Rows and Columns, Arithmetic and Geometric Progressions - Desklib

4.3 Rows and Columns
Method 1
Considering the given rectangle given. To get two rows in a column, 3 toothpicks are required
horizontally. In simple terms, r +1 toothpicks are required. Where, r represented the number of
rows.
Number of toothpicks, n=r +1
For c columns, total number of toothpicks to make the rectangle n=c (r +1)
Similarly, for columns 2 toothpicks are required in to make a row.
Number of toothpicks, n=c +1
Total number of toothpicks in r rows n=r (c +1)
Hence total number of toothpicks to make a rectangle with r rows and c columns is given;
N=c ( r +1 )+r ( c +1 )
From the derived formula we find three rules:
i) Considering the rectangular grid length, there will always be one less toothpick in
comparison to number of columns.
ii) Considering the rectangular grid width, there will always be one less toothpick in
comparison to number of rows.
iii) For all bordering squares, there will always be one toothpick shared between two
squares.
These rules apply to all sets of grids formed. The method used in formulating the formula made
use of the rules.
Justifying the formula:
Using the diagram, we can solve for the number of toothpicks.

Number of rows r =2
Number of columns c=4
Total number of toothpicks N=4 ( 2+1 ) +2 ( 4 +1 ) =22
END OF PROOF!!
5.2 Describe the Rule
1. 4 , 6.5 , 9 ,11.5 ,...
This is an arithmetic progression. We evaluate its formula as follows;
First term: a1=4
Constant difference: d=6.5−4=2.5
Thus, the formula of getting the nth term will be
an=a1 +(n−1)d
Solving the next three terms;
a5=4 + ( 5−1 ) 2.5=14
a6=4 + ( 6−1 ) 2.5=16.5
a7=4 + ( 7−1 ) 2.5=19
Thus, the next three terms are; 14, 16.5 and 19 respectively.
2. 5 , 10 ,20 , 40 ,...

This is a geometric progression.
First term: a1=5
Common ratio: r =10
5 =2
Thus, the nth term will be evaluated by the formula: an=a1 2n−1
Solving the next three rems;
a5=5∗25−1=5∗24=80
a6=5∗26−1=5∗25=160
a7=5∗27−1=5∗26=320
Hence, the next three terms are; 80, 160 and 320 respectively.
3. 20 , 17 , 14 , 11, ...
This is an arithmetic progression series (Nunes, 2017).
First term: a1=20
Constant difference: d=17−20=−3
The nth term will be solved by the formula an=a1 + ( n−1 ) d
Solving the next three terms,
a5=20+ ( 5−1 )∗−3=8
a6=20+ ( 6−1 )∗−3=5
a7=20+ ( 7−1 )∗−3=2
Hence, the next three terms are; 8, 5 and 2 respectively.
4. 1 ,3 ,6 , 10 , 15 , ...
The pattern displayed by the terms;