Meet Archisman Nandy. While most students of his age fear Maths, this soon-to-be-teenager has surprised everyone with his grasp at the subject. A student of class VI of St. Agnes School, Kharagpur, West Bengal, Archisman has recently bagged the Bhagoban Chandra Dey Memorial Award-2018 conferred by the Centre for Pedagogical Studies in Mathematics (CPSM), Kolkata, in recognition for regular participation and outstanding performance in Achievement Cum Diagnostic Test in Mathematics (ADTM), a state-level Mathematics competition organised by CPSM, Kolkata. Right from class III, Archisman has regularly taken ADTM, conducted annually, and has always scored A++ outstanding performance grade (the highest grade of ADTM) every year followed by state level ranks consecutively for three years which has bagged him the CPSM’s ‘coveted’ award. Besides this, he has also achieved Zonal Rank-2, International Rank-5 in Science Olympiad Foundation’s (SOF) International Mathematics Olympiad (IMO-Class IV level), Certificate of Distinction in Australian Mathematics Competition (AMC 2017- Upper Primary Division i.e., V, VI, VII Level) organised by Australian Mathematics Trust (AMT) and Bronze Medal in the Southeast Asian Mathematical Olympiad (SEAMO) 2018.

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Imagine your favourite video game as a kid. Now, imagine that game with the math curriculum built into it. That’s Prodigy — the free, adaptive math game. It helps students practice math and learn new skills as they navigate a fantasy world packed with action and adventure.

**Engagement**

Each student is a hero here. Students build and control their own wizards, defeat creatures and other players in math battles. The experience stays fresh, as the company regularly creates content and designs mechanics to expand the battle system and invigorate the in-game world.

**Adaptivity and Differentiation**

The company promises to provide each player with a different experience. It refines algorithms that automatically assess progress, delivering questions that appeal to distinct learning styles while addressing unique skill deficits and trouble spots.

**Comprehensive Reporting**

Teachers get granular looks at progress and performance. The reports are designed to quantify and measure everything from the place where students play along with the duration of time to how they perform in assignments (noting down the points where mistakes have been made).

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**Maths educator Rupesh Gesota does not believe in teaching his students by simply jotting down the textbook method on blackboard. Instead, he believes in the following teaching mantra– How can students be guided towards discovery by asking them right questions, so that they are able to detect and correct mistakes on their own? **

**In this article, he brings alive his class by detailing about the interactions he had with his students while teaching them one of the dreaded mathematics topics, Arithmetic Progression (AP). We are reproducing it here with the hope that it will help students learn the method in a fun-filled way while at the same time, inspire many educators to invoke curiosity and innovativeness in their teaching methods.**

I have been playing Maths with a bunch of marathi-medium municipal school students after their school-hours from some time now.

As I can now see them approach and solve quite challenging (out of the textbook) problems comfortably, I decided, for a change, to pick up their textbook for a while and see what unfolds… Of course, I cannot make them solve the problems from the school textbook of their age (they have surpassed it long back). So, I directly picked up their SSC (grade-10) Algebra textbook (they have just cleared their grade-7 and grade-8 exams). The first chapter was about Sequence (AP, GP). I was aware that this would be a cakewalk for them. So, I just rushed through the terminologies and asked them to derive the formula for nth term of Arithmetic Progression (AP).*(For those who are unaware or have forgotten what AP is, let me give you a couple of examples, instead of its definition) *

*i) 2, 5, 8, 11, 14,…… *

*ii) 3, -7, -17, -27,….*

*iii) 5, 10, 12, 17, 19,….*

*iv) 2, 20, 200, 2000,…*

*Only (i) and (ii) are APs while (iii) and (iv) are not APs… (common difference (d) = 3 in the 1st case and d = -10 in the 2nd case) *

So, I asked them to come up with a formula for nth term of AP, that is, in a given AP, if you wish to ‘quickly’ find the 200th term, 5369th term, etc., without listing down all the middle terms, then how can you do that using the formula?

They didn’t take more than half-a-minute to derive at the required result. It was correct.

