5 = 2 \times 243 = 486$.

Several important number sequences appear repeatedly in this chapter and throughout mathematics:

Square numbers: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \ldots$
Formula: $a_n = n^2$. These are called square numbers because $n^2$ dots can be arranged in a perfect square.

Triangular numbers: $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots$
Formula: $T_n = \frac{n(n+1)}{2}$. These are called triangular numbers because $T_n$ dots can be arranged in a triangle.

Cube numbers: $1, 8, 27, 64, 125, 216, \ldots$
Formula: $a_n = n^3$.

Fibonacci numbers: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots$
Rule: Each term is the sum of the two preceding terms. $F_n = F_{n-1} + F_{n-2}$.

Powers of 2: $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \ldots$
Formula: $a_n = 2^{n-1}$.

Problem: Find the next three terms in the sequence $1, 3, 5, 7, 9, \ldots$

Solution:

Step 1: Find the difference between consecutive terms.

The common difference is $d = 2$. This is an arithmetic sequence.

Step 2: Apply the rule (add $2$) to find the next three terms.

Step 3: Verify using the general term formula.
The $n$th odd number is $a_n = 2n - 1$.

Answer: The next three terms are $11, 13, 15$.

Problem: Find the next two terms in the sequence $2, 4, 8, 16, \_, \_$

Solution:

Step 1: Check whether the differences are constant.

Step 2: Check the ratios.

The common ratio is $r = 2$. This is a geometric sequence.

Step 3: Find the next terms by multiplying by $2$.

Step 4: Verify. Each term is a power of $2$:

Answer: The next two terms are $32$ and $64$.

Problem: Identify the pattern and find the next three terms: $1, 4, 9, 16, 25, \ldots$

Solution:

Step 1: Observe each term.

These are perfect squares — the squares of consecutive natural numbers.

Step 2: Find the differences between consecutive terms.

The differences are $3, 5, 7, 9$ — consecutive odd numbers! This is a beautiful pattern: the difference between the $n$th and $(n+1)$th square number is always $2n + 1$.

Mathematically: $(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$.

Step 3: The next three terms:

Answer: The next three terms are $36, 49, 64$.

Problem: Find the next two terms in the sequence $1, 3, 6, 10, 15, \ldots$

Solution:

Step 1: Find the differences.

The differences are $2, 3, 4, 5$ — they increase by $1$ each time.

Step 2: The next difference will be $6$, then $7$.

Step 3: Verify using the formula. The $n$th triangular number is:

Why are they called triangular numbers?
- $T_1 = 1$: One dot (a single point).
- $T_2 = 3$: Three dots arranged in a triangle (1 on top, 2 on bottom).
- $T_3 = 6$: Six dots (1 + 2 + 3).
- $T_4 = 10$: Ten dots (1 + 2 + 3 + 4).

In general, $T_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$.

Answer: The next two terms are $21$ and $28$.

Problem: What are the next three terms in $100, 93, 86, 79, 72, \ldots$?

Solution:

Step 1: Find the differences.

The common difference is $d = -7$. This is a decreasing arithmetic sequence.

Step 2: Continue the pattern.

Step 3: Verify with the general term formula.

Answer: The next three terms are $65, 58, 51$.

Problem: Find the next four terms in the sequence $1, -1, 1, -1, 1, \ldots$

Solution:

Step 1: Observe the pattern. The terms alternate between $1$ and $-1$.

The rule is: multiply the previous term by $-1$.

Alternatively, odd-positioned terms are $1$ and even-positioned terms are $-1$.

Step 2: The general term can be written as:

Check: $a_1 = (-1)^2 = 1$, $a_2 = (-1)^3 = -1$, $a_3 = (-1)^4 = 1$.

Step 3: Next four terms: $-1, 1, -1, 1$.

Answer: The next four terms are $-1, 1, -1, 1$.

Problem: Identify the pattern and find the 20th term: $5, 10, 15, 20, 25, \ldots$

Solution:

Step 1: Identify the pattern.

