0:36 Clarification on Differentiating the Sum of an AP Sir, In the first problem, the sum of the "n" terms of the AP is given as a function of "n", and upon differentiating it with respect to n, we get 2 + 6n, which appears to be the n-th term of the AP. Additionally, the coefficient of n represents the common difference in this case. I wanted to know if this has any deeper significance. Can we consider this result as a property of an AP? I know my doubt is trivial, but I’m just curious. Thank you for your time and guidance.
As such, one cannot differentiate a function defined on natural numbers. So I will treat what you suggest as a heuristic (and there is nothing wrong with heuristics). This process does capture the common difference of this AP correctly but note that 2 + 6n is not the n-th term of the given AP. To see this, the first term of the AP is 5, which does not match with the 2 + 6n for n = 1. Overall, it is true that you will get the common difference if you "differentiate w.r.t. n" two times the formula for the sum of the first n terms of an AP, but the reason for this, in my opinion, is rather incidental. It's good that you are experimenting this way. Keep going. But make sure you solve this problem in the "honest to God" way.
Thank you Sir I currently have no idea about the significance of calculus or it's nature therefore I again used differentiation to find the minima in the 5th problem. Is there any other way to solve it without simply putting the values and checking it?
@@AyushmanPandaaa Just by putting values you can get to the correct answer. But the proper way of doing it would be the following. Let S_n denote the sum up to n terms. Then the n-th term is S_n - S_{n - 1}. This way you get a formula for the n-th term. This will tell you all that you need to know about the AP.
Thanks again sir! This really opened my mind about how to visualize a GP. These kind of ideas and proofs and going out of the box is why I love Maths and it makes me aspire to study it even more. I have read from your various comments that you will bring Olympiad lectures as well someday, God! I can't wait for those, I am that excited. You're truly a gem.
Hello sir , this is a 2026 Jee Aspirant and your 460 th subscriber 😁I am writting this to let you know that you teach really really well . Due to some health problems and mid term exams i was not able to properly focus properly on coaching classes and a lot of backlogs has been created ( also while covering the backlogs I forgot the earlier chapters 😓 ) , So I decided to complete my 11 th syllabus in this remaining time from you ❤
0:36 Clarification on Differentiating the Sum of an AP
Sir,
In the first problem, the sum of the "n" terms of the AP is given as a function of "n", and upon differentiating it with respect to n, we get 2 + 6n, which appears to be the n-th term of the AP. Additionally, the coefficient of n represents the common difference in this case.
I wanted to know if this has any deeper significance. Can we consider this result as a property of an AP?
I know my doubt is trivial, but I’m just curious.
Thank you for your time and guidance.
As such, one cannot differentiate a function defined on natural numbers. So I will treat what you suggest as a heuristic (and there is nothing wrong with heuristics). This process does capture the common difference of this AP correctly but note that 2 + 6n is not the n-th term of the given AP. To see this, the first term of the AP is 5, which does not match with the 2 + 6n for n = 1.
Overall, it is true that you will get the common difference if you "differentiate w.r.t. n" two times the formula for the sum of the first n terms of an AP, but the reason for this, in my opinion, is rather incidental.
It's good that you are experimenting this way. Keep going. But make sure you solve this problem in the "honest to God" way.
Thank you Sir I currently have no idea about the significance of calculus or it's nature therefore I again used differentiation to find the minima in the 5th problem.
Is there any other way to solve it without simply putting the values and checking it?
@@AyushmanPandaaa Just by putting values you can get to the correct answer. But the proper way of doing it would be the following. Let S_n denote the sum up to n terms. Then the n-th term is S_n - S_{n - 1}. This way you get a formula for the n-th term. This will tell you all that you need to know about the AP.
@@the_hidden_library Thank you Sir for clarifying my doubt.
Thanks again sir! This really opened my mind about how to visualize a GP. These kind of ideas and proofs and going out of the box is why I love Maths and it makes me aspire to study it even more. I have read from your various comments that you will bring Olympiad lectures as well someday, God! I can't wait for those, I am that excited. You're truly a gem.
never saw this type of GP sum proof , sir you are truly amazing
back on traack again after festivals! That proof technique was so awesome sir never saw that!! thanks sir as always
Hello sir , this is a 2026 Jee Aspirant and your 460 th subscriber 😁I am writting this to let you know that you teach really really well . Due to some health problems and mid term exams i was not able to properly focus properly on coaching classes and a lot of backlogs has been created ( also while covering the backlogs I forgot the earlier chapters 😓 ) , So I decided to complete my 11 th syllabus in this remaining time from you ❤
Thank you for the kind feedback. I hope to complete the JEE Math syllabus by 2026 and then want to start with Physics.
Mind blown 🤯. Awesome, it really gives an insight. Love your lectures, you're great!