‹ Class 7 · Ch 4
Expressions Using Letter-Numbers · Principle 15 of 16

A rule for a pattern

An expression in the step number gives any step at once.

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NCERT: 4.5 Pick Patterns and Reveal Relationships

Think

How many sticks at Step 33?

Matchsticks make a row of triangles. Step 1 has 1 triangle (3 sticks), Step 2 has 2 triangles (5 sticks), Step 3 has 3 triangles (7 sticks).

How can we find the sticks at Step 33 without drawing 33 steps?

What this lesson covers

The idea

The general rule of a pattern can be written as an algebraic expression in the step or position number, which then gives the term at any step.

How many sticks at Step 33?

Matchsticks make a row of triangles. Step 1 has 1 triangle (3 sticks), Step 2 has 2 triangles (5 sticks), Step 3 has 3 triangles (7 sticks).

How can we find the sticks at Step 33 without drawing 33 steps?

  • Find the rule for the pattern
  • Draw every step and count
  • 3 sticks for each triangle: 3 × 33

Build the rule

Step through the pattern. Then build the rule sticks = ? × n + ? so that it matches the table at every step.

A rule that works for every step

Each step adds 2 sticks, and Step 1 has 3. So Step y has a 3 and (y − 1) twos: 3 + 2 × (y − 1).

Step 1 is also 1 + 2 = 3. Seen this way, every step has y twos and a 1: 2y + 1.

The two rules are the same expression: 3 + 2(y − 1) = 3 + 2y − 2 = 2y + 1.

The general rule of a pattern can be written as an algebraic expression in the step number. Put in any step number and the expression gives the term at that step: 2y + 1.

  • Step y | 3 + 2 × (y − 1) | 2y + 1
  • 1 | 3 | 3
  • 5 | 11 | 11
  • 33 | 67 | 67

Notes

The general rule of a pattern can be written as an algebraic expression in the step number: here sticks = 2y + 1. It gives the term at any step.

Check yourself

Use the rule 2y + 1. How many matchsticks does Step 50 need?

Answer: 101

2 × 50 + 1 = 101.

A pattern needs 4, 6, 8, 10 sticks at Steps 1, 2, 3, 4. Which rule gives them?

On a saree border, Designs A, B, C repeat: A B C A B C … Design C is at positions 3, 6, 9, … which is 3n. Design A is at positions 1, 4, 7, … Which expression gives its positions?

Design B is at positions 3n − 1. At which position does Design B appear for the 10th time?

Answer: 29

3 × 10 − 1 = 29.

  • 2n + 2 — correct. Yes! For n = 1, 2, 3, 4 it gives 4, 6, 8, 10.
  • 2n. That gives 2, 4, 6, 8.
  • n + 3. That gives 4, 5, 6, 7.
  • 3n − 1. That gives 2, 5, 8. Those are the positions of Design B.
  • 3n + 1. That gives 4, 7, 10. It misses position 1.
  • 3n − 2 — correct. Yes! For n = 1, 2, 3 it gives 1, 4, 7.
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