# How Many A.Ps? Theory

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. Note by Kalpok Guha
6 years, 6 months ago

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There is a very simple way to obtain the total number of AP's of a fixed length $t$.

Hint: What can you say about $a_t - a_1$?

Staff - 6 years, 6 months ago

Thank you for you hint but sir can you please define $a_t$ and $a_1$ here?

- 6 years, 6 months ago

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

Staff - 6 years, 6 months ago