These are **multiples of $5$**. The general term is $a_n = 5n$.

Step 2: Find the 20th term.

Answer: The 20th term is $100$.

Problem: Find the next two terms: $2, 5, 10, 17, 26, \ldots$

Solution:

Step 1: Find the first-level differences.

First-level differences: $3, 5, 7, 9$ — not constant.

Step 2: Find the second-level differences.

Second-level differences are constant ($= 2$). This means the sequence is quadratic (related to squares).

Step 3: The next first-level differences will be $9 + 2 = 11$ and $11 + 2 = 13$.

Step 4: Verify. Notice $a_n = n^2 + 1$:

Answer: The next two terms are $37$ and $50$.

Problem: In the sequence $1, 1, 2, 3, 5, 8, 13, \ldots$, find the next three terms.

Solution:

Step 1: Check the differences.

The differences are $0, 1, 1, 2, 3, 5$ — these are the same sequence! This is the hallmark of the Fibonacci sequence.

Step 2: The rule is: each term equals the sum of the two terms before it.

Step 3: Apply the rule.

Answer: The next three terms are $21, 34, 55$.

Problem: Create a number sequence that starts with $4$ and has each term $3$ more than the previous term. Write the first $8$ terms and find the general term.

Solution:

Step 1: Start with $a_1 = 4$ and common difference $d = 3$.

The first $8$ terms are: $4, 7, 10, 13, 16, 19, 22, 25$.

Step 2: Find the general term.

Step 3: Verify.

Answer: The sequence is $4, 7, 10, 13, 16, 19, 22, 25$ with general term $a_n = 3n + 1$.

Problem: Show that each square number can be represented as a square arrangement of dots. How many dots are needed for the 6th square number?

Solution:

Square numbers form square arrangements:
- $1 = 1^2$: a single dot
- $4 = 2^2$: $2 \times 2$ arrangement of dots
- $9 = 3^2$: $3 \times 3$ arrangement of dots
- $16 = 4^2$: $4 \times 4$ arrangement of dots
- $25 = 5^2$: $5 \times 5$ arrangement of dots

The 6th square number is:

This is a $6 \times 6$ arrangement of dots ($6$ rows and $6$ columns).

Key insight: To go from the $n$th square to the $(n+1)$th square, you add a new row at the bottom and a new column on the right, plus one corner dot. This adds $n + (n+1) - 1 = 2n + 1$ dots, which is why consecutive square numbers differ by odd numbers.

For example, going from $5^2 = 25$ to $6^2 = 36$, we add $2(5) + 1 = 11$ dots. Check: $36 - 25 = 11$. $\checkmark$

Answer: $36$ dots are needed for the 6th square number.

Problem: Draw the dot arrangements for the first $5$ triangular numbers and find the 10th triangular number.

Solution:

Triangular numbers are arranged in triangular patterns:
- $T_1 = 1$: $\bullet$ (1 dot)
- $T_2 = 3$: Row 1 has 1 dot, Row 2 has 2 dots ($1 + 2 = 3$)
- $T_3 = 6$: Rows have $1, 2, 3$ dots ($1 + 2 + 3 = 6$)
- $T_4 = 10$: Rows have $1, 2, 3, 4$ dots ($1 + 2 + 3 + 4 = 10$)
- $T_5 = 15$: Rows have $1, 2, 3, 4, 5$ dots ($1 + 2 + 3 + 4 + 5 = 15$)

The $n$th triangular number is the sum of the first $n$ natural numbers:

The 10th triangular number:

Answer: $T_{10} = 55$.

Problem: Show that the sum of two consecutive triangular numbers is always a square number.

Solution:

Let us check with examples:

Now let us prove it using the formula:

So $T_n + T_{n+1} = (n+1)^2$, which is always a perfect square. $\square$

Answer: Proven. The sum of two consecutive triangular numbers $T_n$ and $T_{n+1}$ is $(n+1)^2$.

