YouTube Script: Arithmetic Progression
Script: Arithmetic Progression
Topic: Arithmetic Progression (A.P.)
CONCEPT: Introduction to Arithmetic Progression
KT: Need for A.P. in Real-life and Math with Solved Examples
Neha is solving various types of questions to clear her math exam. She solves multiple questions on different concepts each day. However, she is confused about how many questions she should solve to avoid mixing concepts in a variety of questions. Her concern is scoring well in her upcoming exam, so she looks for a good tutor who could guide her in her preparation.
Neha
Hello! Sir? (greeting with a smile)
Professor David
Hey Hi! Please come in! Tell me what brings you here today?(responding with a smile)
Neha
Well! I am scared of my maths exam(concerned)
Professor David
Ok! Why do you feel scared about such an interesting subject? (trying to understand the situation with a polite gesture)
Neha
Well, I feel like I’m solving a lot of questions each day without sticking to a single concept and solving questions based on the same. (tense)
Mr. Khanna
Well, it is good to solve multiple questions, however, when you are not clear of a single concept, it makes you feel scared, so better you stick to one topic first, clear all your doubts, and then switch to the next one? (a positive solution with a smile)
Neha
Right sir! Could you please provide a strategy for preparation please? (requesting)
Professor David
Yes! Of course! (happy to help)
Now, Professor David starts with a step-by-step preparation strategy.
Professor David
Let us say that you are doing chapter 1 — Real Numbers. After understanding the concept, solve two questions on the first day, four questions on the second day, six questions on the third day, 8 questions on the fourth day, and so on.
So, it is like 2, 4, 6, 8,….
Neha
Why did you write 2, 4, 6, 8,…. (curious to know)
(Professor David is trying to explain the step-by-step method she should follow in her preparation journey)
Professor David
Here, you notice that the number of questions you need to solve each day has a difference of 2, which is constant between 2 and 4, 4 and 6, 6 and 8, and so on, so what is this forming? (Asking a question)
Professor David is testing Neha to understand what she answers.
Neha
Well, it is like I have to solve questions in a table of 2?(Hesitant in answering)
Professor David
No Dear!
Now listen! When you have numbers separated by a comma having a common difference between each is called a sequence. So, here 2, 4, 6, 8,…. is a sequence where ‘2’, ‘4’, ‘6’ are terms and ‘2’ is the common difference between each term
As Professor explains the step-by-step preparation strategy to Neha, she comes across a new concept of “Sequence”, however, she mixes this term with “Series”. Professor comes to explain this concept. Neha was not satisfied with her current tutor, so she looked for an online platform where she could switch to another tutor of her choice anytime.
Now presenter (online tutor) addressed the audience
PRESENTER
Hey Guys! You might have seen a set of numbers written in the following manner:
3, 7, 11, 15, 19, 23, 27
5, 10, 15, 20, 25, 30,…..
15, 12, 6, 9, 3,….
In the first series, you see that thee numbers are written in an increasing order and if you see the difference: 7–3 = 4 or 11–7 = 4,… Each time, the difference comes out to be 4.
Similarly, in the second arrangement, you find the difference 10–5 = 5, 15–10 = 5, and so on, the difference is the same, i.e., 5.
So, what does it mean? I know what you are thinking about. Well, in this situation, all three examples stated above are sequences.
So, what is a sequence? When you have terms, like 3, 7, 11,…,27 and the difference comes out to be common, we call it sequence and here you are adding ‘4’ to the current term to get the next term.
Also, 3,..27 has a finite number of terms, so we call it a finite sequence.
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Well, on the other hand, 5, 10, 15, 20,…. It ends nowhere, so we call it an infinite sequence.
Now, in the third sequence, you see that 15–12 = 3, then 12–9 = 3,…Here, the series is endless and arranged in a descending order, so we subtracted the next term from the previous term, as the previous term is greater than the next term.
Thus, in all the three examples, we have used mathematical operators like addition and subtraction and that too progressively.
So, whenever we apply arithmetic operations progressively on a sequence, it becomes an Arithmetic Progression.
Now, let us understand Arithmetic Progression in Detail. We will also understand this concept by solving questions on the same.
Let us take another example: 162, 54, 18, 6, 2
Here,
The first term a1 = 162
The second term a2 = 54
The third term a3 = 18
The fourth term a4 = 6, and
The fifth term a5 = 2
So, we have five terms in total, which means the series is finite.
Here a1 > a2 Now, let us find the difference:
- a1 / a2 = 162/54 = 3
- a2 / a3 = 54/18 = 3
- a4 / a3 = 18/6 = 3, and so on.
You notice that in an descending order sequence, we get such a type of sequence.
However, when the terms are in ascending order, we have the following example:
75, 70, 65, 60,……,5
Here, we see that a1 > a2 Now, let us find the difference:
- a1 — a2 = 75–70 = 5
- a2 — a3 = 70–65 = 5
Here, the common difference ‘d’ = 5.
Now, let us say that the sequence has infinite terms and we need to find the 15th term, for this, we have the following sequence:
6, 12, 18, 24, 30,…….
Suppose that this sequence ends at nth term. The nth term is written as an in A.P., so an is the general term of the sequence. Therefore, the formula to find the nth term will be:
n = 15 and d = a2 — a1 = 12–6 = 6
an = a1 + (n — 1) x d
= 6 + (15–1) * 6
= 6 + 14 * 6
= 6 + 84
an = 90
Hence, the 15th term will be 90.
Also, you notice that when n = 15, then a15 = a1 + 14 d
Similarly, for finding the 16th term, we have:
a16 = a1 + 15 d
For the 17th term:
a17 = a1 + 16 d, and so on, depending on the value of ’n’ you need out.
On the other hand, you have the following sequence that has a last term, but you need to find the number of terms:
6, 12, 18, 24, 30,……., 90
Here, L or the last term = 90
a1 = 6, d = 6
Use the following formula:
L = a1 + (n — 1) d
90 = 6 + (n — 1) * 6
90 = 6 + 6n — 6
=> n = 15
Therefore, we verify that the 15th term was 90 (that we calculated above).
So, this is how progressively, we can determine whether the sequence is in A.P. or not. Also, how parameters like terms, common difference, first term, last term are used in a particular sequence. I hope you were able to understand this concept clearly.
