53-03 Basics of Arithmetic & Geometric Progression

Click here for the previous basics.

In the Previous Blog, we had seen some important concepts of Arithmetic & Geometric Progression. 

E) Find t_n for an AP where t₅  = 19 & t₁₅  = -21.

Solution:

1) Let " a " be the first term & " d " be a common difference.
2) So, t_n = a + ( n - 1 ) d
3) According to the problem, 
            t₅   = a +  ( 4 )  d =  19
            t₁₅ = a + (14 )  d = -21
                  (-)  (-)               (+)
  --------------------------------------------
                            - 10  d  =  40
                                  -  d  =  4
                                     d  =  - 4
4) We know that 
            t₅   =  a +  ( 4 ) d =  19
                    a + 4 * (- 4) =  19
                            a  - 16 =  19
                                  a   =  19 + 16
                                  a   =  35
5) So,  t_n = a + ( n - 1 ) d
            t_n = 35 + ( n - 1 ) * (- 4)
            t_n = 35 - 4 n + 4
            t_n = 39 - 4 n
6) Answer: Here the n^th term of an AP is  t_n =  39 - 4 n

F) Find, how many 3-digit natural numbers are divisible by 3.

Solution:

1) We know that the lowest 3-digit number is 100 & the largest one is 999.
2) So the numbers divisible by 3 between 100 & 999 are
     102, 105, 108, ----, 996, 999.
3) Here a = 102, d = 3 and last term = 999.
4) So, from  t_n = a + ( n - 1 ) d, we have,
                  999 = 102 + ( n - 1 ) * 3
     3 *  ( n - 1 )  =  999 - 102 
     3 *  ( n - 1 )  =  897 
            ( n - 1 )  =  897 / 3
            ( n - 1 )  =  299
                    n    =  299 + 1
                    n    =  300
5) Answer: Here there are 300 three-digit numbers that are divisible by 3.

In the next part, we will see a few examples and some essential formulae.

Click here for the next basics.

Anil Satpute

52-02 Basics of Arithmetic & Geometric Progression

Click here for the previous basics.

In the Previous Blog, we had seen some important concepts of Arithmetic & Geometric Progression. 

In this Blog, we Describe the Important Formulas of an AP & GP.

1) t_n =  a + (n - 1) d 

2) S_n = n * (a + l)/2

3) S_n = n * [ 2 a + (n - 1) d ) ] / 2

Now we will see some examples:

Problems related to t_n =  a + (n - 1) d:

A) Find the n^th term of an AP 3, 5, 7, 9 ...

51-01 Basics of Arithmetic & Geometric Progression

In any Music, March past, or any other rhythmic work, we observe that these tasks are done in proper sequence. Let us take a very simple example of our day-to-day work. At night we sleep, then in the morning we wake up, then we finish our morning tasks. We take our breakfast then go to school/office/business spot/workplace. We take lunch in the afternoon outside those who go out else at home/hotel. Then we come home in the evening then do some entertainment, watch TV or play some computer Games then dinner then go to bed to sleep. This cycle will be maintained on weekdays. This one may be an example of a sequence if we say that each and every task is done very well.  

Harmonic Motion, Sewing needle, and Swing of the Pendulum are also examples of a sequence.

1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, 1, 3, 2....... is also an example of sequence. here the three numbers 1, 3, and 2 are getting repeated several times.

Another example will be 1, -1, 1, -1, 1,1, -1, 1, -1, 1,1, -1, 1, -1, 1,1, -1, 1, -1, 1....... 

50-List of Formulas - 06

Click here for the previous list of formulas.

Today we will see the list of formulas of the Complex Numbers part 2.

12) Polar Form of a Complex Number:

Let P (x, y) be any point at a distance of r units from an origin and make an angle

Anil Satpute

49-List of Formulas - 05

Click here for the previous list of formulas.

Today we will see the list of formulas of the Complex Numbers part 1. This branch of Math is very interesting and takes us to an imaginary world of Numbers.

Click here for the next list of formulas.

Anil Satpute

48-Anilofication: Fun with mathematics

🌟 Starting a Whimsical Adventure with Numbers! 

Today, let’s embark on an exhilarating exploration of the captivating realm of numbers. Numbers resemble stars — bright, limitless, and brimming with fascination. Since my early years, I’ve been delighted to engage with them, discover their hidden patterns, and reveal their mysteries. 

Throughout the years, this passion has blossomed into a voyage through the Extraordinary World of Numbers. And the most exciting part? I’ve enjoyed bringing inquisitive minds like yours along for the ride. 

Together, let’s journey from one mathematical wonder to another, appreciating the beauty, surprise, and inventiveness that numbers contribute to our existence. 

Are you prepared? Let’s set off on this adventure! 🚀

🔢 Unveiling Anilian Numbers  

Join us in delving into an intriguing new mathematical idea — Anilian Numbers — a distinctive pattern combining curiosity and logic!

