SG 
VN 
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, Tn
For any number sequence where each term differs from the successive term by a constant quantity,
Tn=a+d(n−1)
Where, Tn is the nth term in the sequence,
a is the 1st 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:
1stterm 2ndterm 3rdterm 4thterm 5thterm
For this sequence, it starts with 1, then +3 to each term to get the next term.
Tn=1+3(n−1)=1+3n−3=3n−2
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 | 4+3(n−1)=4+3n−3=3n+1 |
Hence, the number of lines in the nth figure, Ln=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, Tn, in Figure n.
Solution:
| Figure Number | Number Of Tiles |
|---|---|
|
1 |
1=1=1×22 |
|
2 |
1+2=3=2×32 |
|
3 |
1+2+3=6=3×42 |
|
4 |
1+2+3+4=10=4×52 |
In Figure n,
Tn=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
- the total number of Space Pods, A, in iteration n, and
- the total number of Space Panels, n, in iteration n.
Solution:
- 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
- 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
B=4+2(n−1)=4+2n−2=2n+2
Primary
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