Arithmetics is the basic foundation for other Branches of Mathematics mentioned in the introduction to this book.
It focuses on the four (4) basic operations, which are: ADDITION (+), SUBTRACTION (-), MULTIPLICATION (×) AND DIVISION ( ÷ or / ).
Let me use simple numbers like '4 and 2' to explain these operations:
*When 4 is added to 2, we get 6 as the result. That's Addition (4+2=6).
*When 4 is subtracted by 2, we get 2 as the result. This is Subtraction (4-2=2).
*When 4 is multiplied by 2, we get 8 as the result. This is Multiplication (4×2=8).
*And when 4 is divided by 2, we get 2 as the result. This is Division (4/2 or 4 ÷ 2 = 2).
A. Sometimes, these operations could be combined together in just a problem (question). Like:
10+(5×2)-6/2
Solution
*First multiply the brackets operation in order to get rid of the brackets (BODMAS)
10+(5×2)-6/2
= 10+10-6/2 (then the division)
= 10+10-3
= 17 (answer)
Note: The BODMAS rule should be applicable when dealing with complicated problems involving different operations.
BODMAS stands for...
*Brackets
*Open
*Division
*Multiplication
*Addition
*Subtraction
The above simply means, when dealing with a problem having different arithmetic operations, BODMAS should be followed word after word... Brackets should be Opened by treating whatever is in them, then followed by Division, next Multiplication, then Addition, and finally Subtraction.
B. Sometimes, the problems could be in an English Sentence form. Little complicationand, right? But it's not a big deal. It would only require that you read carefully and understand the sentence, then the sentence should be interpreted in mathematical format before calculations.
Like...
1. Question. The total age of three kids in a family is 27 years. What will be the total of their ages after three years?
Solution:
*We know about 3 kids' Total age together=27
*After 3 years, the age of each increases by 3, but we don't know the age of each (we only know the result of adding the ages of the 3 kids).
So,
*Sum of the 3 kids ages = 27
*After 3 years, each kid increases by 3 = 3×3
*Therefore 27+(3×3)
=27+9
=36 years (answer)
This means, there would be an increase of 3 years to 27 after the first year, then 6 years to 27 after the second year, and 9 years to 27 after the third year, making 9 years increment to 27.
2. Question. Find the product of all the numbers present on the calculator pad.
*Solution.
Note: Product is another word for Multiplication
*What are all the numbers (not keys or symbols) present on the calculator pad?
0,1,2,3,4,5,6,7,8,9
Product would then be...
0×1×2×3×4×5×6×7×8×9
We would have got a very huge number as an answer, but for the presence of zero (0) which renders everything useless.
= 0 (answer)
This is because any number multiplied by zero, or zero multiplied by any number is still zero (0).
3. Question. Peter ran 12 laps every day for two weeks. How many laps did her run in all?
Solution:
*12 laps a day
*A week is 7 days, two weeks is 14 days
So, 12 laps multiply by 14
=12×14
= 168 laps (answer)
Note: Some problems could be so complicated that certain General formulas had to be derived for clearer, better and easier solving.
You must have heard of certain concepts like Arithmetic Sequence (Progression), Geometric Sequence (Progression), right?
*ARITHMETIC SEQUENCE (or PROGRESSION)
According to encyclopedia, An arithmetic sequence, also known as an arithmetic progression, is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference.
In my words, an Arithmetic Progression is simply a list of numbers following certain patterns.
*Eg. In the following list of numbers,
2,5,8,11,14,17,20, we can say it is an Arithmetic Sequence because they have a Common (or Same) difference, which is 3.
Note: If a list of so-called sequence does not have a Common Difference, then such can't be referred to an arithmetic sequence. It is the Common Difference that validates it to be called an Arithmetic Sequence.
Had it been the above sequence was:
2,5,7,9,12,14,20, then sorry, it becomes invalid, 'cause here the difference varies.
*Why are Arithmetic Progressions Important?
They are a fundamental concept in mathematics with widespread applications in various fields like Finance, Engineering, and Computer Science. Therefore, understanding them helps in problem-solving, planning, and analyzing patterns in real-world scenarios.
*For Mathematical Operations in the form of Arithmetic Progressions/Sequences, we have a General Formula,
an = a + (n - 1)d
{or a_n = a + (n - 1)}
This formula helps in finding the value of a "term" or a figure in a given progression.
Where:
...'an' (it's actually a_n): is the Value of Term asked in a given sequence (Number of AP)
...'a': is the first Term in a given sequence
...'d': is the Common Difference in a given sequence.
...'n': represents the position of the Term you are trying to find (e.g., if you want the 4th term, n=4).
*Let me use a simple arithmetic sequence with a complete information to explain this formula better.
*2,4,6,8,10
1. From the above, we already know the 4th Term is 8, but let's just pretend we didn't know, and imagine we were asked to give the 4th Term.
That means we would assume the question to be: What is the 4th Term of the given sequence, 2,4,6,...,10?
Solution.
Using, an=a+(n-1)d
an = a4 because we are finding the 4th Term
a = 2 because it's the first Term
d = 2 because that's the common difference
n = 4 that's the 4th Position we are asked to find
So, an=a+(n-1)d
a4=2+(4-1)2
a4=2+(3)2
a4=2+6
a4=8 (answer)
*In some problems, we might be asked to find the Sum of Terms in a given sequence.