📘 Explanation: 

An Anilian number is defined as a number where 

The difference between the Left Part and the Right Part

It is precisely equal to the Middle Part.

📌 Additionally, it is crucial that: 

Left Part > Right Part

Example: 

1) In 523, the digit "5" represents the left part, while the digit "3" indicates the right part. The middle part, "2," is obtained by subtracting 3 from 5. In summary, the Middle Part equals the Left Part minus the Right Part. Therefore, 523 qualifies as an "Anilian Number." 

📌 It is also crucial that 

Left Part > Right Part.

2) 25223 qualifies as an Anilian Number since the Left Part (25) minus the Right Part (23) equals the Middle Part (2). The equation 2 = 25 - 23 confirms this, satisfying the definition that the Middle Part equals the Left Part minus the Right Part.

3) 18711. Here, 18 - 11 = 7.

4) 47350 does not qualify as an Anilian number, as 47 - 50 results in -3. The left part is not greater than the right part, so it is not considered an Anilian number.

If we select a number and repeatedly add it to itself in a specific arrangement for a certain number of times, we arrive at what is known as an Anilian number.

Example 01:

Let's consider the number 243. As illustrated below, arrange this number in a zigzag format for 14 repetitions and compute the sum. The result will be an Anilian number.

The figure above depicts the configuration of digit 243. These digits are used 14 times total. We then sum these digits together. This sum equals 18711. Since 18 - 11 equals 7, 18711 is identified as an Anilian number. This technique is referred to as anilofication.

Example 02:

Let's consider the number 241. As illustrated below, organize this number in a zigzag format, repeat the process 12 times, and then sum the results. The final result will be an Anilian number.

As illustrated in the figure, we need to organize 241 exactly 12 times. The sum is 15906; since 15 - 06 equals 9, 15906 is classified as an Anilian number.

Example 03:

Let's consider the number 351. As demonstrated below, organize this number in a zigzag pattern for 14 iterations and calculate the sum. The result we obtain is known as an Anilian number.

In this section, we must organize the number 351 exactly 14 times, as illustrated in the diagram. The sum is represented as 27027, and since 27 - 27 equals 0, we conclude that 27027 is an Anilian Number. With the middle number being 0, we refer to this Anilian number as a standard Anilian number.

Happy Work:
1) Please select the number 503 and organize it as shown for 14 repetitions. Verify if the result is an Anilian number.

You can experiment with any number. This is an excellent concept that I developed in my childhood.

Please respond to me at my email address: anil@7pute.com.

Anil Satpute

47-Solution on perpendicular Bisector: Important key points

Today we will see the answer to the question asked in Blog-19.

The question was:

Draw the perpendicular bisector to a line segment drawn at the bottom of the page as shown in the following diagram.

We know the procedure to draw the perpendicular bisector of a line segment.

1) In a compass,  take a distance of more than half the segment. [Here my question is why do we need to take the distance more than half? ]

Here the Answer is very simple. You will come to know it very soon while discussing the basics of drawing perpendicular bisector on the line segment.

2) Taking Point A as the center and radius as more than half the distance of the segment, Draw the arcs on both sides of the segment.

3) Repeat the same by taking point B as the center and with the same radius.

Now you will get the answer to the question of why we need to take the radius as more than half. The arcs drawn with center A and center B should intersect each other.

4) Then join these two points to draw the line which is the perpendicular bisector of the segment AB.

As per Geometry is concerned, we can draw the perpendicular bisector in the same way.

Here we must understand the Basic Concept of a perpendicular bisector. It is already available in the procedure itself. As we are drawing the arcs with the same radius and the centers as point A and point B on both sides of the segment to get two points which are the point of intersection of these arcs. 

As per the question, so many students say their views are as follows.

Some students said that the question is wrong as the line segment on which the perpendicular bisector is to be drawn is at the bottom of the paper. It must be in the middle of the paper.

Some students said that the perpendicular bisector is not possible as we can't draw the intersecting arcs below the segment. The segment should be a little bit above.

Some students said that some additional blank paper needs to place below the line segment and construct the perpendicular bisector.

I was surprised by listening to these answers. The above problem can be solved simply by applying the basics of the perpendicular bisector.

We must know that every point of the perpendicular bisector is equidistant from the endpoints of the segment. So do the following steps to draw the perpendicular.

1) Take a radius of more than half of the line segment AB.

2) Take point A as the center, and draw an arc above segment AB.

3) Take point B as the center, and draw an arc above segment AB.

4) Name the point of intersection of these arcs as point C.

5) Increase the radius and draw two intersecting arcs with centers A and B above the Point C

6) Name this point of intersection of these arcs as point D.

7) Draw line joining points C and D.

Please see the following diagram to understand more about the concept.

Here Points C and D are at equidistant from the endpoints A and B. So line CD is the perpendicular Bisector of line segment AB.

We all together think properly to understand the basic concepts of Math. We will definitely change the entire world to understand better the basics of Math along with other subjects.

Anil Satpute