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# 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
2 years, 8 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 - 2 years, 8 months ago

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

- 2 years, 8 months ago

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

Staff - 2 years, 8 months ago