Step 1: Seeing the pattern
Do you like patterns? Math loves patterns too.
What pattern do you see in this list?
1, 2, 3, 4, 5, 6, 7, 8, 9
Familiar? These are natural numbers. Right? But what else do you see? Is there a pattern here?
Now, this list:
1, 3, 5, 7, 9, 11
What is the pattern here? All odd numbers?
Let’s examine this list:
1, 5, 9, 13, 17, 21
What’s the pattern here? As you guessed correctly, there are jumps of 4 between two consecutive numbers in the list. Similarly we can make any list that has a pattern (a specified number of jumps) between two consecutive numbers in the list. For example,
2, 7, 12, 17, 22, 27
Can you guess the next number?
Step 2: Introducing Sequence and Arithmetic Progression
These lists are called sequences. An arithmetic progression (AP) is a sequence of numbers that have a pattern that the difference between any two consecutive numbers is constant.
Now the difference can be negative as well. What do you think of this sequence:
30, 26, 22, 18, 14, 10, 6, 2
This is an AP with the constant difference -4.
Step 3: Nomenclature
In an arithmetic progression, the first number is called the “initial term.” And all other numbers are called terms too.
The value by which consecutive terms increase or decrease is called the “common difference.”
In the sequence given below, what is initial term and what is common difference?
0, 5, 10, 15, 20, 25
Initial term (represented as a1) is 0 and common difference (represented as d) is 5.
As you have guessed, if d is positive (>0), the arithmetic progression is an increasing sequence. And if d is negative (<0), the arithmetic progression is a decreasing sequence.
Step 4: Finding value of any term beginning from the beginning
Let’s call any general term (nth term) an. So how would you go about finding the value of nth term without actually writing the whole sequence out? Which by the way, may not be possible as APs go on till infinity (or negative infinity, in case of decreasing APs).
Let’s figure this out. In the sequence above, a1 = 0, d = 5. And we know:
a2 = a1+d
a3 = a2+d
Therefore, a3 = a1+d+d
Similarly, a4 = a1+d+d+d and so on.
We observe that to reach a certain term an, we add (n-1) times d to initial term a1.
an = a1+ (n-1)d
Try out a few times on your own with various sequences.
Step 5: Finding value of any term beginning from anywhere in the middle
Now, what if we do not want to depend on initial term a1 and want to reach an through any random term, let’s say ak?
By the same logic, we will get:
An = ak + (n-k)d
In the example above, try it out.
Some Practice equations can be given to solve in class. My favorite one being: What is the sum of the first 100 positive integers?
Q2. Four years old Michael plays with Lego bricks. He wants to built the construction shown in the Figure 1, with 4 bricks at the bottom.How many Lego bricks does he need?
Videos for Teachers reference: