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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
3 months, 3 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 $$\dfrac{a_{m} -a_{n}}{m-n}$$ · 3 months, 3 weeks ago

Could we have a proof of it as well? · 3 months, 3 weeks ago

Solve terms by substituting Am An in above formula. You will get common difference. You are welcome. · 3 months, 3 weeks ago

Thanks :) · 3 months, 3 weeks ago

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. · 3 months, 3 weeks ago

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. · 3 months, 3 weeks ago