Study S1 Mathematics Maths - Number Patterns - Geniebook

Number Patterns

In this chapter, we will be discussing the below mentioned topics in detail:

  • Common Difference (Direct)
  • Common Difference (Indirect) 
  • General Term
    • Find a formula for general term given a number pattern
    • Find n given a particular iteration in the pattern
    • Determine whether a given iteration is part of a number pattern

 

Formula for the General Term, \(T_n\) 

For any number sequence where each term differs from the successive term by a constant quantity,

\(T_n = a + d \;(n\; – 1)\)

Where, \(T_n\) is the \(n^{th}\) term in the sequence,

\(a\) is the \(1^{st}\) term in the sequence, and 

\(d\) is the common difference in quantity between successive terms.

 

Let’s understand this with the help of some examples:

 

Example 1:

Consider the following number sequence:

\(1^{st}\;term\)                   \(2^{nd}\;term\)                \(3^{rd}\;term\)                \(4^{th}\;term\)                \(5^{th}\;term\)

 

For this sequence, it starts with \(1\), then \(+3\) to each term to get the next term.

\(\begin{align*} T_n &= 1 + 3 (n\; – 1)\\ &=1 + 3 n\; – 3\\ &=3n-2 \end{align*}\)

 

 

Number Patterns

A number pattern is a sequence of figures linked by a specific rule. 

 

Example 1: 

Consider the following number pattern: 

How would the next two figures look like?

 

Solution: 

 

 

Figure Number

Number Of Squares

Number Of Lines

\(1 \) \(1 \) \(4\)

\(2 \)

\(2 \)

\(4 + 3 = 7\)

\(3 \)

\(3 \)

\(4 + 3 + 3 = 10\)

\(4 \)

\(4 \)

\(4 + 3 + 3 + 3 = 13\)

\(5 \)

\(5 \)

\(4 + 3 \;(4) = 16\)

\(6 \)

\(6 \)

\(4 + 3 \;(5) = 19\)

\(n \)

\(n \)

\(\begin{align*} 4 + 3(n – 1) &= 4 + 3n – 3\\                       &= 3n + 1 \end{align*}\)

 

Hence, the number of lines in the \(n^{th}\) figure, \(L_n = 3n + 1\)

 

Example 2:

The diagram shows some patterns made from floor tiles. 

     

Find an expression, in terms of \(n\), for the total number of tiles, \(T_n\), in Figure \(n\).

Solution: 

Figure Number

Number Of Tiles

\(1\)

\(\begin{align*} 1 && && && && =1 && && = \frac{1\times2}{2} \end{align*}\)

\(2 \)

\(\begin{align*} 1+2 && && && =3 && = \frac{2\times3}{2} \end{align*}\)

\(3 \)

\(\begin{align*} 1+2+3 && && =6 && = \frac{3\times4}{2} \end{align*}\)

\(4 \)

\(\begin{align*} 1+2+3+4 &&=10 && = \frac{4\times5}{2} \end{align*} \)

In Figure \(n\)

\(T_n = \frac{n \;(n+1)}2\)

 

Question 1:

The International Space Station (ISS) consists of oval shaped Space Pods and rectangular Solar Panels. The first 3 iterations of the ISS are as shown. 

Find a formula, in terms of \(n\), for

  1. the total number of Space Pods, \(A\), in iteration \(n\), and 
  2. the total number of Space Panels, \(n\), in iteration \(n\).

 

Solution: 

  1. In Iteration 1, there is \(1\) Space Pod. 

In Iteration 2, there are \(2\) Space Pods and so on.

Hence, in Iteration n, the number of Space Pods would be \(A = n\)

 

  1. In Iteration 1, there are \(4\) Space Panels; in Iteration 2, there are \(6\) Space Panels; and in Iteration 3, there are \(8\) Space Panels and so on.

Hence, in Iteration n, the number of Space Panels would be

\(\begin{align*} B &= 4 + 2 \;(n \;– 1)\\ &= 4 + 2n \;– 2\\ &= 2n + 2 \end{align*}\)

 

 

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