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Patterns in Mathematics - Relations among Number Sequences

Grade 6CBSE

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

🔑Concepts

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A number pattern or sequence is a list of numbers that follow a specific rule to find the next terms. For example, in the sequence 3,6,9,12,…3, 6, 9, 12, \dots, each term is a multiple of 33.

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Identifying the rule involves looking at the difference or ratio between consecutive terms. If the difference is constant, such as in 5,10,15,20,…5, 10, 15, 20, \dots, the rule is adding 55 (or 5×n5 \times n where nn is the position).

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Square Numbers are formed by multiplying a number by itself. 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 represented as dots arranged in an equilateral triangle. The sequence starts 1,3,6,10,15,…1, 3, 6, 10, 15, \dots.

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Generalizing a pattern means finding a formula using a variable nn (representing the position of the term) so that any term in the sequence can be calculated without listing all previous terms.

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Relations between sequences: Sometimes two sequences are related. For example, the sequence of even numbers 2,4,6,…2, 4, 6, \dots and odd numbers 1,3,5,…1, 3, 5, \dots are related by the rule Odd Number=Even Number−1\text{Odd Number} = \text{Even Number} - 1.

📐Formulae

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

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

nth term of a sequence with constant difference d=a+(n−1)dn^{th} \text{ term of a sequence with constant difference } d = a + (n - 1)d

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

💡Examples

Problem 1:

Find the 15th15^{th} term of the number sequence 7,14,21,28,…7, 14, 21, 28, \dots.

Solution:

The given sequence is a table of 77. The rule for the nthn^{th} term is 7×n7 \times n. To find the 15th15^{th} term, substitute n=15n = 15: 7×15=1057 \times 15 = 105

Explanation:

Since the difference between each term is constant (14−7=714 - 7 = 7, 21−14=721 - 14 = 7), the sequence follows the multiples of 77.

Problem 2:

Determine the 6th6^{th} triangular number and show the addition pattern.

Solution:

Triangular numbers are formed by the sum of consecutive natural numbers: 1st=11^{st} = 1 2nd=1+2=32^{nd} = 1 + 2 = 3 3rd=1+2+3=63^{rd} = 1 + 2 + 3 = 6 4th=1+2+3+4=104^{th} = 1 + 2 + 3 + 4 = 10 5th=1+2+3+4+5=155^{th} = 1 + 2 + 3 + 4 + 5 = 15 6th=1+2+3+4+5+6=216^{th} = 1 + 2 + 3 + 4 + 5 + 6 = 21 Using the formula: 6(6+1)2=6×72=422=21\frac{6(6 + 1)}{2} = \frac{6 \times 7}{2} = \frac{42}{2} = 21

Explanation:

A triangular number at position nn is the sum of all integers from 11 to nn.

Problem 3:

Observe the pattern and find the missing value: 11×11=12111 \times 11 = 121 111×111=12321111 \times 111 = 12321 1111×1111=?1111 \times 1111 = ?

Solution:

Observing the pattern, the middle digit corresponds to the number of 1s1s in the number being squared. For 1111 (two 1s1s), the middle is 22. For 111111 (three 1s1s), the middle is 33. For 11111111 (four 1s1s), the middle will be 44, with digits increasing to 44 and then decreasing: 1111×1111=12343211111 \times 1111 = 1234321

Explanation:

This is a palindromic number pattern specific to squares of numbers consisting only of the digit 11.