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Introduction to Polynomials - Recognise and generalise linear patterns from tables, sequences, and contexts

Grade 9CBSE

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

🔑Concepts

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A sequence is an ordered list of numbers where each number is called a term, denoted by T1,T2,T3,…,TnT_1, T_2, T_3, \dots, T_n.

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A linear pattern occurs when the difference between any two consecutive terms remains constant. This is known as the common difference (dd).

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Linear patterns can be represented as a first-degree polynomial of the form p(n)=an+bp(n) = an + b, where nn is the position of the term.

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In a table of values representing a linear relationship between xx and yy, the rate of change is constant, which corresponds to the slope mm in the linear equation y=mx+cy = mx + c.

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To generalise a pattern, we identify the first term (aa) and the common difference (dd) to find the nthn^{th} term formula.

📐Formulae

d=Tn−Tn−1d = T_{n} - T_{n-1}

Tn=a+(n−1)dT_n = a + (n - 1)d

y=mx+cy = mx + c

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

💡Examples

Problem 1:

Observe the following sequence: 7,12,17,22,…7, 12, 17, 22, \dots. Find the general rule (nthn^{th} term) and the 20th20^{th} term.

Solution:

First term a=7a = 7. Common difference d=12−7=5d = 12 - 7 = 5. Using the formula Tn=a+(n−1)dT_n = a + (n - 1)d: Tn=7+(n−1)5T_n = 7 + (n - 1)5 Tn=7+5n−5T_n = 7 + 5n - 5 Tn=5n+2T_n = 5n + 2 For the 20th20^{th} term, substitute n=20n = 20: T20=5(20)+2=100+2=102T_{20} = 5(20) + 2 = 100 + 2 = 102

Explanation:

The difference between terms is constant (5), indicating a linear pattern. The general rule is a linear polynomial in terms of nn.

Problem 2:

Given the table below, find the linear equation relating xx and yy: x123y5811\begin{array}{|c|c|c|c|} \hline x & 1 & 2 & 3 \\ \hline y & 5 & 8 & 11 \\ \hline \end{array}

Solution:

  1. Find the change in yy: 8−5=38 - 5 = 3 and 11−8=311 - 8 = 3. So, m=3m = 3.
  2. Use the form y=mx+cy = mx + c. Substitute x=1,y=5x = 1, y = 5: 5=3(1)+c5 = 3(1) + c 5=3+c5 = 3 + c c=5−3=2c = 5 - 3 = 2 The equation is y=3x+2y = 3x + 2.

Explanation:

Since the increase in yy for every unit increase in xx is constant, the relationship is linear. We solve for the constant cc using one pair of values.

Problem 3:

A taxi charges a fixed base fare of ₹50 and an additional ₹15 per kilometer. Represent this as a linear polynomial where xx is the distance in km.

Solution:

Fixed cost = ₹50. Variable cost = 15×x15 \times x. Total cost C(x)=15x+50C(x) = 15x + 50. If a person travels 1010 km, the fare is: C(10)=15(10)+50C(10) = 15(10) + 50 150+50200\begin{array}{r} 150 \\ +50 \\ \hline 200 \end{array} The fare is ₹200.

Explanation:

Contextual problems are generalised by identifying the fixed value (y-intercept/constant) and the rate of change (coefficient of xx).