Problem: A pattern of dots is arranged as follows: Term 1 has $1$ dot, Term 2 has $5$ dots (a plus sign shape), Term 3 has $13$ dots, Term 4 has $25$ dots. Find the number of dots in Term 5.

Solution:

Step 1: Find the differences.

Step 2: The differences increase by $4$ each time.
Next difference: $12 + 4 = 16$.

Step 3: Term 5 $= 25 + 16 = 41$.

Step 4: Verify the pattern. Notice the differences are $4 \times 1, 4 \times 2, 4 \times 3, 4 \times 4$.
The $n$th term $= 1 + 4(1 + 2 + 3 + \cdots + (n-1)) = 1 + 4 \cdot \frac{(n-1)n}{2} = 1 + 2n(n-1) = 2n^2 - 2n + 1$.

Check: $n = 5 \Rightarrow 2(25) - 10 + 1 = 41$ $\checkmark$.

Answer: Term 5 has $41$ dots.

Problem: The first four cube numbers are $1, 8, 27, 64$. Find the next two cube numbers and express each as a sum of consecutive odd numbers.

Solution:

Step 1: Find the next two cube numbers.

Step 2: Express each cube as a sum of consecutive odd numbers. There is a beautiful pattern:

Notice that $n^3$ is the sum of $n$ consecutive odd numbers, starting from the $(n^2 - n + 1)$th odd number, which is $n^2 - n + 1$ when expressed as $2k - 1$ form.

Answer: $5^3 = 125$ and $6^3 = 216$.

Problem: Consider the pattern where L-shaped figures are added to build squares: the 1st L-shape has $1$ square, the 2nd has $3$ squares, the 3rd has $5$ squares, the 4th has $7$ squares. How many squares does the 10th L-shape have? What is the total number of squares after adding $10$ L-shapes?

Solution:

Step 1: The number of squares in each L-shape follows the pattern of odd numbers: $1, 3, 5, 7, \ldots$

The $n$th L-shape has $2n - 1$ squares.

For $n = 10$: $2(10) - 1 = 19$ squares.

Step 2: The total after $n$ L-shapes is the sum of the first $n$ odd numbers.

This is a well-known result! The sum of the first $n$ odd numbers equals $n^2$.

Total after $10$ L-shapes: $10^2 = 100$ squares.

Step 3: Verify with smaller cases.
$1 = 1^2$, $1+3 = 4 = 2^2$, $1+3+5 = 9 = 3^2$. $\checkmark$

Answer: The 10th L-shape has $19$ squares, and the total is $100$ squares.

Problem: Rectangles are drawn with dimensions $1 \times 2$, $2 \times 3$, $3 \times 4$, $4 \times 5$. Find the areas of the first $6$ rectangles. What do you notice about these areas?

Solution:

The rectangles have dimensions $n \times (n+1)$:

The areas are: $2, 6, 12, 20, 30, 42$.

Notice that each area is twice a triangular number:

This makes sense because $n(n+1) = 2 \times \frac{n(n+1)}{2} = 2 T_n$.

Answer: The areas are $2, 6, 12, 20, 30, 42$. Each area equals twice the corresponding triangular number.

Problem: Squares are stacked in a staircase pattern. Step 1 uses $1$ square, Step 2 uses $1 + 2 = 3$ squares, Step 3 uses $1 + 2 + 3 = 6$ squares. How many squares are needed for Step 15?

Solution:

Each step uses a triangular number of squares:

For Step 15:

Answer: Step 15 needs $120$ squares.

Problem: A shape pattern starts with $1$ square, then makes an L-shape with $3$ squares, then a bigger L with $5$ squares, then $7$ squares. How many squares will the 8th figure have?

Solution:

The number of squares in each figure: $1, 3, 5, 7, \ldots$

This is the sequence of odd numbers with general term $a_n = 2n - 1$.

For the 8th figure:

Answer: The 8th figure will have $15$ squares.

Problem: A triangle has $3$ sides, a square has $4$ sides, a pentagon has $5$ sides, a hexagon has $6$ sides. If this pattern continues, how many sides does the 10th polygon in this sequence have? What is it called?

