An Arithmetic Progression is defined as a sequence such that the difference between two consecutive terms is constant. Hence the \(n^{\text{th}}\) term of such a sequence is given by \(a_n=a+(n-1)d\), where \(a\) and \(d\) denote the first term and the common difference respectively. So I was looking at an A.P. with this common difference as the rate of change of a term of an A.P..

If the \(n^{\text{th}}\) term of an A.P. is given by \(a_n=an+b\), then we have

\[a_1=a+b~\text{and}~a_2=2a+b \\ \Rightarrow k=a_2-a_1=a \\ \boxed{k=a}\]

where \(k\) denotes the common difference. This is in sync with

\[k=\dfrac{da_n}{dn}=\dfrac{d}{dn}(an+b)=a \\ \boxed{k=a}\]

But it doesn't go hand in hand here if we look at the sums of \(1,2,3,\dots\) terms of an A.P. from the beginning changing at the rate of the general term \(a_n\). If \(S_n=an^2+bn\) denotes the sum of first \(n\) terms of an A.P. (it will be of this form as an existent formula for the sum of first \(n\) terms of an A.P. which can be rearranged into this form is \(\dfrac{n}{2}(2a+(n-1)d)\), where \(a\) and \(d\) are the first term and the common difference of an A.P. respectively), then we have

\[a_1=S_1=a+b~\text{and}~a_1+a_2=S_2=4a+2b \\ \Rightarrow a_2 = S_2-S_1=3a+b \\ \Rightarrow k=a_2-a_1=2a \\ \Rightarrow a_n=(a+b)+(n-1)(2a)=b-a+2an \\ \boxed{a_n=b-a+2an}\]

And if we go with derivatives, we have

\[a_n=\dfrac{dS_n}{dn}=\dfrac{d}{dn}(an^2+bn)=2an+b \\ \boxed{a_n=2an+b}\]

Can someone explain to me the disappearance of the \(-a\)? One interesting observation is that this does hold true.

\[k=\dfrac{da_n}{dn}=\dfrac{d}{dn}(a_n)=\dfrac{d}{dn}\left(\dfrac{dS_n}{dn}\right)=\dfrac{d^2S_n}{dn^2} \\ k = \dfrac{d^2}{dn^2}(an^2+bn)=2a \\ \boxed{k=2a}\]

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## Comments

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TopNewestYou can't differentiate it that way. It's the same reason why you can't differntiate \(x^2 = x + x + x +\ldots + x \) to get \(2x = 1 + 1 + 1 + \ldots + 1 \).

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Sorry, lot of questions. I'm very much new to differentiation. 1. Why can't we differentiate in your example? 2. Is the fact that the other two things hold true a coincidence?

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What I have is a discrete case. The equation is only true for discrete number of \(x\). You need to make sure it's continuous first before differentiating. No, it's not a coincidence, it's just the way it was set up to mislead people.

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