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Arithmetic Progression Problems' Compendium

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.

Note by Swapnil Das
5 months, 4 weeks ago

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A 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 \(\dfrac{a_{m} -a_{n}}{m-n}\) Harsh Shrivastava · 5 months, 4 weeks ago

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@Harsh Shrivastava Could we have a proof of it as well? Swapnil Das · 5 months, 4 weeks ago

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@Swapnil Das Solve terms by substituting Am An in above formula. You will get common difference. You are welcome. Sai Nath · 5 months, 4 weeks ago

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@Sai Nath Thanks :) Swapnil Das · 5 months, 4 weeks ago

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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. Ameya Daigavane · 5 months, 3 weeks ago

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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. Ashish Siva · 5 months, 3 weeks ago

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