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Patterns in Mathematics - Visualising Number Sequences

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A number pattern is a sequence of numbers that follow a specific rule. For example, in 2,4,6,8,…2, 4, 6, 8, \dots, the rule is to add 22 to the previous term.

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Visual patterns use shapes like dots or matchsticks to represent numbers. Common visual patterns include Square Numbers and Triangular Numbers.

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Square Numbers are numbers that can be arranged in a square grid of dots. The sequence is 1,4,9,16,25,…1, 4, 9, 16, 25, \dots, which corresponds to 12,22,32,42,52,…1^2, 2^2, 3^2, 4^2, 5^2, \dots.

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Triangular Numbers are numbers that can be arranged in the shape of an equilateral triangle. The sequence is 1,3,6,10,15,…1, 3, 6, 10, 15, \dots.

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Rules for patterns can be expressed using variables. If 'nn' represents the position of the term, the rule for even numbers is 2n2n and for odd numbers is 2n−12n - 1.

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Growing patterns are sequences where each term is larger than the previous one by a fixed amount or a changing amount, such as 1,3,7,15…1, 3, 7, 15 \dots.

📐Formulae

nth Square Number=n×n=n2n^{th} \text{ Square Number} = n \times n = n^2

nth Triangular Number=n(n+1)2n^{th} \text{ Triangular Number} = \frac{n(n + 1)}{2}

nth Even Number=2nn^{th} \text{ Even Number} = 2n

nth Odd Number=2n−1n^{th} \text{ Odd Number} = 2n - 1

Sum of first n odd numbers=n×n=n2\text{Sum of first } n \text{ odd numbers} = n \times n = n^2

💡Examples

Problem 1:

Observe the sequence 7,12,17,22,…7, 12, 17, 22, \dots. Find the rule and the 10th10^{th} term.

Solution:

nth term=5n+2n^{th} \text{ term} = 5n + 2; 10th term=5210^{th} \text{ term} = 52

Explanation:

The difference between consecutive terms is 55 (12−7=512-7=5, 17−12=517-12=5). To get the first term (77) when n=1n=1, we use 5(1)+25(1) + 2. So the rule is 5n+25n + 2. For the 10th10^{th} term, substitute n=10n=10: 5(10)+2=50+2=525(10) + 2 = 50 + 2 = 52.

Problem 2:

Using the formula for triangular numbers, find the 6th6^{th} triangular number.

Solution:

2121

Explanation:

The formula for the nthn^{th} triangular number is n(n+1)2\frac{n(n + 1)}{2}. For n=6n = 6: 6×(6+1)2=6×72=422=21\frac{6 \times (6 + 1)}{2} = \frac{6 \times 7}{2} = \frac{42}{2} = 21.

Problem 3:

If a pattern of squares made with matchsticks follows the rule 3n+13n + 1, where nn is the number of squares, calculate how many matchsticks are needed for 1212 squares.

Solution:

3737

Explanation:

Substitute n=12n = 12 into the given rule: 3(12)+1=36+1=373(12) + 1 = 36 + 1 = 37. Therefore, 3737 matchsticks are required.

Problem 4:

Find the sum of the first 88 odd numbers without adding them individually.

Solution:

6464

Explanation:

The sum of the first nn odd numbers is given by the formula n2n^2. Here n=8n = 8, so the sum is 8×8=648 \times 8 = 64.