Patterns in Mathematics — NCERT Solutions, AI Tutor & Practice

The counting numbers increase by $1$ each time.

• After $5$: $5 + 1 = 6$
• Then $6 + 1 = 7$
• Then $7 + 1 = 8$

So the next three numbers are $6, 7, 8$.

Rule: each term is $1$ more than the previous term.

The odd numbers go up by $2$ each time.

• $9 + 2 = 11$
• $11 + 2 = 13$

So the next two terms are $11$ and $13$.

Pattern: start at $1$ and keep adding $2$. These are numbers that are not divisible by $2$.

Compute each sum:

• $1 = 1$
• $1 + 3 = 4$
• $1 + 3 + 5 = 9$
• $1 + 3 + 5 + 7 = 16$

The results are $1, 4, 9, 16$ — these are the square numbers: $1 = 1^2,\ 4 = 2^2,\ 9 = 3^2,\ 16 = 4^2$

Observation: the sum of the first $n$ odd numbers equals $n^2$.

Triangular numbers are made by adding counting numbers one at a time:

• $1$
• $1 + 2 = 3$
• $1 + 2 + 3 = 6$
• $1 + 2 + 3 + 4 = 10$
• $1 + 2 + 3 + 4 + 5 = 15$

To get the next ones, keep going:

• $15 + 6 = 21$
• $21 + 7 = 28$

So the next two triangular numbers are $21$ and $28$.

They are called triangular because that many dots arrange into a neat triangle.

Add each pair:

• $1 + 3 = 4$
• $3 + 6 = 9$
• $6 + 10 = 16$
• $10 + 15 = 25$

The results are $4, 9, 16, 25$, which are the square numbers $2^2, 3^2, 4^2, 5^2$.

Conclusion: adding two neighbouring triangular numbers always gives a perfect square.

Each term is double the previous term (multiply by $2$).

• $16 \times 2 = 32$
• $32 \times 2 = 64$
• $64 \times 2 = 128$

So the next three terms are $32, 64, 128$.

Rule: multiply by $2$ each time. These are the powers of $2$: $2^0, 2^1, 2^2, \dots$

Each term is the sum of the two terms before it:

• $1 + 1 = 2$
• $1 + 2 = 3$
• $2 + 3 = 5$
• $3 + 5 = 8$

Continuing:

• $5 + 8 = 13$
• $8 + 13 = 21$
• $13 + 21 = 34$

So the next three terms are $13, 21, 34$.

Cube numbers come from multiplying a counting number by itself three times:

• $1 = 1 \times 1 \times 1 = 1^3$
• $8 = 2 \times 2 \times 2 = 2^3$
• $27 = 3 \times 3 \times 3 = 3^3$
• $64 = 4 \times 4 \times 4 = 4^3$

The next one uses $5$: $5^3 = 5 \times 5 \times 5 = 125$

So the next cube number is $125$.

Pair numbers from the two ends:

$(1+10) + (2+9) + (3+8) + (4+7) + (5+6)$

Each pair adds up to $11$, and there are $5$ pairs:

$5 \times 11 = 55$

So $1 + 2 + 3 + \dots + 10 = 55$.

This is also the $10$th triangular number.

The figures hold square numbers of dots:

• 1st: $1^2 = 1$
• 2nd: $2^2 = 4$
• 3rd: $3^2 = 9$
• 4th: $4^2 = 16$

The 6th figure has: $6^2 = 36 \text{ dots}$

To go from the 5th figure ($25$ dots) to the 6th, we add the next odd number: $25 + 11 = 36$

So each square figure is built by adding an odd number of dots to the previous square.

Find the differences between terms:

• $5 - 2 = 3$
• $10 - 5 = 5$
• $17 - 10 = 7$
• $26 - 17 = 9$

The differences are $3, 5, 7, 9$ — consecutive odd numbers. The next difference is $11$:

$26 + 11 = 37$

So the blank is $37$.

(Notice each term is one more than a square: $1^2{+}1, 2^2{+}1, 3^2{+}1, \dots$, and $6^2 + 1 = 37$.)

Each answer is a square number:

• $1 = 1^2$
• $4 = 2^2$
• $9 = 3^2$
• $16 = 4^2$

The square equals the largest number in the middle of the row squared. In the next row the middle number is $5$, so: $1+2+3+4+5+4+3+2+1 = 5^2 = 25$

We can check by adding: $1+2+3+4+5+4+3+2+1 = 25$. ✓