Tn = T1 + (n-1) d

So now, I immediately asked them to derive the formula for Sum of first nth terms of AP (Sn).

The book was in my hand and hence I saw its method. I was wondering if this method will click to them or will they do something else, but I was very sure they will crack it. And in almost a minute, all hands went up.

No wonder, they had used a different method… But the main thing to wonder was they all had used the same method.

I forgot to take the snap of their work. So I am typing down what they did:

Sn = T1 + T2 + T3 +T4 +…

= T1 + (T1 + d) + (T1 + 2d) + (T1 + 3d) +….

= nT1 + d + 2d + 3d +….

= nT1 + d (1 + 2 + 3 +….)

= nT1 + d (n-1) n / 2

= n [T1 + d (n-1)/2]

If you observe, they have used the formula for sum of first few (n-1) natural numbers to derive expression for Sn.

However, the textbook derivation does not use this method and formula (in fact, the formula for sum of first n natural numbers is derived later in the book using the formula for Sn)… And that’s also fine.

So, I first appreciated them for their accuracy and then challenged them to get the expression for Sn without using the formula for sum of first few natural numbers. I asked them – what if you didn’t know the formula that you have used in this derivation?

To this, one of them replied, “Sir, we would have derived it the way we had done it long back.”

“Hmm… And how did you do that?”

“By pairing up, first and last numbers, second and second-last, and so on.”

“Yes, then why don’t you use the same idea here too?”

They looked at me, puzzled…

So after a minute, I just showed them the method given in the textbook.

They studied it and then said, “Sir, isn’t this the same pairing method?”

I then gave them some textbook problems just for practice (were they problems?). Luckily, the last one among them was a little better.

**Problem: 4 terms are in AP. Product of 1st and 4th terms is 45. Sum of 2nd and 3rd terms is 18. Find the terms. **

I would suggest you to (try to) solve this problem before reading further…

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….

….

….

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Since the textbook was in my hand, I saw the method. However, let me also confess that I would have used the same method because this is what was directly taught to me through ‘drill and kill’.

Thankfully, I recovered from being an instructor and didn’t do this damage anymore.

And you know what? This not only saves my effort but also helps me learn interesting methods from my students.

**Method-1: By Jitu**

He argued that since 3 x d x t1 and 45 are multiples of 3, hence (t1)^{2} should also be multiple of 3. Hence, it can take only two values: 9 and 36 to get the required sum 45. Substituting (t1)^{2} as 36 does not satisfy the equation and substituting as 9 does satisfy. He solved further to arrive at the required AP.

He was the first to complete the problem. To engage him, I gave him another similar problem, but of 3 consecutive numbers – Sum of 3 numbers is 27 and their product is 504; and he was quick to solve this one too.

**Method-2 – By Sahil (original 4 numbers problem)**

He too argued similarly about the addition equation. Since ‘2a’ and ‘3d’ sum up to an even number, and ‘2a’ is already even, so ‘3d’ too should be even. Hence, ’d’ should be even. Did you understand this?

He then found the possible values for d. He also said that ‘d’ cannot be 0 since that would make all terms equal, which is in turn not possible as 45 (given as product of two terms) is not a perfect square. He also said that ’d’ cannot be ‘6’ because that would make a=0 leading to the product of 1st and 4th terms as 0, thus contradicting the given information.

He then found the corresponding 2 values for a, substituted each pair in the equation.

(a, d) = (6, 2) didn’t satisfy but (3, 4) did satisfy. He then got the required sequence.

**Method-3 – by Kanchan**

She first told me that Sum of 2nd and 3th terms would be same as the Sum of 1st and 4th terms.

When I asked her why, she said – it’s obvious. 1st term is less than 2nd term by the same amount as the 4th term is more than 3rd term.

Did you get this?

I was happy she could visualise this relation; however, I challenged her to prove it while she was writing on the board, and if you notice, she has done it partially.

What drove me crazy was her next move. She went into Quadratic Equation.

“We know the sum of two numbers; we know the product of same two numbers… So we can make an equation, and find its roots…”

Hope you will take some time to study her solution.

