Hello young mathematicians.

Arithmetic Progressions, famously called *AP* is one of the most alluring topics in high school mathematics. I have made this note in order to compile all *proof based* problems on *AP*. You need to follow the following rules:

Post a problem of

*AP*and also its solution. The problem can only be proof based.The posted problem might be original, or borrowed from variety of resources.

These problems altogether will help us prepare for various competitive examinations.

*Thanks.*

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## Comments

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TopNewestA simple but.nice property :

If \(a_{n}\) and \(a_{m}\) are nth and mth term of an AP, then common difference of this AP is \(\dfrac{a_{m} -a_{n}}{m-n}\)

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Could we have a proof of it as well?

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Solve terms by substituting Am An in above formula. You will get common difference. You are welcome.

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Will dedicate a problem to whoever solves this:

If \( s_1, s_2, s_3, \ldots \) is a sequence of strictly increasing positive integers such that the sequences, \[ s_{s_1}, s_{s_2}, \ldots \] and \[s_{s_1 + 1}, s_{s_2 + 1}, \ldots \]

are arithmetic progressions themselves, then show that \( s_1, s_2, s_3, \ldots \) is also an arithmetic progression.

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Question:-

In a non-constant arithmetic progression, \(n\) times the \(n^{\text{th}}\) term is equal to \(m\) times the \(m^{\text{th}}\) term, prove that the \((m + n)^{\text{th}}\) term is \(0\).

Solution:-

Then \(n(a + (n - 1)d) = m(a + (m - 1)d)\\ n(a + dn - d) = m(a + dm - d)\\ an + dn^2 - dn = am + dm^2 - dm\\ an - am + dn^2 - dm^2 - dn + dm = 0\\ a(n - m) + d\left(n^2 - m^2 - n + m\right) = 0\\ a(n - m) + d\left(n^2 - m^2 - (n - m)\right) = 0\\ a(n - m) + d\left(\left(n - m\right)\left(n + m - 1\right)\right) = 0\\ (n - m)\left(a + (n + m - 1)d\right) = 0\\ a + (n + m - 1)d = 0\\ a_{n + m} = 0\)

My own original question.

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