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.

#NumberTheory

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Comments

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 $a_t - a_1$ ?

Calvin Lin Staff - 6 years, 4 months ago

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

Kalpok Guha - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$