Calvin Lin inspired me to this problem.In his problem Arithmetic Progressions.I tried to generalize the problem taking it as 'How many arithmetic progressions of length \(t\) are there, such that the terms are integers from \(1\) to \(n\)?'

I found that when common difference is \(1\),total A.Ps will be \(n-(t-1)\)

When common difference is \(2\),total A.Ps will be \(n-2(t-1)\)

The common difference will be continued till \([\frac {n}{t}]+1\)

{ where\([\frac {n}{t}]\) is greatest integer contained in \(\frac {n}{t}\) ]

Let \([\frac {n}{t}]=p\).

So I found that the total number of arithmetic progressions will be \(n+(p+1)[2(n-t+1)-p(t-1)]\)

For example \(70+(7+1)[2(70-9+1)-7(9-1)]=70+8*[2*62-56]=614\)

So the generalization formula will be \(n+(p+1)[2(n-t+1)-p(t-1)]\)

Please my friends if you find this useful Re-share and like.If you find any fault please tell me.

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## Comments

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TopNewestThere is a very simple way to obtain the total number of AP's of a fixed length \(t \).

Hint:What can you say about \( a_t - a_1 \)?Log in to reply

Thank you for you hint but sir can you please define \( a_t\) and \(a_1\) here?

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The first and last term of the arithmetic progression of length t.

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