*(let me mention that quadratic equation is formally taught in 10th, that too in a dry way… she has not only used quadratics while in 8th but has also applied that knowledge in another topic altogether!)*

And her method, gave us both the possible sequences – 3, 7, 11, 15 and 15, 11, 7, 3 which was not the case with other methods.

Finally, I asked them if they wanted to know how it’s solved in the textbook. And they all were curious to know.

So, I simply copy pasted this method on the board, leaving for them to analyze and understand…. And they could easily do so….

I waited for them to study this approach. And then –

“So if I give you another problem, but this time of 3 consecutive numbers in AP, then how will you solve it?”

“We will then take those terms as a-d, a, a+d”

“Oh! You will use this textbook method?”

“Yes sir, this is an interesting and even an efficient one!”

I could see them smiling.

“So I should have directly showed this method to you, isn’t it?”

“No…. then how would we think and discover our own methods?” Vaishnavi reacted.

*(All the above conversations happened in Marathi. These students are from Marathi medium municipal school, based in Navi-Mumbai. They are part of a Maths Enrichment programme – MENTOR. To know more about the programme, check the website **www.supportmentor.weebly.com**)*

*The writer can be reached at *rupesh.gesota@gmail.com*. You can read the above article and many such articles on his blog, *www.rupeshgesota.blogspot.in

**DISCLAIMER: Views expressed above are that of the author and do not reflect the views of the website. ****The Peeper Times**** does not assume any responsibility or liability for the same.**

**Do you feel strongly about something? Have a story to share? Write to us at ****info@thepeepertimes.com**** or connect with us on ****Facebook**** ****or ****Twitter **** **

CBSE has decided to re-conduct exams for Economics (Class 12th) and Mathematics (for Class 10). Dates for fresh examinations will be announced soon. The news, however, has left both students and parents fuming.

**Students express unhappiness with CBSE decision:**

Most of my friends slept for a total 7 hours(out of 48) to study, cramp, revise and prepare for their Economics exam. They had to and they did. They wrote their exams and the curtains fell… Or they thought. #CBSE wants them to write the exam again. Man, how dense can you be!

— Mohd Aadil Faizal (@mohdaadilf) March 28, 2018

Today I came home happily finishing y #CBSE class 10 math paper & was surprised to find in the news that the paper was leaked & we have to rewrite our exams now, like literally I don’t understand what are the officials doing.

— Pranav Viju (@pranav_viju) March 28, 2018

#CBSE I came running home and told my mother that the exam was easy and my hard work paid off. Was very happy but then the news of this paper leak. Cbse itself is the cause of stress why to blame studies. Many people already planned to go somewhere after long studies, what to do?

— Ayush Kushwaha (@Ayushdotnetbabu) March 28, 2018

**Just when you thought you have done well, comes the news of re-examination:**

#CBSE I hate you !!! Like is it our problem paper was leaked !!

Economics was so nice and i did well in it !! Why this reschedule ???

Accountancy paper was also leaked this way ??

Conduct Accounts instead of Eco !!!

Really pissed off !! High time !— Satyam Chaudhary (@Zeev_sk) March 28, 2018

**Why only Economics?**

#CBSE @CBSEWorld Good that U r holding re-examination For12th board exams for Economics( All India )due to paper leakage ! Please look into the news of Accountancy paper leakage also . Honest students should not suffer please @narendramodi @PrakashJavdekar @TimesNow @IndiaToday

— Shivaani K Talwar (@ShivaaniKTalwar) March 28, 2018

The Chemistry exam should be also reconducted..it is unfair for science students..The chemistry paper was also leaked#CBSE

— Aman Jain (@amanjain_29) March 28, 2018

**Why should students suffer? Let CBSE handle…**

Students should not go for retest…Let #CBSE do what they want

— TheCancerian (@AmTrehan) March 28, 2018

**Some term it torture:**

Kids will never forgive you #CBSE. This is nothing less than torture.

