Rows and Columns, Arithmetic and Geometric Progressions - Desklib

This document explains the formulas and rules for calculating the number of toothpicks required to make a rectangle with given rows and columns. It also provides solved examples of arithmetic and geometric progressions with formulas for finding the nth term. The document includes examples of both increasing and decreasing patterns in progressions.

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;

a2−a1=3−1=2
a3−a2 =6−3=3
a4−a3=10−6=4
From the pattern, the formula for nth term will be;
an=an−1 +n
Solving the next three terms;
a6=a5 +6=15+6=21
a7=a6 +7=21+7=28
a8=a7 +8=28+ 8=36
Hence, the next three terms are; 21, 28 and 36 respectively.
5. 1
12 , 1
6 , 1
4 , 1
3 ,…
This is an arithmetic progression. We evaluate as follows:
First term: a1= 1
12
Common difference: d= 1
6 − 1
12 = 1
12
The formula for getting nth term will be;
an=a1 + ( n−1 )∗1
12
Solving for the next three terms;
a5= 1
12 + ( 5−1 ) ∗1
12 = 5
12

a6= 1
12 + ( 6−1 )∗1
12 =1
2
a7= 1
12 + ( 7−1 ) ∗1
12 = 7
12
Hence, the next three terms are; 5
12 , 1
2 , 7
12 respectively.
6. 12 , 6 ,3 , 3
2 , …
This is a geometric progression series (Thomas Bond, 2014).
First term: a1=12
Common ration: r = 6
12 = 1
2
The formula for obtaining nth term:
an= 1
2 rn−1
The next three terms will be;
a5=12∗( 1
2 )5−1
= 3
4
a6=12∗( 1
2 )
6−1
= 3
8
a7=12∗( 1
2 )7−1
= 3
16
Thus, the next three terms are; 3
4 , 3
8 , 3
16 respectively.
7. −12 ,36 ,−108 , 324 , …
This is a geometric progression series. It is evaluated as follows;

First term: a1=−12
Common ratio: r = 36
−12 =−3
The formula for obtaining nth term becomes;
an=−12∗ (−3 )n−1
Solving for the next three terms;
a5=−12∗ (−3 )5−1=−972
a6=−12∗ (−3 )6−1=2916
a7=−12∗ (−3 )7−1=−8748
Hence, the next three terms are; -972, 2916 and -8748 respectively.
8. 1 ,3 ,8 , 16 , 27 , …
The following pattern is showed by the series;
a2−a1=3−1=2
a3−a2 =8−3=5
a4−a3=16−8=8
a5−a4=27−16=11
From the pattern, we can see that difference between successive terms form a pattern of
increasing by 3. We can formulate a formula as follows;
a6−a5 =11+3=14
a6=a5 +14=27+ 14=41
Solving for seventh term;
a7=a6 +14 +3=41+17=58
Solving for eighth term;

a8=a7 +17 +3=58+ 20=78
The next three terms are; 41, 58 and 78 respectively.

References
Nunes, R. M., 2017. ON THE LEAST SQUAREFREE NUMBER IN AN ARITHMETIC PROGRESSION.. In:
s.l.:London Mathematical Society, pp. 53-99.
Thomas Bond, C. H., 2014. A-level Mathematics Challenging Drill Questions (Yellowreef). In:
s.l.:Yellowreef Limited, pp. 123-154.