Solution:

The number of sides follows the pattern: $3, 4, 5, 6, \ldots$

The $n$th polygon has $(n + 2)$ sides.

For $n = 10$: $10 + 2 = 12$ sides.

A $12$-sided polygon is called a dodecagon.

The sequence of polygon names:
- $n = 1$: Triangle ($3$ sides)
- $n = 2$: Quadrilateral ($4$ sides)
- $n = 3$: Pentagon ($5$ sides)
- $n = 4$: Hexagon ($6$ sides)
- $n = 5$: Heptagon ($7$ sides)
- $n = 6$: Octagon ($8$ sides)

Answer: The 10th polygon has $12$ sides and is called a dodecagon.

Problem: A triangle has $0$ diagonals, a quadrilateral has $2$, a pentagon has $5$, and a hexagon has $9$. Find the number of diagonals in a heptagon ($7$ sides).

Solution:

Step 1: List the data.
| Polygon | Sides ($n$) | Diagonals ($D$) |
|---------|-------------|------------------|
| Triangle | $3$ | $0$ |
| Quadrilateral | $4$ | $2$ |
| Pentagon | $5$ | $5$ |
| Hexagon | $6$ | $9$ |

Step 2: Find the differences.

Differences: $2, 3, 4, \ldots$ (increasing by $1$).

The next difference is $5$, so the heptagon has $9 + 5 = 14$ diagonals.

Step 3: Verify with the formula. The number of diagonals of an $n$-sided polygon is:

Answer: A heptagon has $14$ diagonals.

Problem: Using matchsticks, $1$ triangle needs $3$ matchsticks, $2$ triangles in a row need $5$ matchsticks, $3$ triangles need $7$ matchsticks. How many matchsticks are needed for $10$ triangles?

Solution:

Step 1: Observe the pattern.
| Triangles | Matchsticks |
|-----------|-------------|
| $1$ | $3$ |
| $2$ | $5$ |
| $3$ | $7$ |
| $4$ | $9$ |

The common difference is $2$.

Step 2: Find the general term.

Step 3: For $n = 10$:

**Why $2n + 1$?** The first triangle uses $3$ matchsticks. Each additional triangle shares one side with the previous triangle, so it only adds $2$ new matchsticks.

Answer: $10$ triangles need $21$ matchsticks.

Problem: How many matchsticks are needed to make a row of $n$ squares? Find the answer for $n = 15$.

Solution:

Step 1: Count for small cases.
| Squares | Matchsticks |
|---------|-------------|
| $1$ | $4$ |
| $2$ | $7$ |
| $3$ | $10$ |
| $4$ | $13$ |

The common difference is $3$.

Step 2: Find the general term.

**Why $3n + 1$?** The first square uses $4$ matchsticks. Each subsequent square shares one side, adding only $3$ new matchsticks.

Step 3: For $n = 15$:

Answer: $15$ squares need $46$ matchsticks.

Problem: An arrow points right, then up, then left, then down, then right again. Which direction does the arrow point at the 23rd position?

Solution:

The pattern repeats every $4$ steps: Right, Up, Left, Down.

To find the direction at position $23$, divide $23$ by $4$:

The remainder is $3$, so the arrow is in the same position as the 3rd term.

Position 3 = Left.

Answer: The arrow points left at the 23rd position.

Problem: A pattern of squares grows as follows: Step 1 has a $1 \times 1$ square, Step 2 adds squares to make a $2 \times 2$ square, Step 3 makes a $3 \times 3$ square. How many new small squares are added at Step $n$?

Solution:

Total squares at Step $n$: $n^2$.

New squares added at Step $n$:

Let us verify:
- Step 1: $1$ new square ($2 \times 1 - 1 = 1$). $\checkmark$
- Step 2: $3$ new squares ($2 \times 2 - 1 = 3$). Total: $4$. $\checkmark$
- Step 3: $5$ new squares ($2 \times 3 - 1 = 5$). Total: $9$. $\checkmark$
- Step 10: $2 \times 10 - 1 = 19$ new squares.

