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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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