— Thanos (@ivivek_nambiar) March 28, 2018

**And some want to take out a protest:**

ITS TIME THAT WE ORGANIZE A PROTEST IN FRONT OF #CBSE HEADQUARTERS #CBSEPaperLeak #CBSEMaths #cbseretest #CBSEBoardExams

16.38 +11.86 L is almost 28 lakhs students. imagine this amount of students and the amount of power they could have if they come together.— ayyyyyyyyaan (@lilspongeboi) March 28, 2018

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In a society where poor scores in exams often lead to students committing suicide, meet Rajkaran Barua. He has failed in his maths exams continuously for the last 19 years, yet hasn’t lost hope.

A resident of Madhya Pradesh’s Jabalpur area, Barua has been working as a domestic help since his childhood. He completed his graduation in 1993 by borrowing books from raddi-wallahs. And in 1997 Barua, a student of MSc in Mathematics in the Rani Durgavati University, first attempted his post-graduation exams. Since then, every year, he sits for the exams, but without luck. This year too, the result has been no different.

Does he not feel demotivated because of so many unsuccessful attempts? Does he not feel like giving it up?

“Every time I failed, I only became stronger. I will continue to study and appear in exams till I become a postgraduate in mathematics. It doesn’t matter how long it takes,” said Barua in a report published in The Times of India.

It is this determination that keeps him going despite the setbacks. He hopes to achieve his masters degree in his 20^{th} attempt next year. And until he achieves his goal, he won’t rest.

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Vedic Maths involves mental calculation techniques that, if practised well, can reduce your calculation time drastically. Many institutes teach this as a part of their curriculum while preparing for entrance examinations. We recommend you to try this once and judge for yourself the effectiveness of these techniques. ((break))

In Vedic maths, there are various methods of multiplication. Regular practice will enable you to choose the one that best suits a particular problem. Below are some simple multiplication methods that can really improve your calculation speed.

**Method 1:**

How to multiply when one number ends with 9?

**Example 1: 33 x 29**

Step 1: Since 29 ends with 9, we will focus on this number.

29 is succeeded by 30. So we will multiply the given number (in this case 33) by 30.

33 x 30 = 990

Step 2: The difference between 30 and 29 is 1. This 1 needs to be multiplied with 33 which gives us 33 as answer. We will now subtract 33 from 990 to get the correct answer.

990 – 33 = 957

957 is our answer (33 x 29 = 957)

Isn’t is simple?

Try this with other examples. Below is an example when one number is a two-digit and the other a three-digit.

**Example 2: 59 x 437**

Step 1: 60 x 437 = 26220

Step 2: 26220 -437 = 25783

Answer: 59 x 437 = 25783

**Method 2:**

How do we multiply two-digit numbers where the sum of the numbers at tens digit is 10 and the units digit is same?

**Example 1: 37 x 77**

In the above example, the units digits is the same (7) and the sum of the numbers at tens digits (3 + 7) is 10.

Step 1: Multiply the tens digits

3 x 7 = 21

Step 2: Add units digit to the answer obtained in step 1

21 + 7 = 28

Step 3: Multiply the units digits

7 x 7 = 49

(We need a two-digit answer in this step. In case, the answer is not a two-digit number, then we will add zero in the tens place)

Step 4: Combine the answers obtained in step 2 and 3. The combined figure is the answer.

2849 (37 x 77 = 2849)

**Example 2: 83 x 23**

Step 1: Multiply the tens digits

8 x 2 = 16

Step 2: Add units digit to the answer obtained in step 1

16 + 3 = 19

Step 3: Multiply the units digits

3 x 3 = 9 Since we need a two-digit answer in this step, we will add zero before 9. So, 3 x 3 = 09

Step 4: Combine the answers obtained in step 2 and 3. The combined figure is the answer.

1909 is the answer (83 x 23 = 1909)

Does these not sound too simple? Vedic Maths offers several such techniques. We will discuss some more techniques in our subsequent articles. Till then, practice them well.

**SEE ALSO:**

Vedic Maths: For Fast Calculations – Part 2

Vedic Maths: For Fast Calculations – Part 3

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