Answer: At Step $n$, exactly $2n - 1$ new squares are added.

Problem: A Koch snowflake starts with an equilateral triangle of side $1$. At each step, the number of sides is multiplied by $4$. The starting shape has $3$ sides. How many sides does the shape have after $4$ steps?

Solution:

Step 0: $3$ sides
Step 1: $3 \times 4 = 12$ sides
Step 2: $12 \times 4 = 48$ sides
Step 3: $48 \times 4 = 192$ sides
Step 4: $192 \times 4 = 768$ sides

The general formula: after $n$ steps, the number of sides is $3 \times 4^n$.

Check: $3 \times 4^4 = 3 \times 256 = 768$. $\checkmark$

Answer: After $4$ steps, the Koch snowflake has $768$ sides.

Problem: The number of petals in many flowers follows the Fibonacci sequence: $1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots$. Lilies have $3$ petals, buttercups have $5$, delphiniums have $8$, marigolds have $13$, and daisies have $21$ or $34$. What are the next two Fibonacci numbers after $34$?

Solution:

Using the Fibonacci rule ($F_n = F_{n-1} + F_{n-2}$):

Answer: The next two Fibonacci numbers are $55$ and $89$.

Problem: In a calendar, if the 3rd of a month is a Monday, what day is the 31st of the same month?

Solution:

From the 3rd to the 31st: $31 - 3 = 28$ days.

$28 = 4 \times 7$, so $28$ days is exactly $4$ complete weeks.

Since $28$ days later lands on the same day of the week:

The 31st is also a Monday.

Answer: The 31st is a Monday.

Problem: A floor is being tiled with square tiles. If you need $1$ tile for a $1 \times 1$ floor, $4$ tiles for a $2 \times 2$ floor, $9$ tiles for a $3 \times 3$ floor, how many tiles do you need for a $12 \times 12$ floor?

Solution:

The number of tiles follows the square number pattern: $n^2$.

For a $12 \times 12$ floor:

Answer: You need $144$ tiles.

Problem: A bacterial colony doubles every hour. If there are $100$ bacteria at $12:00$ PM, how many bacteria will there be at $6:00$ PM?

Solution:

The population follows a geometric sequence with first term $a = 100$ and common ratio $r = 2$.

After $n$ hours: $P = 100 \times 2^n$

From $12:00$ PM to $6:00$ PM is $6$ hours.

Answer: There will be $6400$ bacteria at $6:00$ PM.

Problem: At a party, everyone shakes hands with everyone else exactly once. If there are $8$ people, how many handshakes take place?

Solution:

Step 1: Each person shakes hands with $7$ others. That gives $8 \times 7 = 56$ handshakes, but each handshake is counted twice (once for each person).

Step 2: Actual handshakes:

Step 3: Verify the pattern.
| People | Handshakes |
|--------|------------|
| $2$ | $1$ |
| $3$ | $3$ |
| $4$ | $6$ |
| $5$ | $10$ |
| $6$ | $15$ |
| $7$ | $21$ |
| $8$ | $28$ |

These are triangular numbers! $\frac{n(n-1)}{2}$ for $n$ people.

Answer: $28$ handshakes take place.

Problem: Cannonballs are stacked in a pyramid: $1$ ball on top, $4$ in the second layer ($2 \times 2$), $9$ in the third layer ($3 \times 3$), $16$ in the fourth ($4 \times 4$). How many cannonballs are in a pyramid with $6$ layers?

Solution:

Each layer $k$ has $k^2$ cannonballs. The total for $n$ layers is:

For $n = 6$:

Verification: $1 + 4 + 9 + 16 + 25 + 36 = 91$. $\checkmark$

Answer: A $6$-layer pyramid has $91$ cannonballs.

Problem: In the multiplication table from $1$ to $10$, how many times does the product $12$ appear? List all the ways to get $12$.

Solution:

$12$ can be written as a product of two numbers from $1$ to $10$ in these ways: