Identify linear relationships in patterns where the difference between consecutive terms is constant.
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Introduce the visual pattern from Fig 2.4.
Let’s observe a growing pattern of square tiles: Stage 1 has 1 tile, Stage 2 has 3 tiles, Stage 3 has 5 tiles, and Stage 4 has 7 tiles.
Extracting the formula 2n - 1.
To predict the number of tiles at any given stage, we need to connect the stage number to the tile count. We observe that the number of squares at each stage is one less than twice the term number.
Worked example based on Example 7.
Problem. Bela has ₹100 for pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?
Faded example based on Example 8.
An auto-rikshaw fare starts at ₹25 and remains the same for the initial 2 km. Then it increases by ₹15 per km. We want to determine the fare for a travel of 10 km and generalize this for km.
For the initial 2 km, the fare is ₹25, leaving 8 km to be charged at the per kilometre rate. Therefore, the total fare for a travel of 10 km will be calculated as , which equals ₹.
To find the total fare for a travel of km, we use the expression for . Expanding and simplifying this gives the linear expression .
This demonstrates that the fare for a specific distance covered is a linear function of the distance km.
Order the steps to define a linear pattern.
Drag the steps into the correct logical order for finding a linear pattern equation.
Guided problem based on Exercise 2.3 Q1.
Problem: Model the savings account balance as a linear polynomial.
Base balance: ₹500 Monthly addition: ₹150 Time: months
List the starting amount and the constant monthly difference.
State the general formula for linear growth.
Substitute the known values into your formula.
Arrange the expression in the standard polynomial form (ax + b).
State your final answer clearly.
Test your expression for n=1 to see if it matches reality.