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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, \(\mathrm{T_n}\) 

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

\(\mathrm{T_n = a + d(n - 1)}\)

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

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

\(\mathrm{d}\) 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:

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

 

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

\(\begin{align*} \mathrm{T_n} &= \mathrm{1 + 3 (n-1)}\\[2ex] &=\mathrm{1 + 3n - 3}\\[2ex] &=\mathrm{3n-2} \end{align*}\)

 

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
\(1 \) \(1 \) \(4\)
\(2 \) \(2 \) \(4 + 3 = 7\)
\(3 \) \(3 \) \(4 + 3 + 3 = 10\)
\(4 \) \(4 \) \(4 + 3 + 3 + 3 = 13\)
\(5 \) \(5 \) \(4 + 3 \;(4) = 16\)
\(6 \) \(6 \) \(4 + 3 \;(5) = 19\)
\(\mathrm{n}\) \(\mathrm{n}\) \(\begin{align} \mathrm{4 + 3(n - 1)} &= \mathrm{4 + 3n - 3}\\ &= \mathrm{3n + 1} \end{align}\)

 

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

 

Example 2:

The diagram shows some patterns made from floor tiles. 

     

Find an expression, in terms of \(\mathrm{n}\), for the total number of tiles, \(\mathrm{T_n}\), 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 \(\mathrm{n}\)

\(\begin{align} \mathrm{T_n = \frac{n \;(n+1)}2} \end{align}\)

 

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 \(\mathrm{n}\), for

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

Solution: 

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

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

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

  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 \(\mathrm{n}\), the number of Space Panels would be

\(\begin{align} \mathrm{B} &= \mathrm{4 + 2(n -1)}\\ &= \mathrm{4 + 2n - 2}\\ &= \mathrm{2n + 2} \end{align}\)

Continue Learning
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Basic Algebra And Algebraic Manipulation 1 Approximation And Estimation
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