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 tt are there, such that the terms are integers from 11 to nn?'

I found that when common difference is 11,total A.Ps will be n(t1)n-(t-1)

When common difference is 22,total A.Ps will be n2(t1)n-2(t-1)

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

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

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

So I found that the total number of arithmetic progressions will be n+(p+1)[2(nt+1)p(t1)]n+(p+1)[2(n-t+1)-p(t-1)]

For example 70+(7+1)[2(709+1)7(91)]=70+8[26256]=61470+(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(nt+1)p(t1)]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 tt .

Hint: What can you say about ata1 a_t - a_1 ?

Calvin Lin Staff - 6 years, 6 months ago

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Thank you for you hint but sir can you please define at a_t and a1a_1 here?

Kalpok Guha - 6 years, 6 months ago

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

Calvin Lin Staff - 6 years, 6 months ago

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