And instead of getting the value of each Term before adding them together, which might take a whole lot of time and stress. And imagine if the Last Term of a given sequence was not given in a question, wouldn't it be complicated?
This is why there are two General Formulas for the Sum of Terms in a given sequence.
A. If the Last Term is known:
Sn = n/2 * (a + an)
B. If the Last Term is not known:
Sn = n/2 * [2a + (n-1)d]
Where:
*Sn...Sum of Terms in a given progression (Sum of AP)
*a... First Term
*d... difference
*n... number value of a Term
*MORE EXAMPLES ON APs:
1. Question
If the first term of an AP is 5 and the common difference is 3, what is the 10th term?
*Solution:
a=5
d=3
an=10th Term
*Using an=a+(n-1)d
an=5+(10-1)3
an=5+(9)3
an=5+27
an=32 (answer)
2. Question
If the 3rd term of an AP is 10 and the 7th term is 22, what is the first term and the common difference?
*Solution
Here, the results of Two different Terms (3rd and 7th) are given, 10 and 22. But no more information given to know how these values came about. However, these unknown information happened to be what the question is requesting for:
1. First Term of the AP
2. Common Difference of the AP
First, we have to present the information given to us in the AP formula we already know:
*3rd Term of AP (a3) is: a+(3-1)d=10
*7th Term of AP (a7) is: a+(7-1)d=22
Now, in problems like this, that leaves us with Equations having Two unknown values, we have to apply or borrow a different approach.
... Let's hold on here for a moment...
*EQUATIONS
I never wanted to talk about this just yet, but I have to, because even though we were concerned about Arithmetics, the question got complicated that we would have to borrow some knowledge from Algebra (though elementary) in order to complete the solution to the Problem we put on hold.
When we get to ALGEBRA properly, in a different Chapter, I would, of course, go much deeper into this concept.
However, (it is assumed we are stuck here and need this knowledge). Therefore, let me give a summary of this.
You see, an Equation is simply a mathematical statement that shows an expression or a value is equal to another.
Examples:
(i) 2+3=5
(ii) 3+5=10-2
Most times, one or more of the values would be represented by an alphabet (s) and we would be asked to find the value.
From the second example above, 3+5=10-2, let's assume it was, 3+5=X-2. Now, let's find the value of "X":
3+5=X-2
8=X-2
8+2=X (separating like -terms or similar-terms)
10=X or X=10 (answer)
Now that we know what an equation is, understand that equations come in different forms and patterns.
Sometimes the value of an unknown could just be in the "Power of 1" (eg. X-2=6).
Other times, am unknown could carry a "square-power" value (eg. 2+X^2=3×4)
Note: The symbol (^) is used to represent "Power of a given number or value"
So, X^2 is simply "X" raised to the power of "2". In handwriting form, we usually ignore the Power Sign and just write the "2" a bit smaller at the top-right corner of the number or value or variable in question.
*Let me just go straight to the point for the sake of our pending problem in Arithmetics.
In Mathematics, there are Equation Concepts we call Polynomials:
A polynomial is a mathematical expression consisting of variables, coefficients, and the operations of addition, subtraction, and multiplication, with non-negative integer exponents.
Now,
1. Linear Equation: 2x + 3 = 0 (power/degree 1)
2. Quadratic Equation: x² - 4x + 4 = 0 (power/degree 2)
3. Cubic Equation: x³ + 2x² - 5x + 1 = 0 (power/degree 3)
4. Quartic Equation: x⁴ - 3x³ + 2x² - x + 7 = 0 (power/degree 4)
5. Quintic Equation is when the power/degree is up to 5... and so on.
My point is, sometimes two Equations appear in a single problem like the one we put on hold:
*3rd Term of AP (a3) is: a+(3-1)d=10
*7th Term of AP (a7) is: a+(7-1)d=22
We call these Simultaneous Linear Equations, since both of them are Linear in nature:
a+(3-1)d=10... equation 1
a+(7-1)d=22... equation 2
Personally, I know of 3 different approaches to this: 1. factorization by Substitution
2. factorization by Elimination
3. graphical or graphing approach
So, I will be using the Substitution Method.
Here, we solve one equation for one variable in terms of the other.
*From equation....1
a+(3-1)d=10 (let's solve for "a")
a+(2)d=10
a+2d=10 (now, we separating like -terms)
a=10-2d (so, for now, the value for "a" is "10-2d", and we are going to input it into the second equation as the value of "a")
a+(7-1)d=22... (equation 2)
(10-2d)+(7-1)d=22
10-2d+6d=22
-2d+6d=22-10 (separating like -terms)
4d=12 (now, we divide both sides of the equation by the coefficient of the unknown "d" which is "4")
4d/4 = 12/4 (the "4" standing with "d" cancels out the "4" under, and "12" is divided by "4")
d = 3 (answer for Difference according to the question)
Now, we substitute "3 which is the value for d" into any of the Two Equations.
Let's use equation....1
a+(3-1)d=10 (equation....1)
a+(3-1)3=10
a+(2)3=10
a+6=10
a=10-6
a=4 (answer for First Terms according to the question)